Math 230 Final

A function f(x, y) satisfies ∇f(0, 1) = <3, −4> (a) Find the polar coordinates (r, θ) of the point (0, 1)

(Draw on xy plane) (r,θ) = (1,π/2)

Let S be the sphere given by the equation x^2 − 8x + y^2 + 24y + z^2 − 12z = −171 (a) (5 points) Find the center and radius of the sphere S.

(a) Complete the square to find center and radius

x^2 + y^2 + z^2 = 4 x^2 - 6x + y^2 - 8y + z^2 -24z + 153 = 0 (a) Calculate the distance between the centers of the two spheres. (b) Find the distance between the spheres.

(a) Complete the square to find the equation of sphere 2. Use distance equation to find the distance between centers. (b) Distance between spheres = distance between centers - radius1 - radius2

Let P be the plane given by 2x − y + 2z = 1, and let L be the line given by x = 7t + 3 y = 2t + 1 z = 3t − 2 (a) (7 points) Show that L is not parallel to P, and find the point of intersection of the line and the plane.

(a) For L to be parallel to P, it must be orthogonal to the normal vector of the plane. Dot direction vector of L with normal vector of P. Because dot product != 0, vectors not parallel. Plug in parametric equations into plane equation to solve for t and find point of intersection.

Let S be the sphere given by the equation x^2 − 8x + y^2 + 24y + z^2 − 12z = −171 (b) (5 points) The sphere S intersects the plane z = 2. Find the area enclosed by this intersection.

(b) Plug in z = 2 to get equation of circle. Use new radius for A = πr^2 to find area

A snowball with mass 0.4 kg is thrown northward into the air with a speed of 20m/s at an angle of 45◦ from the ground. A wind applies a steady force of 4N to the ball in a westerly direction. The magnitude of the acceleration due to gravity is given by 10 m/s^2 (d) How fast is its speed changing at the instant when the ball reaches the highest point?

5√2 m/s^2 because at the apex, the rate of change of speed is equal to the tangential acceleration.

Let L1 be the line parametrized by r(t) = <1, 2, 3> + t<1, 1, 1> and let L2 be the line give by the equations: x(t) = 7t y(t) = 3t z(t) = 4 (a) (6 points) Determine if the lines L1 and L2 are parallel, intersecting or skew.

A direction vector for L1 is v1 = h1, 1, 1i, and a direction vector for L2 is v2 = <7, 3, 0>. The system of equations given by v1 = cv2 has no solution since the first coordinates would give c = 7, but the second coordinates give c = 3. Therefore, the lines are not parallel. Now, we determine if they intersect: 1 + t = 7s 2 + t = 3s 3 + t = 4 The third equation gives t = 1, Putting this into the second equation gives s = 1. Plugging these values of t and s into the first equation would give 2 = 7. The lines L1 and L2 are skew.

Consider the surface whose equation in spherical coordinates is ρ = 4 sin(θ) sin(φ).

Always multiply both sides by ρ. ρ^2 = x^2 + y^2 + z^2 x = rsinθcosφ y = rsinθsinφ z = rcosθ

(b) (6 points) A stray branch from the tree catches you in the face as you round this curve at t = π/2,and you reflexively let go of your innertube to protect yourself. If the normal component of your acceleration is greater than 12, you go flying off of your innertube. What is your fate; i.e., do you stay on the innertube or do you fall into the stream?

An = |<r'(t) x r''(t)>|/|r'(t)| Find An at π/2 and see how that compares to 12 (FLI At = |<r'(t) · r''(t)>/|r'(t)|)

Let L1 be the line parametrized by r(t) = <1, 2, 3> + t<1, 1, 1> and let L2 be the line give by the equations: x(t) = 7t y(t) = 3t z(t) = 4 (b) (9 points) Find an equation for the plane that contains L1 and is parallel to L2.

Any point on L1 is on the plane, so we may take P = (1, 2,3) as a point on the plane. We now just need to find a normal vector for the plane. Since L1 is in the plane, the direction vector v1 = <1, 1, 1> is perpendicular to the normal vector, n. Furthermore, L2 is parallel to the plane if and only if its direction vector v2 is perpendicular to n. v1 x v2 = <1,3,-4> Plane given by <1,3,-4> · <x-1,y-2,z-3> = 0 x + 3y - 4z = -5

Find the length of the following curve: x = -2t + t^3/3, y = √(2t^2), 0 <= t <=2

Arclength forumula L(t) = t0->t1∫√((x'(t))^2 + (y'(t))^2) Solve integral for length

Bender: r1(t) = <5t, 25t^2 + 3> Robot Santa: r2(t) = <2t, 8t3 − 4t + 3> In both of these parametrizations, t represents the number of hours since the beginning of the journey, and distance is measured in miles. (a) (3 points) Who will reach the destination first, Bender or Robot Santa?

