MATH 150B exam4

Ex7: finding force vectors

A child pulls a wagon with a force of |f|=20lbs at an angle of Θ=30° to the horizzontal. Find the force vector f. F=<|F|cosΘ,|F|sinΘ>=<2θcos30°,2θsin30°>=<10√3,10>.

Ex4: Calculating work

A force F=<3,3,2> (in newons) moves and object along a line segment from P(1,1,0) to Q(6,6,0) in meters. What is the work done by the force. INterpret the resulat. Slution: The displacement of the object is d-PQ→=<6-1,6-1,0-0>=<5,5,0>. Therefore, the work fone by the force is w=F8d=<3,3,2>*<5,5,0>=3oj. To interpret this result, notice that the angle between the force and the displacement vectro satisfies cosθ=F*d/|F||d|=<3,3,2>*<5,5,0>/|<3,3,2>||<5,5,0|=30/√22√50=.905. therefore, θ=uvrad250. The magnitude of the force |f|=√22=4.7N, but only the componet of that force in the directio of motion |f|≈cosθ≈√22cos.440≈4.2N, contributes to the work.

limit of a vector-valued fuction

A vector-valued function r approaches the limit l as t approacehs a, written lim t→a r(t)=a, provided limt→a (r(t)-L)=0.

Arc length for vector functions

Consif=der the parameterized curve r(t)=<f(t),g(t),h(t)>, where f',g' and h' are continuous and the curve is traversed for a≤t≤b. HTe arc length of the curve between (f(a),g(a),h(a)) and (f(b),g(b),h(b)) is L=∫(b:a)√f'(t)^2+g'(t)^2+h'(t)^2dt=∫(b:a)|r'(t)|dt

identifu=ying equations

Describe the set of points that satisfy the equatuion x²+y²+z²²-2x+6y-8z=-1. Solution: We simplify the equation by completing the square and factoring: (x²-2x)+(y²+6y)+(z²-8z)=-1 (x²-2x+1)+(y²+6y+9)(z²-8z+16)=25 (x-1)²+(y+3)²+(z-4)²=25 the equation describes a sphere with radus 5 and with center (1,-3,4).

defintie integrals

Evaluate ∫(π:0)(i+3costj0-4tk)dt. Solution: ∫(π:0)(i+3cost/2j-4tk)dt=ti|(π:0)+(sin t/2)j|(π:0)-2t²k|(π:0)=πi+6j-2π²k.

initial conditions

INitial velocity at t=0; v(0)=<u₀,v₀> and initial position at t=0: r(0)=<x₀,y₀>.

rigth- handed coordinate system

If the curled fingers of the right hans are rotated from the positive x-axis, the thumb points in the positive direction of the z-axis.

equal vectors

Two vectors are equal if they have equal length and a point in the same direction.

Summary: two- dimensional motion in a gravitational field

Cder an object moving in a gravitational plane iwth a horizontal x-axis, a vertical y-axis. given the initial velocity v(0)=<u₀,v₀> and the initial position r(0)=<u₀,v₀>, the velocity object for t≥0 is v(t)=<x'(t),y'(t)>=<u₀,-gt+v₀> and the postion vectoe ir r(t)=<x(t),y(t)>=<U₀t+x₀,-1/2gt^2+v₀t+y₀>.

Theorem 13.1 Dot product

Given two vectors u=<u₁,u₂,u₃> and v=<v₁,v₂,v₃>, u*v=u₁v₁+u₂v₂+u₃v₃

vector operations in R3

Let c be scalar,u=<u1,u2,u3> and v=<v1,v2,v3>. u+v=<u1+v1,u2+v2,u3+v3> : vector addition u=v=<u1-v1,u2-v2,u3-v3>: vector subtratcion cu=<cu1,cu2,cu3> : scalar multipliction

E5: vectors in R3

Let u=<2,-4,1> and v=<3,0,-1>. Find the componets of the following vectors and draw them in r3. a. 1/2u b. 2v c. u+2v Solution:a. Using the definition of scalar multiplication, 1/2u=<2,-4,1>=<1,-2,1/2>. The vector 1/2u has the same direction u with half the length of u. b. Using scalar multiplication, -2v=-2(3,0,1)=<-6,0,2>. The vector -2v has the direction opposite that of v and twice the length of v. c. Using vector addition and scalar multiplication, u+2v=<2,-4,1>+2<3,0,-1>=<8,-4,1>. The vector u+2v is drawn by applying the parallelogram rule to u and 2v.

Summary: properties of vector operations

Suppose u v and w are vectors and a and c are . The the following properties hold for vectors i any number of dimensions: 1. u+v=v+u= ccommutatiative property of addition 2. (u+v)+x=(u+v)+w=associative property of addition. 3. v+0=v: Additive identity 4. v+(-v)=additive invere 5. c(u+v)=cu+cv= distributive property1 6. (a+c0+v=av+cv= distributive property2 7. 0v=0= Multiplication by sero sclar 8. c0=0= multiplication by zero ector. 9. 1v=v= MUltiplicative idetity 10. a(cv)=(ac)v= Associative property of sclaar multipication.

straight- line motion

Written in the form r(t)=<x(0)+at,y(0)+bt,z(0)+ct, for t≥0, where X₀,Y₀ and Z₀ are constant. This function describes a straigth- line trajectory with an initial position of <x₀,y₀,z₀> and direction given by hte vector <a,b,c>. the direction of the trajectory and the acceleration is a=<0,0,0>. The motion associateed with this funsction is uniform (constant velocity) or straight- line motion.

Ex4: flight of a baseball

A baseball is hit form 3 ft above home plate with an initial velocity of v(0)=<u₀,v₀>. Neglect all forces oher that gravity. a. Find the postiion and velocity of the ball between the time it is hit and the time it first hits the ground. b. Show that the trajectory of the ball is a segment of a parabola. c. Assuming a flat playig field, how far does the ball travel horizonatlly> PLot the thrajectory of the ball/ d. What is the maximum height of the bal? e. Does the ball clear a 2oft fence that is 380ft from home plate (directly under the path of the ball?) Solution: Assume the origin is located at home plate. Because the distances are measured in feet, we use g=32ft/s^2. a. Substituting x₀=0 and y₀=3 into the equation for r, the position of the ball is r(t)=< x(t)y(t)>=<8ot,-16t^2+80t+3>, for t≥0. We then compute v(t)=r'(t)=<80,-32t+80. . Equation 3 says that the horizontal position is x=80t and the vertical position is y=-16t^2+8ot+3. Substituting /80 into the equation for y gives y=-16(x/80)^2+x+3=x^2/400+x+3, which is the equation of a parabola. c. the ball lands on the ground at the value of t>0 at which y=09. Solving for y(t)=-16t^2+80t+3=0, we find the t=0, and t=5.04 s. the first root is not relevant for the problem at hand, so we conclude that the ball land swhen t=5.04s. The horizontal distance traveled by the ball is x(5.04)=403ft. the path of the ball in the xy-coordinate system on the time interval [0,5.04 ] is shown in figure 14.21. d. the ball reaches its maximum height at the time its velocity is zero. y'(t)=-32t+80=0, we find that t=25s. the height of that time is y'(2.5)=103ft. e. THe ball reaches a horizontal distance of 380ft (the distance to the fence) when x(t)=80t=380. Solving for t, we find that t=4.75 s. HTe height of the ball at that time is y(4.75)=22ft.So, indeed, the ball clears a 20ft fence.