Bender will arrive at the restaurant when 5t = 2 ⇒ t =2/5 Robot Santa will arrive at the restaurant when 2t = 2 ⇒ t = 1

Let f(x, y, z) be a differentiable function and g(u, v) = f(u − v, u^2 − 1, 3v − 3) If fx(0, 0, 0) = 2, fy(0, 0, 0) = 3 and fz(0, 0, 0) = 3, compute gu(1, 1) and gv(1, 1)

Both partial derivatives follow from the Chain Rule. Note that we have x = u − v,y = u^2 − 1 and z = 3v − 3, which gives the following partial derivatives: xu = 1, xv = −1 yu = 2u, yv = 0 zu = 0, zv = 3 gu(1, 1) = fx(x(1, 1), y(1, 1), z(1, 1))xu(1, 1) + fy(x(1, 1), y(1, 1), z(1, 1))yu(1, 1) + fz(x(1, 1,), y(1, 1), z(1, 1))zu(1, 1) = fx(0, 0, 0)xu(1, 1) + fy(0, 0, 0)yu(1, 1) + fz(0, 0, 0)zu(1, 1) = 2(1) + 3(2) + 3(0) = 8 gv(1, 1) = fx(x(1, 1), y(1, 1), z(1, 1))xv(1, 1) + fy(x(1, 1), y(1, 1), z(1, 1))yv(1, 1) + fz(x(1, 1,), y(1, 1), z(1, 1))zv(1, 1) = fx(0, 0, 0)xv(1, 1) + fy(0, 0, 0)yv(1, 1) + fz(0, 0, 0)zv(1, 1) = 2(−1) + 3(0) + 3(3) = 7

Find an equation for the plane containing the three points P(1, 2, 0), Q(−1, 3, 1) and R(0, 1, 4)

Cross PQ and PR. Use weights of cross product as coefficients, use point P for center.

*Determine whether the following function is continuous at the point (0, 0) f(x,y) = {(x^4*y + x^3*y^3 + x^2 + y^2)/(x^2 + y^2) | (x,y) != (0,0) { 0 | (x,y) = (0,0)

Determine limit as (x,y) -> (0,0). If limit = 0 (as dictated by function definition), then function is continuous. Otherwise, not continuous.

Direction of gradient vector

Direction of greatest increase

Let C be the curve in R3 parametrized by r(t) = <9− 5t^2, 3t^2 + 8t, t^4 + 2 ln(t)> (a) (10 points) Find all points on C where the tangent line to the curve is parallel to the line given by l(s) = <5s + 13, 1 − 7s, 11 − 3s>

Directional vector at C given by r'(t). For directional vector to be parallel to line direction vector, r'(t) = c*l(s) r'(t) = <−10t, 6t + 8, 4t^3 +2t> v = <5,-7,-3> -10t = 5c 6t + 8 = -7c 4t^3 + 2/t = -3c  First eq gives c = -2 Plug c =-2 into second eq gives t = 1 Plug c = -2 and t = 1 into third eq gives 6=6 so r(1) = <4,11,1> is the only point where r is parallel to l

A curve represented by a vector-valued function r(t) lies entirely on a surface. If r(0) = <1, 2, 3>, then the tangent vector r'(0) can be taken as a normal vector for the tangent plane to the surface at (1, 2, 3).

F

T/F Every ellipse has constant curvature.

F

T/F Let i, j, and k be the unit vectors along the three-dimensional rectangular coordinate axes. The vectors i and k satisfy (i×i)×k = i×(i×k)

F

T/F The line x = 3 + 2t, y = 4 − t, z = 1 + 3t intersects the y-axis.

F. There is no value of t where x = 0 AND z = 0.

T/F The point (x, y, z) = (√3, 3, 6) lies on the surface described in spherical coordinates by φ = π/3.

F. x = rcosθsinφ, y =rsinθsinφ, z = rcosφ

T/F If u and v are unit vectors, then u × v is also a unit vector.

False

T/F The equation x = z in R3represents a line.

False (plane)

T/F Two non-parallel lines in the three dimensional space must intersect.