Tightening a bolt

A force of 20N is applied to a wrench attached to a bolt, in a direction perpendicular to the bolt, which produces more torque: applying the force at an angle of 60° on a wrench that is .15 m long or applying the force at an angle of 135° on a wrench that is.25 m long? IN each case, what is the direction of the torque? Solution: the magnitude ot= the torque in the first case is |t|=|r||f| sinθ=(.15m)(20N)sin60°=2.6n/m. In the second case, the magnitude of the torque is |t|=|r||f|sinθ=(.25m)(20N)sin135°=3.5n/m. the second instance gives greater torque. In both cases, the torque is orthogonal to r and f, parallel to the shaft of the bolt.

flight of a golf ball

A golf ball is driven down a horizontal fairway, with and initial speed of 55m/s at an initial angle of 25° (from tee with a negigible height)> Neglect all forces except gravity, assume the ball's trajectory lies in a palne. a. When the ball first touches the ground, how far has it traveled horizontally and how long has it been in the air? b. What is the maximum height of the ball? c. At what angle should the ball hit the ground to reach a green that is 300m from the tee? Solution:a. Using hte range formula, with α=25° and |v₀|=55m/s, the ball travels |v₀|²sinα/g=2(55m/s)sin25°/9.8m/s^2=4.7s b. THe maximum height of the ball is (|v₀|sinα)²/2g=/(55m/s)(sin25°)/2(9.8m/s^2)=27.6m c. Letting r denote the range and solving for the range formula for sin2α, we find that sin2α=Rg/|v₀|²=(300m)(9.8m/s^2)/(55m/s^2)=.972. For the ball to travel a horizontal distance of exactly 300m, the required angles are α=1/2sin^-1.972=38.2° or 51.8 °

Fligth in crosswinds

A plane is flying horizontally due north in calm air at 300 mi/h when it encounters a horizontal crosswind blowing southeast at 40mi/h and a downdraft blowing vertically downward at 30mi/h. What are the resulting speed nad velocty of the plane relative to the ground? Slution: Let the unit vectors i, j, nd k point east, orth and upward, respectively. The velocity of the plane relative to the air (300 mi/h due orth) is Va=300i. the crosswind blows at 45° south of east , so its componet is 40cos45°=20√2 (in the i direction, and its componet to thesouth is 40cos45°=20√2 in the j irection). therefore, the crosswind may be expressed as w=20√2-20√√2. Finally, the downdaft in the negative k-direction is d=-30k. The velocity of the plane relative to the ground is the sum of Va, w and d: V=Va+w+d=300i+(20√2i-20√2j_-30k=20√2i+(300-20√2j)-30k. Figure 13.42 shows the velocity of the vector near the plane. A quick calculation shows that the speed of the plane is |v|=225mhr. The direction of the plane is slightly east of north and downward.

Force on a proton

A proton with a mass of 1.7x10^-27 kg and a charge of q=+1.6x10^-19 Columbs(c) moves along the x-axis with a speed of |v|=9x10^5m/s. When it reaches (0,0,0),a uniform magnetic field is turned on. the field has a constant strength of 1 Tesla(1T) and is directed long the negative z-axis. a. Fid the magnitude and direction of the force on the proton at the instant it centers the magnetic field. b. Assume the rpoton loses no energy, and the force ad=cts as a centripital force with magnitue=de |F|=m|v|²/R that keeps the proton in a circular orbit of radius R, Sind the radius of the orbit. Solution: a. expressed as vectors, we have v=9*10^5i and B=-k. therefore, the force on the proton newtons is F=q(vxb)=1.6x10^-19(9X10^5)IX(-K))=1.44X10^-13J. As shown, when the proton enters the magnetic field in the positive x-direction, the force acts in the positive y-direction, which changes the path of the proton. b. the magnitude of the force acting on the proton remains 1.44x10^-13 N at all times (from part (a)). Equating tis force to the centripetal force |f|=m|v|²/R, we find that m|v|^2/||=1.7x10^-7kg)(9x10^5m)/1.44x10^-13N=.01m: Assuming no energy oss, the proton moves in a circular orbit of radius .01m.

projectile motion

A small projectile is fired due east over horizontal ground with an initial speed of |v₀|=300 m/s at an angle of α=30° above the horizontal.A crosswind blows from south to north, producing an acceleration of the projectile or .36 m/s^2 to the north. a. Where does the projectile land? How far does it alnd from its launch site? b. In order to correct the crosswind and make the projectile land, at what angle must the projectile be fired. Assume the initial speed, |v₀|=300 m/s and the and the angle of elevation over 30° are the same as in part a. Solution: a. Letting g=9.m/s, the equations of motion are a(t0=v'(t)=<0,0.36,-9.8>. Proceeding as in the two- dimensional case, the indefinite integral of the acceleration is the velocity function v(t)=(300cos30°,0,300sin30°>.=<150√3,0,150>.Subsitiuting t=0 and using hte initial condition, we find that c=<150√3,0.36t,-9.8t+150>.Integrating the velocity function produces the position function r(t)=<150√3t, 81t^2, -4.9t^2+150t>+c. Using the initial condition r(0)=<0,00>, we find that c=<0,0,0> and the position function is <x(t),y(t),z(t)>=(150√3t,.18t^2,-4.9t^2=150t>. The projectile lands when z(t)=-4.9t^2+150t=0. Solving for the positive root, which gives the time of flight is t=150/4.9=30.6s. The x- and y- coordinates at that time are x(T)≈79.53 m and y(T)=169m north of the firing site. Because the projectile started at <0,0,0> , it traveled a horizontal distance of √(7953)^2+(169)^2=7955m. b. Keeping the initial speed of the projectile equal to |v₀|=300m/s, we decompose the horizontal componet of the speed, 150√3 m/s into an east componet, u₀=150√3√ cosθ, and a north componet, v₀=150√3 sinθ, where θ is the angle relative to due east: we must determine the correction of angle θ (figure 14.25, pg.892). THe x- and y- componets of the function are x(t)=(150√3 cosθ)t and y(t)=.18t^2+(150√3 sinθ)t.These changes in hte initial velocity affect the x- and y- equations, but ot the the z- equation. The time of flight is still T=150/4.9=30.6s. The aim is to choose θ so that the projectile lands on the x-axis. (due east from ht elaunch site), which means that y(t)=0. Solving y(t)=.18t^2+(150√3sinθ)t=0, with T=150/4.9, we find that sinθ=-.0212; therefore, θ=-0.0212 rad =-1.215°. I other words, the projectile mmust be fired a horizontal angle of 1.215° to the south of east to correct for the not=rtherly crosswind (figure 14.25b, pg. 892). The landing location of the projectile is x(t)=7952m and y(T)=0

spheres and balls

A sphere centerd at (a,b.c) with radius ris the set of points satisfu=ying the equation (x-a)^2+(y-b)^2+(z-c)^2=r^2 A ball centered at (a,b,c) with radius r is the set of points satisfying hte inequality (x-a0^2+(y-b)^2_(z-c)^2≤r^2.

three-dimensional rectangular coordinate system, or xyz-coordinate system

A three-dimensional coordinate system created by adding a new axis, the z-axis to the xy-coordinate system. The new z-axis is inserted through the origin perpendicular to the x-and y-axes.

lengths of plnetary orbits

According to Kepler's first law, planetd revolve aroudn the sun in an eliptical orbit. A vector function that describes an elipse in th exy-plane is r(t)=<acost,bsint>, where 0≤t≤2π. Ifa>b>0, then 2a is the length of the major axis and 2b is the length of the minor axis (figure 14.26, pg. 897). verify the lengths of the planetary orbits given in tabl 14.1. Distances are given in terms of astronomical units (AU), which is the length of the semimajor axis of Earth's orbit, or about 93 million miles. PLanet; semmajor axis:a(AU); Semiminoraxis(b(AU); α=b/a; Orbit Length (AU) mercury; .387; .379; .979; 2.407 venus; .723; .732; 1.000; 4.643 earth; 1.00; .999; .999; 6.280 mars; 1.524; 1.517; .995; 9.554 jupiter; 5.203; 5.179; .995; 32.616 saturn; 9.539; 9.524; .998; 59.888 uranus; 19.182; 19.161; .999; 120.458 neptune; 30.058; 30.057; 1.000; 188.857 Using the arc length formula, the length of a general eliptical orbit is L=∫(2π:0)√x'(t)^2+y'(t)^2dt=∫(2π:0)√(asint)²+(bcost)²=∫(2π:0)√a^2sin^2t+b^2cos^2tdt. Factoring a62 out of the square root and letting a=b/a, we have L=∫(2π:0)√a^2(sin^2t+(b/a)^2cos^2tdt=a∫(2π:0)√sin²t+α²cos²tdt=4a∫(π/2:0)√sin²t+α²cos²tdt. UNfortunately, ,an antiderivative for this integral cannot be found in terms of elementary functions, so we have two options: the integral is well knnown and values have been tabulated for various results for α. Alternatively, we may use a calcualtor to approximate the integral mumerically (see sectio 8.8). Using numerical integration, the orbit lengths in table 14.1are obtained.For example, the length of mercury's orbit with α=.387 and α=.979 is L=4a∫(π/2:0)√sin²t+α²cos²tdt=1.548∫(π/2:0)√sin²t.959os²t=2.407. The fact that g is closse to 1 for all orbits shoe=wn in hte table nearly equal to 2π, which is the lenght of a circular orbit with radius a.

path on a sphere

An object moves on a trajectory described by r(t)=<x(t),y(t),z(t)>=<3cost,5sint,4cost>, for 0≤t≤2π. a. Sjow that the object moves on a sphere and find the radius of the sphere. b. Find the velocity of the sphere. c. Consider the curve r(t)=<5cost,5sint,5sin2t>, which is the roller coaster curve from example 3 in section 14.1, with different coefficients. Show that this curve does not lie on a shpere. How could r be modified so that it describes a curve that lies on a sphere of radius 1 centered at the origin? Solution: a.|r(t)|²=x(t)²+y(t)²+z(t)²=(3cost)²+(5sint)²+(4cost)²=25cos²t+25sin²t=25(sin²t+cos²t)=25. Therefore, |r(t)|=5, for 0≤t≤2π and the trajectory lies on a sphere of radius 5, centered at the origin. b. v(t)=r'(t)=<-3sint,5cost,-4sint>=|v(t)|=√(v(t)*v(t))=√9sin²t+25cos²t+16sin²t)=√(25(sin²t+cos²t)=5; the speed of the object is always 4. You should verify that r(t)*v(t)=0, for all t, implying that r and v are always orthogonal. c. We first compute the distance from the origin to the curve |r(t)|=√((5cost)²+(5sint)²+(5sint2)²)=√(25(cos²t+sin²t+sin²2t))=5√(1+sin²2t). ITS CLEAR the r(t)is not constant and therefore, the curve does not lie on a sphere. One way to modify the curve so that it does lie on a sphere is to divide each output vector r(t) by its length. IN fact, s long as |r(t)|≠0, on the interval of interest, we can force any path onto a sphere, (centered at the origin) with thihs modification. The function u(t)=r(t)/|r(t)|=<cost/√(1+sin²t),sint/√(1+sin²2t>,sin2t/√(1+sin²2t)> descirbes a curve on which u(t) is constant because r(t)\|r(t)| is a unit vector. We cpnclude that the new curve, which is remniscent of the seam on a tennis ball, lies on the unit sphere, centered at the origin.

two- dimensional motion

Assume an object traveling over horizontal ground, acted on ony by the gravitational force, has an initial position <x₀,y₀> =<0,0> and, innitial velcity <u₀,v₀>=<|v₀|cosα,|v₀|sinα>. the trajectory, which is a segment of a parabola, has the following properties. time of flight=T=2|v₀|sinα/g range'=|v₀|²sin2α maximum height=y(T/2)=(|v₀|sinα)²/2g.

Ex3: Derivative rules

Cmpute the following derivatives, where u(t0=ti+t^2j+t³k and v(t)=sinti+2costj+costk. a. d/dt (v(t²) b.d/dt(t²v(t))) c.d/dt(u(t))*v(t)) Solution: a. Note that v'(t)==costi-2sintj-sintk. Using the chain rule, we have d/dt(t²)=/(cost²i-2sint²j-sint²k)(2t)/(2t) b. d/dt(t²(v(t)))=d/dt(t²)v(t)+t²d/dt(v(t)=2tv(t)+t²v'(t) = 2t(Sinti_+2csotj+costk)+t²(costi-sintj-sintk)= (2tsint+t²cost)i+(4tcost-2t^2sint)j+(2tcost-t^2sint)k c. d/dt u(t)*v(t)=u'(t)*v(t)+u(t)*v'(t)=(i+2tj-3t^2k)*(sinti+2costj+costk)+(ti+t^2j+t^3k)*(costi-2sintj-sintk)=)sint+4cost-3t^2cost)+(tcost-2t^2sint+t^3sint)=(1-2t+t^3)sin(5t-3t^2)cost. Note that the result is a scalar. the result is obtained if you first comute u*v then differentiate.

circular motion

Cnsider the two- dinemsional circular path r(t)=<Acost,Asint>, for 0≤t≤2π, where A is a nonzero constant. The velocity and acceleration vectors are v(t)=<-Asint,Acost) and a(t)=<-Acost,-Asint>. NOtice that r and a are parallel, but in opposite directions. Futhermore, r*v=a*v=0; therefore, the position and acceleration vectors are orthogonal to the velocity vectors at any given point. Finally r, v an a have a constant magnitude and A variable directions. The conclusion that r*v=0 applies to any motion for which r is a conatant. That is, to any motion on a circle or a sphere.

Ex1: derivative of functions

Compute the derivative of the following functions. a. r(t)=<t³,3t²,t³/3> b. r(t)=<e^-ti+10√t j +2cos3tk. Solution: a. r'(t)=<3t²,6t,t²/2>; note that r is differentiable for all f and r'(0)=0. b. r'(t)=<-e^-ti+5/√t j-6sin3tk; the function is differentiable for all t>0.