False (they can be skew)

T/F If fx(0, 0) = fy(0, 0) = 0 and fxx(0, 0) is positive, then f has a local minimum at (0, 0).

False, can also be a saddle point

T/F There exists such a function f whose linear approximation at (0, 0) is L(x, y) = 4x + 2y − 3 and whose quadratic approximation at (0, 0) is Q(x, y) = 4x^2 + 2y^2 − 3.

False, not possible.

T/F Two vectors which are parallel must have their dot product equal to zero.

False, their cross product equals zero.

T/F Two lines which are not parallel must intersect.

False, they can be skew.

T/F If fx(0, 0) = fy(0, 0) = 0 and fxx(0, 0) · fyy(0, 0) < [fxy(0, 0)]^2 then f has a local minimum at (0, 0).

False, we don't know the sign of fxx

Find all of the critical points of the function f(x, y) = 3x^2*y − 3x^2 − 6y^2 + 1and determine whether each is a local maximum, a local minimum, or a saddle point.

Find Fx=0, Fy =0 to find critical points. Plug x/y value found by one equation into the other. Use determinant (fxx*fyy - fxy^2 ? 0) If D(x) = fx*fy -fxy^2 > 0 If fxx > 0 minimum If fxx < 0 maximum If D(x) = fx*fy - fxy^2 < 0 saddle point

Consider the surface z(x^2 + y^2) = y^3 Find an equation for the plane tangent to the surface at (-2,2,1)

Find gradient vector for the surface. Use Fx, Fy, Fz as weights and the point (-2,2,1)

A leprechaun is walking along the curve given by r(t) = 0<= t < ∞ where the components are measured in meters. If his pot of gold is located 10 meters along the path from his starting location, at what time will he reach it?

Find r'(t) = <-sint,cost,1> s(t) = 0->t∫|r'(t)| s(t) = 0->t∫√(-sint)^2 + (cost)^2 + 1^2)du s(t) = 0->t∫√(2)du s(t) = √(2) t 10 = √(2) t t = 10/√(2)

Find all the points on the space curve r(t) = <t^3/3− 3t, 3t4,5t^6/3> where the tangent line is parallel to the line x = −1 + 2t, y = 12t, z = −2 + 10t.

Find r'(t), that becomes v. Set v = a*w (w= line), because v can be any multiple of w. Set each component equal to each other, and find values of t that solve each equation. Plug back t's into original equation to find points.

*A satellite circles around the earth along the orbit: *r(t) = <3 sin t, cos t, 7> (b) Find the curvature of the orbit at time t = 0.

Find v(0) and a(0) ϰ = |v(t) x a(t)| / |v(t)|^3

Calculating rate of change in a certain direction

First calculate gradient vector at point, then dot it with vector (must be calculated) in direction of point given. Greatest rate of change will always be in direction of gradient vector, rate of increase is magnitude of gradient vector

*Let f(x, y) = x^2 + 2y^2 − 2x − 4y. Find the absolute maximum and minimum values of f over the closed triangular region with vertices (0, 0), (0, 2), and (4, 2).

First, find critical points for the gradient (inside bound)-> f(1,1) = -3 Next, find the max/min along the boundaries (triangle) f(0, t) = 2t^2 − 4t on [0, 2] f'(0) = 4t − 4 = 0 ⇒ t = 1 f(0) = 0, f(1) = -2, f(2) = 0 f(t) = f(t, 2) = t^2 − 2t on [0, 4] f'(t) = 2t − 2 = 0 ⇒ t = 1 f(0) = 0, f(1) = −1, f(4) = 8 f(t,t/2)=3t^2/2− 4t on [0, 4] f'(t,t/2) = 3t - 4 = 0 ⇒ t = 4/3 f(0) = 0, f(4/3)= -8/3, f(4) = 8 Max: 8 Min: -3

Consider the function f(x, y) = e^(2x+y) Find the directions in which the directional derivative of f at (0, 0) has the value 1.

First, the gradient vector at (0,0) must be calculated. We also know that the directional derivative Duf = ∇f ∙ u where u is a unit vector. Let's say u = <a,b>. To be a unit vector a^2 + b^2 = 1 So, <2,1> ∙ <a,b> gives 2a + b = 1 (Duf has value 1). b = 1 - 2a plugged back gives a^2 + (1-2a)^2 = 1. a(5a - 4) = 0. a = 0 or 4/5. So u = <0,1> or <4/5,-3/5>

Let P be the plane given by the equation x + y − 2z = 2. (a) Find parametric equations for the line L through (5, 9, 0) that is perpendicular to the plane P.

From the equation of the plane we immediately see that the normal vector is n =< 1, 1, −2 > Then the parametric equation of the line is r(t) =< 5, 9, 0 > +t < 1, 1, −2 >.

How to classify critical points

If D(x) = fx*fy -fxy^2 > 0 If fxx > 0 minimum If fxx < 0 maximum If D(x) = fx*fy - fxy^2 < 0 saddle point

Consider the two planes x − y + z = 3 and x + 2y + z = 3 Are they parallel? If so, find the distance between them. If not, find parametric equations for the line where they intersect.

If direction vectors for each are not multiples of each other (they are not in this case... <1,-1,1> vs <1,2,1>) then not parallel. Because these are two planes, they will intersect in a line. This line is given parametrically by a point on both the planes (set x = y = 0 in this case to give (0,0,3)), and the cross product of the plane tangent vectors to give a direction vector for the line.

Let C be the curve in R3 parametrized by r(t) = <9− 5t^2, 3t^2 + 8t, t^4 + 2 ln(t)> (b) (10 points) Find all points where the tangent line to C is parallel to the plane given by 3x+ 5z= 11π

In order for the tangent of the curve line to be parallel to the plane, it must be perpendicular to the normal vector to the plane. r'(t) · n = 0 n = <3, 0, 5> −30t + 20t^3 +10t = 0 2t^4 − 3t^2 + 1 = 0 (2t^2 − 1)(t^2 − 1) = 0 t = +/- 1, +/- 1/√2 domain is r>0 so.. t = 1,1/√2 C, where tangent vector of line is parallel to curve is at r(1) and r(1/√2)

A projectile has total acceleration (including the gravitational acceleration) of at the time t after it is launched. The projectile is launched from the ground with an initial speed of 6√2 at an angle of π/4 relative to the horizontal. (a) Find formulas for the velocity and the position vectors.

Integrate a(t) and remember constants. Set v(0) = initial velocity (<6√2cos(π/4)=6,6√2sin(π/4)=6>) to solve for first round of constants. Integrate again and set s(0) = initial position (<0,0>) to solve for second round of constants.

r(t) = <2t, cos(t), 65 − 3 cos(2t) − 2t>(a) (10 points) At time t =π/2, the stream takes you around a particularly big tree. Find the curvature of the stream at that point.

K(t) = |r'(t) x r''(t)|/|r'(t)|^3

Find an equation of the plane containing the points (3, 2, 1), (6, 7, 2), (16, 11, 2). Find the area of the triangle formed by the three points in part (a).

Label points as P,Q,R, calculate PQ and PR. Cross PQ and PR, use that as normal vector for plane. Use P as point. The area of the triangle is half the area of the parallelogram spanned by PQ and PR. Area =12|PQ×PR|

Bender: r1(t) = <5t, 25t^2 + 3> Robot Santa: r2(t) = <2t, 8t3 − 4t + 3> In both of these parametrizations, t represents the number of hours since the beginning of the journey, and distance is measured in miles.

Length given by 0->t∫√((dx/t)^2 + (dy/dt)^2) L1 = 0->2/5∫√((5)^2 + (50t)^2) L1 = 0->1∫√((2)^2 + (24t^2-4)^2)

A tow truck pulled a broken-down car horizontally 10 m. The chain attaching the car to the truck is 60◦ above the horizontal. In moving the car, the tow truck performs 100 Joules of work. Supposing the truck applied a constant force along the chain to perform this work, what was the magnitude of that force (in Newtons)?

Let F be the force vector applied by the tow truck in the direction of the chain, and let D be the displacement vector of the car. Note that kDk = 10 and the angle between F and D is 60◦. The work done by the tow truck in moving the car is W = F · D = |F||D| cos(60◦) = |F|(10)(1/2) = 5|F| Since W = 100, we see that the magnitude of the force is 100 = 5|F| |F| = 20 N

The following equation in spherical coordinates describes a sphere in R3. Give an equation in rectangular coordinates that describes the surface. Find the radius and center of the sphere: ρ = 2 cos(θ) sin(ϕ) − 4 cos(ϕ)

Multiply entire equation by ρ ρ^2 = x^2 + y^2 + z^2 x = rsinθcosφ y = rsinθsinφ z = rcosθ Move everything over to left side, complete the square to find the center and radius

Find the maximal and minimal values of f(x, y) = 2xy in the region R given by x^2 + y^2 ≤ 1

Only critical point where fxx = 0 and fyy = 0 is (0,0) f(0,0) = 0 Solve rest w/Lagrange multipliers to find test points on boundary -> (+-√(1/2),+-√(1/2)) = 1, -1, -1, 1 max = 1, min = -1

Consider the curve C defined by x = 1 − t4, y = t2 + 2, z = cos(πt). (a) (4 points) Show that C passes through the point P = (0, 3, −1) more than once. Find all t values when the curve passes through P.