Ex1: dot products

Compute the dot products of the followinng vectors. a. w=2i-6j,v=12k b. w=<√3,1> and v=<0,1>. Solution: the vector u lies in the xy-plane and the vectorv is perpendicular to the xy-plane. Therefore, θ=π/2, u, πv are orthogonal an u*v=0. b. u and v form two sides of a 30-60-90 triangle in the xy-plane, with and angle of π/3 between them.Because |u|=2, |v|=1, and cosπ/3=1/2, the dot product is u*v=|u||v|cosθ=2*1=1/2=1.

higher-order derivative

Compute the first, second, and third derivatives of r(t)=<t²,8lnt,3e^-2t>. Solution: Differentiating once, we have r'(t)=(2t,8/t,-6e-2t>. Differentiating again produces r''(t)=<2,-8/t^2,2e^-2t>. Differentiating once more, we have r³(t)=<0,16/t³,-24e^-2t>.

indefintite integrals

Compute ∫t/√t₂+2 i+e^-3t j+(sin4t+1)k dt. Solution: We compute the integral of each componet. ∫t/√t₂+2 i+e^-3t j+(sin4t+1)k dt= (√(t²+2)+C₁)i+(-1/3e^-3t+c₂)j+(-1/4cost4t+t+C₃)k=√(t^2+2) i-1/3e^-3tj+(t-1/4cost)k+c. The constants c₁,c₂ and c₃ re combined to form one vector c at the end of the calcualtion.

arc lelngth parameterization

Conside rthe helix r(t)=<2cost,2sint,4t>, for t≥0. a. Find the arc length funcion s(t). b. Find another description of the helix that uses arc length as a parameter. Solution:a. NOte that r'(t)=<-2sint, 2cost,4> and |v(t)|=|r'(t)|=√((2sint)²+(2cost)²+4²)=√(4(sin²t+cos²t)+4²)=√(4+4²)=√20=2√5. Th√erefore, the relationship between the arc length and the parameter is s(t)=∫(t:a)|v(u)|+∫(t:0)2√5=2√5t. An increse of1/2√5 in the parameter t corresponds to an increase of 1 in the arc length. Therefore, the curve is not parameterized by arc length. b. Substituting t=s/2√5 in the √origionial parametric escription of the helix, we find that the description with arc length as a parameter is (using a different function mname), r₁(s)=<2cos(1/2√5),2sin(s/2√5)2s/√5>, for s≥0. This description has the property that an increment of ∆s in the parameter corresponds to an increment of exactly ∆s in the arc length.

limits and continuity

Consider the function r(t)=cosπti+sinπj+e^-tk, fro t≥0. a. Evaluate limt→2 r(t) b. Evaluate limt→∞ r(t) c. At what point is r continuous? Solution: a. We evaluate the limit of each componet of r: lim t→2 r(t)=limt→2(costi+sintj+e^-tk)=e_-2k. b. Note that although limt→∞ e^-t=o. lim t→∞ cost and lim t→∞ do not exist. therefore, lim t→∞ r(t) does not exist. As shown in figure 14.47 (pg. 873), the curve is a coil that approaches the unnit circle in the xy-plane. c. Because the componets of r are continuous for all t, r is continuous for all t.

Magnitude and unit vectors

Consider the points P(1,-2) and Q(6,10). a. Find PQ→ and wo unit vectors parallel to PQ→. b. Find two vectors of length 2 parallel to PQ→. c. Express PQ→ as the product of its magnitue and a unit vector. Solution: a. PQ→=<6-1,10-(-2)>=<5,12>, or 5i+12j. Because |PQ→|=√5²+12²=√169=13, a unit vector parallel to PQ→ is PQ→/|PQ→|=<5,12>/13=<5/13,12/13>=5/13i+12/13j. The unit vectro parallel to PQ→ with the opposite direction is <-5/13,-12/13>. b. To obtain two unit vectors of length 2 parallel to v, we mutiply the unit vector 5/13i+12/13j by 2: 2(5/13i+2/13j)=10/13i+24/13j and -2<5/13i,12/13j>= -10.13i-24/13j. c. The unit vector <5/13,12/13> points in the same direction of PQ→, so we havePQ→=|PQ→| PQ→/PQ→=13<5/13,12/13>.

magnitudes of unit vectors

Consider the points P(5,3,1) and Q(-7,8,1). a. Express PQ→ in terms of the unit vectors i,j, and k. b. Find the magnitude of PQ→. c. Find the position vector of magnitude 10 in the direction of PQ→. Solution: a. PQ→ is equal to the position vector <-7-5,8-3,1-1>=<-12,5,0>. Therefore, PQ→=-12i+5j. b. |PQ→|=|-12i+ 5j|=√((-12)²+(5)²)=√(169)=13. c. c. the unit vector in the direction of PQ→ is PQ→|PQ→|=1/13<-12,5,0>. Therefore, teh vector in the direction of u with a magnitude of 10 is u=10/13<-12,5,0>.

Equation of a sphere

Consider the pointsP (1,-2,5) and Q(3,4,-6). Find an equation of teh shpere for which the line segment PQ is a diameter. Solution: The center of the shpere is the midpoint of PQ: (1+3/2,-2+4/2,5-6/2)=(2,1,-1/2). the diameter of the shpere is the distance |PQ|, which is √((3-1)^2+(4+2)^2+(-6-5)^2) =√(161). Therefore, the sphere's radius is ±√(161), its center is (2,1,-1/2) and it is described by (x-2)^2+(y-1)^2+(z+1/2)^2=(1/2 √(161)²=161/4.

Ex1: velocity and acceleration for circular motion

Consider the two- dimensional motion given by the position vector r(t)=<x(t),y(t)>=<3cost,3sint., FOR 0≤t≤2π. a. Graph the trajectory of the object. b. Find the velocity and speed of the object. c. Find the acceleration of the object. d. Sketch the pposition, velocity and acceleration vectors for t=0,π/2 and t=3π/2. Solution: x(t)²+y(t)²=9(cos²t+sin²t), which is an equation centered at the origin with radius 3.The object moves on thiscircle in the counterclockwise direction. b. v(t)=<x'(t),y'(t)>=(-3sint,3cost>=|v(t)|=√x'(t)²+y'(t)²=√(-3sint)²+(3cost)²=√9(sin²t+cos²t)=3. the velocity vector has a constant magnitude and a continuously ca=haging direction. c. Differentiating the velocity, we find that u(t)=v'(t)=<-3cost,3sint-r(t).In this case, the acceleration vector is negative of the position vector at all times. d. the relationships among r,v, and a at four points in this time are shown in figure 14.3 (pg,.885). the velocity vector is always tangent to the trajectory, and has lenght 3, while the acceleration vector and the position vector each have length 3 and point in opposite directions. At all times, v is orthogonal to r and a.