Plug (0,3,-1) into x,y,z and find t values that satisfy.

lim(x,y,z)->(-1,0,0) (3x^3 + x^2y^2-z^4)/(x^2 + y^2 + z^4 + 2)

Plug in (-1,0,0) into x,y,z -> lim(x,y,z)->(-1,0,0) = -1

Find an equation of the plane that contains both the point (1, 3, 4) and the line x = 2t, y = 3 + t, z = 1 + 4t.

Plug in 0 into line for another point. Find vector between two points, cross that vector with direction vector of line, use that as normal line and (1,3,4) as point.

Let P be the plane given by the equation x + y − 2z = 3 (b) Find the point of intersection of the line L with the plane P.

Plug in parameterizations of x y and z into plane equation, solve for t, then use that t value in the parameterizations to find the point.

How to find critical points

Points where fx = 0, fy=0.

Quadratic approximation formula

Q(x,y) = f(x,y) + fx(a,b)(x-a) + fy(a,b)(y-b) + .5fxx(x-a)^2 + fxy(x-a)(y-b) + .5fyy(y-b)^2

Let C be the curve parametrized by r(t) = <2cost, 2sint, e^t>, 0 ≤ t ≤ π. Find the point on C where the tangent line to C is parallel to the plane √3x + y = 1

Recall that a line with direction vector v is parallel to a plane with normal vector n if and only if v and n are perpendicular, which is true if and only if v · n = 0. r'(t) · <√3, 1, 0> = 0 r'(t) = <−2sin(t), 2cos(t), e^t> r'(t) · n = −2√3 sin(t) + 2 cos(t) 0 = −2√3 sin(t) + 2 cos(t) cost = √3 sin(t) t = π/6 C = r(π/6) = <√3, 1, e^π/6>

Let P be the plane given by the equation x + y − 2z = 3 (c) Find the distance from the point (4, −1, 3) to the plane P.

Same thing as distance between (4,-1,3) and point of intersection. Use distance formula.

lim(x,y)->(0,2) (x^4 + x^2*y^2)/(x^2 + (y-2)^2)

Set x = 0, find lim as y->2 Set y = 2, find lim as x->0 If they differ, then lim DNE.

A projectile has total acceleration (including the gravitational acceleration) of at the time t after it is launched. The projectile is launched from the ground with an initial speed of 6√2 at an angle of π/4 relative to the horizontal. b)What is the horizontal distance traveled by the projectile from the time it is launched until it hits the ground?

Set y component of position vector equal to 0. -4t^2 + 6t = 0, solve for t. Plug t value into x component of position vector, result is horizontal distance.

T/F If z = h(x + y) for some differentiable function h(u), then ∂z/∂x and ∂z/∂y must be equal for all values of x and y.

T

T/F Let g(x, y, z) be a continuous function of three variables. The level surface g = 1 must not intersect the level surface g = 2.

T

T/F The function f(x, y) = √(x2 + y2) is continuous on the entire xy-plane.

T

Bender: r1(t) = <5t, 25t^2 + 3> Robot Santa: r2(t) = <2t, 8t3 − 4t + 3> In both of these parametrizations, t represents the number of hours since the beginning of the journey, and distance is measured in miles. (b) (5 points) If they leave the office at the same time, will Bender and Robot Santa crash into each other after they leave the office?

They will crash into each other only if r1(t) = r2(t) for some t. This gives the following system of equations: 5t = 2t 25t^2 + 3 = 8t^3 − 4t + 3 The first equation gives t = 0 as the only solution. Therefore, Bender and Robot Santa DO NOT crash into each other after leaving the office!

The trace of this surface in the plane z = 1 is a curve described by the equation x^2 + y^2 = y^3. This equation implicitly defines y as a function of x. Compute y'(x) when x = −2 and y = 2

To find dy/dx, take derivatives of both sides of the equation d/dx(x^2 + y^2) = d/dx(y^3) 2x + 2y · dy/dx = 3y^2 · dy/dx dy/dx = 2x/(3y^2 - 2y) dy/dx |(-2,2) = -1/2

T/F c · (~a ×~b)| = |a · (b × c)|

True

The plane x + y − 2z = 1 is perpendicular to the plane x + y + z = 1.

True (dot products = 0)

T/F A circle has constant curvature.

True, ellipses have variable curvature.

T/F The domains of f(x, y) = √(x^2 − y^2) and g(x, y) =ln(x^2 − y^2) are different.

True, one can be zero, two cannot.

T/F Two curves which are tangent to each other at a point may have different curvatures at that point.

True, they don't necessarily have to have the same incident angle.

T/F If a particle moves at constant speed then the tangential component of the acceleration is zero.

True.

T/F The mixed partial derivatives fxy(a, b) and fyx(a, b) are equal.

True.

Find the points on the surface x2 − y^2 +z^2/4= −1 that are closest to the point (0, 0, 1).

Under the assumption that x^2 - y^2 + z^2/4 = -1 we have x^2 + y^2 + (z-1)^2 = 2x^2 + z^2/4 + (z-1)^2 + 1 = 2x^2 + 5z^2/4 - 2z + 2 = 2x^2 + (5/4)(z-4/5)^2 + 6/5 Minimum value when x = 0 and z = 4/5

Consider the curve C defined by x = 1 − t^4, y = t^2 + 2, z = cos(πt). (b) (8 points) Find the equations of all tangent lines to C at the point P.

Use P as position vector. Take derivative of original, plug t values into derivative and use that as directional vector.

Let P be the plane given by the equation x + y − 2z = 2. (c) Find the distance from the point (5, 9, 0) to the plane P : x+y−2z = 2

Use distance formula between point of intersection of line and plane and (5,9,0)

Let P be the plane given by the equation x + y − 2z = 3 (a) Find parametric equations for the line L through (4, −1, 3)that is perpendicular to the plane P.

Use normal vector to plane as direction vector of the line and point as position vector x = 4 + t, y = -1 + t, z = 3 - 2t

*Two balloons are flying in the sky. Their paths (for t ≥ 0) are given by r1(t) = r2(t) = <1, t, t^2 + 2t> (b)Do the balloons collide? If so, at what time ?

Use parameter 's' for r2. If r1(t) = r2(s) at a point where t = s, then the balloons collide.

After a daring leap from a tree branch while chasing a squirrel, Marcel the cat is left dangling from a clothes line hanging between a house and the tree. The segment of rope between Marcel and the tree makes an angle of 45◦ with the horizontal. The segment of rope between Marcel and the house makes an angle of 30◦ with the horizontal. If Marcel weighs 15 lbs, find the tension vectors T1 and T2 in each piece of rope and their magnitudes.

W = <0,15> T1 = |T1| = <-T1cos45,T1sin45> T2 = |T2| = <T1cos30,T2sin30> T1 + T2 + W = <0,0> T2√3/2−T1√22 = 0 T2/2+T1√2/2− 15 = 0 First eq gives T1 = T2√3/√2 Plug into second eq for T2 + T2√3 = 30 T2 = 30/(1+√3) T1 = 15(3-√3)/√2 T1 = <-15(3-√3)/2,15(3-√3)/2> T2 = <15(3-√3)/2,15(√3-1)/2>

A wind blows in the northeast direction (i.e., in the direction of the vector v = <1, 1>, exerting a force of 30 Newtons on the boat. The professor sets the coconut leaf sails due north, intending to reach a nearby island, which is 75 meters away. How much work will the wind need to do to carry them to the island?

W = F · D = |F||D| cos(45◦) = 30(75)√2/2= 1125√2

They reach the island, collect more coconuts, and wait for the wind to change direction so they can head back. Eventually the wind begins to blow in a direction east of due south and they sail back, carried by a wind exerting a force of 20 Newtons. If the wind does 750 Joules of work to carry them back, in which direction was the wind blowing?

W = F · D = |F||D| cos(θ) 750 = 20*75*cos(θ) cos(θ) = 1/2 θ = π/3 60◦ E of S

r(t) = <t3 + 4t, 16t, 4t^2 + 100> l(t) = <15t + 20, 9t + 28, 25t + 14>, will the Reavers' ship collide withthe Serenity

We need to determine if the system of equations given by r(t) = l(t) has a solution. This system of equations is 1. t^3 + 4t = 15t + 20 2. 16t = 9t + 28 3. 4t^2 + 100 = 25t + 14 2. Only t = 4 solves eqation. Solve for r(4) = (80, 64, 164) Solve for l(4) = (80, 64, 114) r(4) != l(4) Ships do not collide

Let P be the plane given by the equation x + y − 2z = 2. (b) Does the line intersect the plane? If so, find the point of intersection.