computing trajectories

Constder the trajectories, describeed by hte positive functions r(t)=<t,t²-4,t^3/4-8>, for t≤0 and R(t)=<t^2,t⁴-4,t⁶/4-8>, for t≤0, where t is measured in the same time unit for both functions. a. graph and compare trajectories using a graphing utility. b. Find the velocity vetors associated with the position function. Solution:a. Plotting the position functions of selected values of f results in the trajectories shown in figure 14.14. Because r(0)=<0,-4,-8>, both curves have the same initial point. For t≥0, the two curves consist of the same points, but they are traced out differently. For example, both curves pass through the point <4,12,8>, but the point corresponds to r(4) on the first curve and r(2) on the second curve. In general, r(t²)=R(t), for t≥0. b. The velocity vectors are r'(t)=<1,2t,3t²/4> and R(t)= <2t,4t³,3/2t^5>. The difference in the motion on the two curves is revealed by the graphs of the speedss associated with the trajecories. the object on the trajectory reaches the point <4,12,8> at t=4, where its speed is |r'(4)|=|<4,12,8>|=14.5. THE object on the second trajectory reches the oint 94,12,8) AT T=2, where its speed is |(r'(2)|=|<4,32,38>=57.8

corss products of unit vectors

Evaluate all the cross products amon ghte cordinate vectors i,j,k. Solution: these vectors are mutually orthogonal, which meand the angle ebtween any two distinct vectors is θ=π/2 and sinθ=1. Futermore, |i|=|j|+|k|=1. therefore, the cross product of any two distinct vectors has a magnitude of 1. By the right-hand-rule, wehn the fingers of the right hand curl form i to j, the thumb points in the direction of the positive z- axis. The unit vector in the positive z-direction is k, so ixj=k. Similar calcualtions show that jxl=i and kxi=j. By property 10 of theorem 13.4, jxi=-k, so jxi and ixj point in opposite directions. Similaryl, kxj=-i and ixk=-j. these realtionships are easily remembered with the circle diagram. Finally, the angle between any unit vector and itself is θ=0.aTerefore, ixi=jxj=kxk=0.

LInes and vector-valued functions

Find a vector function for the line that passes through the points P(2,-1,4) and q(3,0,6). Solt=ution: Recall from section 13.5 that the parametric equations of the line parallel to the vector v=<a,b,c> and passing through the point P₀<x₀,y₀,z₀> are x=x₀₀+at,y=y₀+bt,z=z₀+ct. The vector v=PQ→<3-2,0-(-1),6-4>=<1,1,2> is parallel to the line, and we let p₀=P(2,-1,4). Therefore, the parametric equations for the line are x=2+3t,y=-1+t,z=4+2t, and the corresponding vector fun=ctions for the line are r(t)=(2+t,-1+t,4+2t), with a domain of all real numbers. As t increases, the line is generated in the directionof PQ→. JUst as we did with parametric equations, we can restrict the domain to a fiite interval to produce a vector for a line segment.

vectors of orthogonal vectors

Find a vector orthogonal to vectors u=i+6k, and v=2i-5j-3k. Solution: A vector orthogonal to u an v is parallel to uxv. One such orthogonal vector is uxv=|i,j,k/-i,0,6/2,-5,-3|=(0+30)i-(3-12)j+(5-0)k=30i+9j+5k.Any scalar multiple of this vector is orthogonal to u.

orthogonal projections

Find projvu and scalvu for the following vectors and illustrate each result. a. u=<4,1>,v=<3,4> b. u=<-4,-3>,v=<1,-1> solution: a, the scalra componet of u in the direction of v is scalvu=u*v/|v|=<4,1><3,4>/|<3,4>|=16/5. Because v/|v|=<3/5,4/5<, we have projuv=scaluv=(v/|v|=16/5<3/5,4/5>=16/25<3,4> b. Using anotehr formula for projvu, we have projvu=(u*v/v*v)=(<-4,-3>*<1,-1>/(1,-1>*<1,-1>)*<1,-1>=-1/2(1,-1>> The vectors v and r and projvu point in opposite directions because π/2<θ≤π. this fact is reflected in the scalar componet of u in the direction of v, which is negative: scalvu=<-4,-3>*<1,-1>/|<1,-1>|=-1/√2.±

Area of a triangle

Find the area of the triangle with vertices )(0,0,0),P(2,3,4) and Q(3,2,0) Solution: First consider the [arallelor=gram, two of whose sides are OP→ and OQ→.By theorem 13.3, the area of the=is [arallelogram is |OP→xOQ→|. Computing the cross product, we find that OP→xOQ→=|i,j,k/2,3,4/3,2,0|=}3,4/2,0|i-|2,4/3,0|j+|2,3/3,2|k. thereofer, the area of teh parallelogram is |OP→xOQ→|=|-8i+12j-5k|=√233=15.26.The triangle with vertices o,p and Q fr=orms half of the area of its parallelogram, so its area is √233/2=7.63

A cross product

Find the magnitude and the direction of uxv, where u=<1,10> and v=<1,1,√2>. Solution: Because u is one side of a 45-45-90 triangle and v is the hypotenuse, we have θ=π/4 and sinθ=1/√2. Also, |u|=√2 and |v|=2, so the magnitude of uxv is sinθ=√2*2*1/√2=2.√The direction is given by hte right- hand rule: uxv is orthogonal to u and v.

Ex2: Unit tangent vectors

Find the unit tangent vectors for the following parameterized curves. a. r(t)=<t²,4t,4lnt> for t>0. b. r(t)=<10,3cost,3sint>, for 0≤t≤2π. Solution: a. A tangent vector of r'(t)=<2t,4,4/t>, which has a magnitude of √((2t)²+4²+(4/t)²)=√(4t²+16+16/t²)=√(2t+4/t²)²=2t+4/t². As shown in figure 14.10 (pg. 878), unit tangent vectors change direction along the curve, but maintain unit length. b. IN this case, r'(t)=<0,-3sint,3cost> and |r'(t)|=√((0)²+(-3sint)²+(3cost)²)=√9(sin²t+cos²t_=3. therefore, the unit vector for a particular value of t is T(t)=1/3<0,-3sint, 3cost>=<0,-sint,cost>.The direction of t changes along the curve, but its length remains.

scalar multiples and parallel vectors

Given a scalar and a vector v, the scalar multiple cv is a vector whose length is |c| multiplied by the length of v. Ifc>0, then cv has the same direction as v. If c<0, then cv and v point in opposite directions. Two vectors are parallel if they are scalar multiples of each other.

dot product

Given two nonzero vectors u and v in two or three dimensions, their dot product is u*v=|u||v| cosθ, where θ is the angle between u and v with 0≤θ≤π. If u=0 or v=0, then u*v=0 and θ is undefined.

cross product

Given two nonzero vectors u and v inR3, the corss product uxv is a vector with magnitude |uxv|=|u||v| sinθ, where 0≤θ≤π is the angle between u and v. The direction of uxv is given by hte right-hadn rule: When you out the vectors tail-to tail and let the fingers of your right hand curl form u to v, the direction of uXv is the direction of your thumb orthogonal to both u and v. When uxv=0, hte die=rectionof uxv is undefined.