Yes, because the line is orthogonal to the plane. Plug the parameterization of the line into x, y, and z and solve for t to find the point of intersection.

Travis Pastrana wants to jump his bike across a 14 m gap at a 45 degree angle and the wind makes his acceleration <4,0>. How fast does he have to go to make the jump?

a(t) = <4,0> + <0,-10> integrate a + v(0) = velocity v(t) = <4t,-10t> + <Vcos45,Vsin45> v(t) = <4t + V√2/2,-10t + V√2/2> integrate v + r(0) = position r(t) = <2t^2 + Vt√2/2, -5t^2 + Vt√2/2> + <0,0> r(t) = <2t^2 + Vt√2/2, -5t^2 + Vt√2/2> Find t where h=0 5t^2 = Vt√2/2 Put this into equation: 2t^2 + Vt√2/2 = 14 and solve for velocity

V (x, y, z) = 5x^2 − 3xyz (a) Find the rate of change of the potential at P(0, 1, 2) in the direction of the vector v = i + j − k

a) Calculate ∇V(0,1,2) ∇V (x, y, z) = <10x − 3yz, −3xz,−3xy> ∇V (0, 1, 2) = <−6, 0, 0> Calculate unit vector in direction of v u = v/|v| = <1/√3,1/√3,-1/√3> DuV (0, 1, 2) = ∇V (0, 1, 2) · u = −6/√3

Picture of 3 vectors where v + w = u |u| = 6, |v| = 3, u · v = 12 (a) (2 points) Express w in terms of u and v. (b) (3 points) Find u · w (c) (3 points) Find v · w (d) (4 points) What is the magnitude of w?

a) v + w = u w = u - v b) u · w = u · (u - v) u · w = u · u - u · v u · w = 36 - 12 = 24 c) v · w = v · (u - v) v · w = v · u - v · v v · w = 12 - 9 = 3 d) |w| = ? |w|^2 = w · w (Remember: w = u-v) |w|^2 = w · (u - v) |w|^2 = w · u - w · v |w|^2 = 24 - 3 = 21 |w| = 21

(a) Find parametric equations for the line of intersection of the two planes x + 2y + z = 1 and 3x + y + z = 1. (b) (3 points) Find the cosine of the angle between those planes in part (a).

a) Need to find a point on both planes. Plug in x = 0 for both, you get y = 0, z = 1. Cross two directional vectors. Cross product becomes new directional vector for line, use (0,0,1) as position vector of line. b) cosθ = (n1∙n2)/|n1||n2| = √6/√11

*Two balloons are flying in the sky. Their paths (for t ≥ 0) are given by r1(t) = r2(t) = <1, t, t^2 + 2t> (a)Do the paths of the balloons intersect? If so, where?

a) Use parameter 's' for second equation, and set vector components of r1 and r2 equal to each other. If r1(t) = r2(s) at some point, they intersect.

A snowball with mass 0.4 kg is thrown northward into the air with a speed of 20m/s at an angle of 45◦ from the ground. A wind applies a steady force of 4N to the ball in a westerly direction. The magnitude of the acceleration due to gravity is given by 10 m/s^2 (c) Find the tangential component of the ball's acceleration at the highest point, that is, find the component of the acceleration vector along the tangent vector you computed in part (b).

aT = (a·v)/|v| = (<-10,0,-10> · <-10√2,10√2,0>)/ |<-10√2,10√2,0>| = 100√2/(10√2 · √2) = 5√2 m/s^2

V (x, y, z) = 5x^2 − 3xyz . P(0, 1, 2). v = i + j - k (b) In which direction does V increase most rapidly at P?

b) V is increasing fastest in the direction of the gradient vector ∇V (0, 1, 2) = −6i

V (x, y, z) = 5x^2 − 3xyz . P(0, 1, 2). v = i + j - k (c) What is the maximum rate of change at P?

c) The maximum rate of increase is |∇V (0, 1, 2)| = 6.

Let P be the plane given by 2x − y + 2z = 1, and let L be the line given by x = 7t + 3 y = 2t + 1 z = 3t − 2 (b) (5 points) Find two points on the line L which are a distance of 24 units from the plane.

d =|tv · n|/|n| d =18|t|/3 d = 6|t| 24 = 6|t| t = +/- 4 Plug t back into line equation to find two points.

A function f(x, y) satisfies ∇f(0, 1) = <3, −4> b) Compute ∂f/∂r at (0,1)

f is in terms of x and y, x and y are in terms of r and θ. x = rcosθ, y = rsinθ ∂x/∂r = cosθ = cos(π/2) = 0 ∂y/∂r = sinθ = sin(π/2) = 1 ∂f/∂r = ∂f/∂x * ∂x/∂r + ∂f/∂y * ∂y/∂r ∂f/∂r = (3) * (0) + -4(1) = -4

A partial differential equation of the form utt = a 2uxx, where a is some real number, is called a wave equation. Suppose that f(x, t) = cos(2x + 6t) − 2 sin(2x − 6t) Verify that ftt = 9fxx

fx = −2 sin(2x + 6t) − 4 cos(2x − 6t) ft = −6 sin(2x + 6t) + 12 cos(2x − 6t) fxx = −4 cos(2x + 6t) + 8 sin(2x − 6t) ftt = −36 cos(2x + 6t) + 72 sin(2x − 6t) From the calculation, we see that 9fxx = 9(−4 cos(2x + 6t) + 8 sin(2x − 6t)) = −36 cos(2x + 6t) + 72 sin(2x − 6t) = ftt Therefore, ftt = 9fxx, so f is a solution of the given differential equation.

Points where gradient vector equal zero

maxima/minima/saddle points

Find an equation for the tangent plane to the quadric surface x^2 +y^2/4 − z^2/9 = 1 at the point P(1, 2, 3).

normal vector to use is gradient vector at (1,2,3) = <2x, y/2, -2z/9> = <2,1,-2/3> Use point as center so... 2(x-1) + (y-2) -2/3(z-3) = 0

r(t) = <t3 + 4t, 16t, 4t^2 + 100>. (a) (8 points) What is the curvature of the Serenity's path at the point (0, 0, 100)?

r(t) = (0, 0, 100) only when t = 0. Therefore, we need to calculate the curvature at t = 0. K(0) = |r'(t) x r''(t)|/|r'(t)|^3 K(0) = 32√17/(4√17)^3 K(0) = 1/34

Suppose a triangle in R3 has vertices (0, 0, 0), (1, 1, 1), and (a, a, 1), where a != 1 is some real number. If the triangle has area 5√2, find all the possible values for a.

u = <1,1,1> j = <a,a,1> Area = (u x j)/2 5√2 = |<1-a,a-1,0> 5√2 = (|a - 1|√2)/2 |a - 1| = 10 a = -9,11

A snowball with mass 0.4 kg is thrown northward into the air with a speed of 20m/s at an angle of 45◦ from the ground. A wind applies a steady force of 4N to the ball in a westerly direction. The magnitude of the acceleration due to gravity is given by 10 m/s^2 (a) Find the ball's acceleration and initial velocity vectors. Hint: You may find it convenient to set up a three-dimensional coordinate system with the x-axis pointing east and the y-axis pointing north.

v(0) = <0, 20 cos 45◦, 20 sin 45◦> = <0, 20√2/2, 20√2/2> = <0, 10√2, 10√2> g = <0,0,-10> awind = -<4N/.4kg,0,0> awind = <-10,0,0> atotal = g + awind atotal = <-10,0,-10>

*A satellite circles around the earth along the orbit: *r(t) = <3 sin t, cos t, 7> Find the velocity, the speed, and the acceleration of the satellite at time t.

v(t) = s'(t) speed(t) = |v(t)| a(t) = v'(t)

A snowball with mass 0.4 kg is thrown northward into the air with a speed of 20m/s at an angle of 45◦ from the ground. A wind applies a steady force of 4N to the ball in a westerly direction. The magnitude of the acceleration due to gravity is given by 10 m/s^2 (b) Find parametric equations for the line tangent to the path of the ball when it reaches the highest point.

v(t) = v(0) + 0->t∫a(u)du = <0,10√2,10√2> + 0->t∫<-10,0,10> du = <-10t,10√2,10√2 - 10t r(t) = <0,0,0> + 0->t∫<-10u,10√2,10√2 - 10u>du = <-5t^2,10√(2)t,10√(2)t,10√(2)t-5t^2>

The equation x^2 + y^2/4 − z^2/9 = 1 defines y implicitly as a function of x and z. Find ∂y/∂x.

∂y/∂x = -Fx/Fy = -2x/(y/2) = -4x/y