Roller coaster curve

Graph the cure r(t)=costi+sintj+.4sin2tk, for 0≤t≤2π. Solution: Withput the z-componet, the resulting function r(t)=costi+sintj describes a circle of radius 1 in the xy-plane. The z-componet of the function varies between -.4 and .4 with a period of π units. Therefore, on the interval [0,2π], the z-coordinates of the points on the curve oscilate twice between -.4 and.4, while the x-andy- coordinates describe a circle. the result is a curve that circles the z-axis once in the counterclokwise directionwith two peaks and two valleys. the space curve in htis example is particularly complicated, but visualizing a given curve is easier when determining the suraces on which it lies. Writing the vector function r(t)=costi+sintj+.4sin2tk in parametric form, we have x=cost,y=sint,z=-.4sin2t, for 0≤t≤2π. NOting that x²+y²=cos²t+sin²t=1, we conclude that the curve lies on the cylinder x²+y²=1. IN this case, we can also eliminate the parameter by writing z=.4sin2t=4(2sintcost)=.8xy whhich implies that the curve lies on the hyperbolic paraboloid z=.8xy. In fact, the roller coaster curve in which the surfaces x²+y²=1 and z=.8xy intersect.

Ex2: A spiral

Graph the curve described by the equation r(t)=4costi+sintj+t/2πk, where (a). 0≤t≤2π and (b)-∞<t<∞. Solution: We begin by setting z=0 to determine the projection of the curve in the xy-plane. The resulting function implies that x=4cost and y=sint: these equations describe and elipse in the xy-plane, whose positive direction is counterclockwise. Because z=1/2π, the value of z increases from 0 to 1 as t increases form 0 to 2π. therefore, the curve rises out of the xy-pplane to create an eliptical spiral (or coil). Over the interval [0,2π], the spiral begins at (4,0,0), circles the z-axis once, and ends at (4,0,1). b. Letting the parameter vary over the interval -∞<t<∞ generates a spiral that winds around the z-axis endlessly in oth directions. the positive orientation is upward (increasinf z-direction). Noticing once more that x=4cost and y=sint are x- and y- componets of ‖r, we see that the sprial lies on the eliptical cylinder (x/4)^2+y^2=cos²t+sin²²t=1.

Arc length as a function of a parameter

Let r(t) describe a smooth curve, for t≥a. The arc length is given by s(t)=∫(t:a)|v(u)|du, where |v|=|r'|. Euivalently, d/dt=|v(t)|. If |v(t)|=1, for all t≥a, then the parameter corresponds to arc length.

indefintie integral of a vector-valued function

Let r(t)=f(t)i+g(t)j+h(t)k be a vector-valued function, and let R(t)=f(t)i+G(t)j+H(t)k, where F,G, and H are antiderivatives of f,g, and h. The indefinite integral of r is ∫r(t)=R(t)+c, where c is and arbitrary constant vector. Alternatively, in componet form, ∫<f(t),g(t),h(t)> dt=<F(t),G(t),H(t)>+<C₁,C₂,C₃>.

derivative and tangent vectors

Let r(t)=f(t)i+g(t)j+h(t)k, where f,g, and h are differentiable functions on (a,b). Then r has a derivative ( or is differentiable) on (a,b) and r'(t)=f'(ti)+g'(tj)+h'(tk), provided r'(t)≠0 or r'(t) is a tangent vector corresponding to r(t).

Definite integral of a vector-valued function

Let r(t)=f(t)i+g(t)j+h(t)k, where f,g, and h are integrable on the interval (a,b). The definite integral of r on [a,b] is ∫(b:a)r(t)dt=(∫b:a(f(t)dt))i+(∫b:a(g(t)dt))j+(∫(h(t)dt))k

position, velocity, speed, acceleration

Let the position of an object moving in 3-dimensional space be given by r'(t)=(x(t),y(t),z(t)>, for t≥0. the velocity of the object is v(t)=r'(t)=<f'(t),g'(t),h'(t)>. The speed of the object is the scalar function √(x'(t)²+y'(t)²+z'(t)²). The acceleration of the object is a(t)=v'(t)=r''(t).

Theorem 14.1: Derivative rules

Let u and v be differentiable vector-valued functions and let f be a differentiable scalar- valued function at all points t. Let c be a constant vector. The following rules apply: 1/d/dt(c)=0: constant rule 2. d/dt(u(t)+v(t))=u'(t)+v'(t): sum rule 3. d/dt(f(t)u(t))= f'(t)u(t)+f(t)u'(t): product rule 4. d/dt(u(f(t))=u'(f(t))f'(t): chain rule 5. d/dt(u(t)*v(t))=u'(t)v(t)+u(t)v'(t): dot product rule 6. d/dt(u(t)xv(t))=u'(t)xv(t)+u(t)xv'(t): cross product rule

Geometry of the cross product

Let u and v be two nonzero vector in r3. 1. the vectors u and v are parallel (θ=0 or=π) if and only if uxv=0. 2. If u and v are two sides of a parallel parallelogram, then the area of the parallelogram is |uxv|=|u||v|sinθ

properties of the cross product

Let u,v and w be nonzero vectors in r3 and let a and b be scalars: 1.uXv=-(vxu): anticomnnutative property 2. (au)x(bv)=(ab)(uxv): associative property 3. ux(v+w)=(uxv)+w:Distributive property 4. (u+v)*w=(U*W)+(V*W): dISTRIBUTIVE PROPERTY

vector operations

Let u=<-1,2> and v=<2,3> a. Evaluate|u+v| b. Simplify 2u-3 c. Fid two vectors half as log as u nd parallel to u. Solution: a. Because u+v=<-1,2>+<2+3>=<1,5>, we have |u+v|=√(1^2+5^2)=√(26). b. 2u-3v=2<-1,2>-3<2,3>=<-2,4>-<6,9>=<-8,5> c. Teh vector 1/2u=1/2<-1,2>=<-1/2,1> and -1/2u=-1/2<-1,2>=<1/2,-1> have half the length of u and are parallel to u.

dot products and angles

Let u=<√3,10> and v=<1,√3,0> and w=<1,√3,2√3>. a. Compute u*v b. Find the angle betwenn u and v. c. Find Find the angle between u and w. Solution: a. u*v=<√3,1,0>*<1,√3,0>=√3+√3+0=2√3. b. Note that w=√((<√3,1,0>)^(√3,1,0)>))=2 and similarly, |v|=2. Therefore, cosθ=u*v/|u||v|=2√3/2*2=√3/2. Because 0≤θ≤π, it follows that θ=cos^-1√3/2=π/6. c. cosθ=u*w/|u|\w|=<√3,1,0>*<1,√3,2√3>/|<√3,1,0>||1,√3,2√3||=2√3/2*4=√3/4. It follows that θ=cos^-1√3/4=1.12 rad=64.30

Evaluating the cross procudt

Let u=u₁i+u₂j+u₃k and v=v₁i+v₂j+v₃k. then uxv=||(i,j,k)/(u₁,u₂,u₃/v₁,v₂,v₃|=|u₂,u₃/v₂,v₃|i-|u₁,u₃/v₁,v₃|j+|u₁,u₂/v₁,v₂|k

motion wih a constant |r|

Let |r| describe a path on which r is constant (motion ona circle, or sphere centered at the origin). Then r*v=0, which means that the position vector and the velocity are orthogonal at all times for which the functions are defined.

properties of the dot rpoduct

Suppose u,v, and w are vectors and let c be a scalar. 1. u*v=v*u: Commutaative property 2. c(u*v)=(cu)*v=u*(cv): Associative property 3. u*(v+w)=uv+uw: Distribuive property.

solve for velocity

The velocity is an antiderivative of the acceleration, which we have v(t)=∫a(t)dt=∫<0,-g>dt=∫<0,-gt>+c. where c is an arbitrary constant vector. the arbitrary constant is determoned by substituting t=0 and using the initial condition v(0)=<U₀,V₀>. We find that v(0)=<0,0>+C=<U₀,V₀>. Therefore, the velocity is v(t)=<0,-gt>+<U₀,V₀>=<U₀,-gt+V₀>. Notice that the horizontal cmponet of the velocity ins simply hte initial horizontal velocity U₀₀ for all time. the vertical componet of velocity decreases linearly from its initial value of V₀.

slnky curve

Use a graphing utility to graph the curve r(t)=(3+cso15t)costi+(3+cos15t)sintj+sn15tk, for 0≤t≤2π and discuss its properties. Solution: the factor A9T)=3+cos15t that appears in the x- and y- componets is a varying amplitude for costi and sintj. Its effect is seen in the graph of the x- componet a(t)=cost. For 0≤t≥2π, the curve of one period 3cost with 15 oscillations superimposed on it. As a result, the x-componet of r varies from -4 to 4 with 15 small oscillations along the way. A similar behavior is seen in the y- componet or r. Finally, the z-componet, which is 15sint, oscillates between -1 and 1 15 times over[0,2π]. COmbining these effects, we discover a coil-shaped curve that circles the z-axis in the counterclockwise direction nad closes on itself. Figure 14.6b and 14.6 c(pg.872) show two views, one looking along the xy- plane and the other from overhead on th zz-axis. It can be shown that eliminating the parameter of the parametric equations defining r leads to the standard equation of a torus in Cartesian coordinates- in this case,(3-√x²+y²)²+2²=1 and thereofre, the curve lies ion htis torus, as shown in figure 14.6b.

scaler multiplication

c=scalar c and a vector v can be combined using scalar vector multiplication, or scalar multiplication.

plotting points in the xyz-plane

plot the following points: a. (3,4,5) b. (-2,-3,5). So,ution: a. Starting at (0,0,0), we move 3 units in the x-direction to the point (3,0,0) then 4 units in the y-direction to the point (3,4,0) and finally 5 units in the z-direction to the point (3,4,5). b. We move -2 units in the x-direction to the point (-2,0,0), -3 units in the y-direction to (-2,-3,0) and 5 units in the z-direction to (-2,-3,5).

componets of a force

A 10 lb block e=rests on a plane that is inclined at 30° above the horizontal. Find the componets of the gravitational force parallel to and normal (perpendicular) to the plane. Solution: The gravitational force F acting on the block equals the weight of the block (10lb): We regard the blov=ck as a point mass. Using the coordinate system, the force ats in a negatic=ve y- direction: therefore, F=<1,-10>. the direction down the plane is given by the unit vector v=<cos(-30°),sin(-30°)>=<√3/2,-1/2>.(Check that |v|=). The componet of the gravitational force parallel to the plane is projvu=(f*v/v*v)v=(<0,-10>*<√3/2,-1/2>)=5<√3/2,-12/>. Let the componet of F normal to the plane be N. Nte that F=projvFN, so N=F*projvF=<0,-10>-5<√3/2,-1/2>=-5<√3/2,1/2>.

Ex 8: Balancing forces

A 400lb engine ids suspended fro two chains that for 60° angles with a horizontal ceiling. How much weight does each chain support? Solution: Let F₁ and F₂ denote the forces exerted by the chains on the engine, and let F₂ be the downward force due to the weight of the engine. Placing the vectors in a standard coordinate sysem, w find that F₁=<|f₁|cos60°, |F₁|sin60°>F2=<-|F₂|cos60°,|F₂|sin60°> and F₃=(0,-400). Because the engine is in equilibrium( the chains and engine are stationary), the sum of the force is zero; that is, F₁+F₂+F₃=0 or F₁+F₂=-F₃. Therefore, <|F₁|cos60°-|F₂|cos60°,|F₁|sin60°+|F₂|sin60°=<0,400>. Equating corresponding componets, we obtain two equations to be solved for |F₁| and |F₂|:|F₁|cos60°-|F₂|cos60°=0 and |F₁|sin60°=400. Factoring the first equation, we find that (|F1|-|F2|)cos60°=0, which implies that |F1|=|F2|. Replacing |f2\ with |fF1| in the second equation gives 2|F1|=400. NOting that sin60°=√3/2 and solving for |F1|, we find that |F1+=400/√3≈231. Each chain must be able to support a weigh o f approximately 231lbs.

position vectors and vector compoenets

A vector v with its tail at the origin and head at the point (v1,v2) is called a position vector (or is said to be in standard position) and is written <v1,v2>. The real numbers v1 and v2 are the x- and y- componets of v, respectively. the position vectors, u=<u1,u2> and v=<v1,v2> are equal if and only if u1=v1 and u2=v2.

flight of an eagle

An eagle rises at a rate of 100 vertical ft/min on a vertical path given by r(t)=<250cost, 250sint, 100t>, where r is measured in feet and f is measured in minutes. How far does it ravel in 10 min? Solution: The speed of the eagle is |v(t)|=√(x'(t)²+y'(t)²+z'(t)²=√(-250sint)²+(250cost)²+100²=√250(sin²t+cos²t)+100²=√250²+100²=269. The constant speed makes the arc length easy to evaluate:L=∫(10:0)|v(t)|dt≈∫(10:0)269dt=2690. The eagle travels approximately 2690 ft in 10 min.

time of flight

Assuming an object is launched from the origin over horizontal gorund whe y(t)=-gt^2/2t+(|v₀|sinα)t=0. Solving for t, the time of flight is T=2(|v₀|sinα)t

parallel planes

Determine the equation of the plane parallel to the xz- plane passing through the point (-2,-3,7). Solution: points on a plane paarallel to the xz-plane have the same y-coordinate. Therefore, the plane passing through the poiint (2,-3,7) with a y-coordinate of -3 has the equation y=-3.

Finding one antiderivative

Find r(t)such that r'(t)=(10,sint,t) and r(0)=j Solution: the required function r is an antideravitive of (10,sint,t); r(t)=∫<10,sint,t>dt=<10t,-cost,t²/2>+C, where C is an arbitrary constant vector. The condition r(0)=j allows us to determine C; substituting =0 implies that r(0)=<0,-1,0>+, where j is< 0,1,0>. Solving for C, <0,1,0>-<0,-1,0>=<0,2,0>. Therefore, r(t)=<10t,2-cost,t^2/2>.

calculating componets and magnitude

Given the points O(0,0) and P_-3,4) and Q(6,5), find the componets and magnitude of the following vectors. a. OP→ b. PQ→ Solution: a. Teh vector OP→ is the positon vetor whose head is located at P(-3,4). Tehrefpre, OP→=<-3,4> and its magniu=tude is |OP→|√((-3)^2+(4)²)=5. b. PQ→=<6-(-3),5-4>=<9,1> and | PQ→|=√(9²+1²)=√(82).

magniutde of a vector

Given the ppoints p(x1,y1) and Q(x2,y2), the magnitude, or length of PQ→=<x2-x1,y2-y1> denoted |pq→|, is the distance between P and Q. |PQ→|=√((x₂-x₁)²+(y₂-y₁)²) Teh magnitude of postition vector v= <v₁,v₂> is |v|=√(v₁²+v₂²).

work

Le t a constant force f be applied to an object producing a displacement d. If the angle between F and d is θ, then the work done by the force is W=|F||d|cosθ=F*d.

unit tangent vector

Let r(t)=f(t)i+g(t)j+h(t)k be a smooth, parameterized curve for a≤t≤b. The unit tangent vector for a particular value of t is T(t)=r'(t)/|r'(t)|.

scalar and zero vectors

Quantities having magnitude, but no direction are called scalars. One exception is the zero vector, denoted 0: It has length 0 and no direction.

Vector operations in r^2

Suppose c is a scalar, u=<u1,u2> and v=<v1,v2>. u+v=<u1+v1,u2+v2>= vector addition u-v=<u1-v1,u2-v2>= vector subtraction cu=<cu1,cu2>= scalar multiplication

Ex6: Speed of a boat in a current

Suppose the water in a river moves southwest (45° wwest of south) at 4mi/h and a motorboat travels due east at 15 mi/h relative to the shore. Determine the speed of the boat and its headline, relative to the moving water. Soultion: To solve this problem, the vectors are paced in a coordinate system. Because the boat mioves east at 15 mi/h, the velocity realtive to the shore is v=<15,0>. To obtin the componets of w<wx,wy>,mobserve that |w|=4 and the lengths of teh sides of the 45-45-90 triagel are |wx|=|wy|=|w|cos45°=4/√2=2√2 Given the orientation of w (southwest,w=<-2√2,-2√2>. Because vg=vw+w: vw=vg=w=<15,0>-<-2√2,-2√2>=<15+2√2,2√2>. The magnitude of vw is |vw|= √((15+2√2)²+(2√2)²≈18. Therefore, the speed of the boat relative to the water is approximately 18 mi/h. The heading of the boat is given by the angle θ between vw nd the postivie x-axis. The x- componet of vw is 15+2√2 and the y- componet is 2√2. Therefore, tan6-1(2√2/15+2√2)≈9°. the heading of the boat is 9 north of east and its speed is relative to the water is 18 mi/h.

Disatenc eand midpoint formulas in xyz-space

The distance formula between the poi ts p (x,y,z) and Q(x,y,z) is √((X2-x1)²(y2-y1)^2+(z2-z1)^2) midpoint= (x1+x2/2, y1+y2/2, z1+z2/2)

Special cases of dot products

The dot product of two vectors itself is called a scalar. Two special caese immediately arise: * u an v are parallel (θ=0 or=π) if and only if u*v=±|u||v|. * u and v are perpendicular (θ=π/2) if and only if u*v=0.

magnitude of a vector

The magnitude (or length) of the vector PQ→=<x2-x1,y2-y1,z2-z1> is the distance fro P(x1,y1,z1) to Q(x2,y2,z2) |PQ→|=√((x2-x1)^2+(y2-y1)^2+(z2-z1)^2)

orthogonal projections of u onto v

The orthogonal projections of u pnto v, denoted proj vu, where v≠0, is porjvu=|u|cos(v/|v|). The orthogonal projections may also be computed with the foe=rmuals projvu=scalvu=(v/|v|)=(u*v/v*v)V, where the scalar componet of u in the the direction of v is scalvu=|u|cosθ=u*v/|v|

range

The range of the pbject, which is the horizontal distance it travels, is the x-coordinate trajectory when t=T:x(T)=(|v₀|cosα)T=(|v₀|coxα)2|v₀|sinα/g=2|v₀| sinαcosα/g=|v₀|sin2α/g. Note that the interval 0≤α≤π/2 sinα has a maximum value of 1 when α=π/4, so the maximum range is |v₀|²/g.IN OTHER WORDS, IN AN IDEAL WORLD, firing an object form the ground at an angle of π/4(45°) maximizes its range. otice that the ranges obtainde with the angles α and π/2 are equal.

vector-valued function

Three dependent variab;es (x,y and z) are the components of r, and each component varies to a single independent variable t (that often represents time).

orthogonal vectors

Two vectorsu and v are orthogonal if and only if u*v=0. THe zero vector is orthogonal to all vectors. IN two or three dimensions, two nonzero orthogonal vectors are perpendicular to eachother.

vector operations

Use figure 13.10 to write the following vectors as sums of scalar multiples of v and w. a. OP→ b. OQ→ c.QR→ Solution: a. Using the triagle rule, we start at 0, move three lengths in the dirction of v and then two lengths of w in the direction of w to reach P. therefore, OP→=3v+2w. b. The vector coincides with the diagonal of a parallelogram having adjacent sides equal to 3v and -w. y hte parallelogram rule, OQ→=3v-w. c. The vector QR→ lies on the diagonal of a parallelogram having adjacent sides equal to va nd 2v. Therefore, QR→=v+2w.

parallel vectors

Using figure 13.6a, write the vectors in terms of u or v. a. PQ→ b. QP→ c. QR→ d. RS→ Solution: a. The vector PQ→ has he same direction and length asa u: therefore, though they have PQ→=4. These two vectors are equal even though they have different locations. b. Because QP→ and u have equal length but opposite directions, QP→ (-1)u==4. c. QR→ points in the same direction as v and is twice as long as v, so QR→=2v. d. RS→ points in the opposite direction of u with 3 times the length of u. Consequently, RS→=-3u.

smooth

We say a function is smooth on an interval if f,g, and h are differentiable and r'(t)=0 on that interval. Smooth curves have no cusps or corners.

cross products of coordiante vectors

ixi=-(jxi)=k, kxi=-(ixk)=j, jxk=-(kxj)=i, ixi=jxj=kxk=0.

maximum height

of an object is reached when the vertical velocity is zero, or when y'(t)=-gt+|v₀|sinα=0. Solvinfg for t, the maximum height is reached at t=|v₀| (sinα)/g=T/2, which is half the time of the flight.Hte object spends equal amounts of time descending. the maximum height is y(T/2)=(|v₀|sinα)²/2g. Finally, by eliminating t from the equation for x(t) and y(t), it can be shown htat the trajectory of the object is a segemnt of a parabola.

octants

the coordinate plane containing the x-axis and the y-axis is still called the xy-plane. We now have two new planes: the xz-plane containing the x-axis a nd the z-axis and the yz-plane, containing the y-axis and the z-axis. taken together, these three coordinate planes divide xyz-space in to eight regions called octants.

domain of r

the largest values of t, of which all f, g and h are defined.

solve for the position

the positon is yan antiderivative of the velocity, given by r(t)=∫v(t)dt=∫<u₀,-gt+v₀>dt=<u₀t,-1/2gt^2+v₀t>+c, where c is an arbitrary constant. Substituting f=0, we have r(0)=<0,0>+c=<x₀,y₀>, which implies that c=<x₀,y₀>. Therefore, the position of the object for t≥0 is r(t)=<u₀t,-1/2gt^2+v₀t>+<x₀,y₀>=2 +<u₀t+x₀,-1/2gt^v₀t+y₀>.