Calc 3 Final test explain

Find the shortest distance dmin from the point P(1,4,5) to the line through the points Q(-5,5,3) and R(4,-4,1). Test 1 #15 1.1.3 #10

1. Shortest distance from P to the line will be the length of the orth comp of QP WRT QR. 2. The orth comp will be perpendicular in this case. 3. We define the 3 pts given. 4. By finding vectors QP and QR, we form a triangle that will give us the orth comp of QP WRT QR. 5. We find the projection of QP onto QR. 6. We find the difference between QP and the projection since the sum of the projection comp of the orth comp is QP. 7. Once we have the coordinates, we calc the magnitude.

Find the shortest distance dmin from the point P(4.6,0) on the x-axis to the ellipse x^2/8^2+y^2/7^2=1. Test 2 #13 2.3.1 #3

1. We find the function, f, by deriving it from the distance formula (f=Sqrt[(x-x1)^2+(y-y1)^2]) 2. This measures the distance from P, to the ellipse. 3. We know the shortest distance occurs at a pt {x,y} such that the level curves g(x,y)=0 and f(x,y)=C touch. 4. For the level curves to touch, the must share the same tangent line. 5. General eq of this tangent line for the level curve f(x,y)=C at a pt {x0,y0}: fx(x0,y0)(x-x0)+fy(x0,y0)(y-y0)=0. 6. For the ellipse g(x,y)=0 at the same pt: gx(x0,y0)(x-x0)+gy(x0,y0)(y-y0)=0 7. We find the partial derivatives of each function to find the gradient of each tangent line. 8. If the tangent lines are the same, the gradients will be proportional, where z is the proportionality constant. 9. The shortest distance is the value of the distance function (f) at the critical points obtained after solving for x, y, and the proportionality constant. 10. The smallest value is the shortest distance.

Use the quadratic approximation to estimate each number. Test 2 #5 2.1 #5

1. We first establish that this is a Taylor Series centered about x=4, indicated by the problem. 2. Function will be defined by the number that changes as the variable, x. 3. This makes functions: x^4, 4^x, x^x, respectively. 4. Taylor Series takes the form: f(x)~f(a)+f'(a)(x-a)+f''(a)(x-a)^2/2, where a is the pt the series is centered around. 5. We evaluate the change in f(x) in the Series to get the answer.

Find the area of the ellipse x^2/5^2+y^2/9^2=1. Test 3 #3 3.1.2 #1

1. Area of an ellipse is equal to the area swept by the position vector of a particle going along the ellipse once. 2. Parameterize the ellipse by f(t)={acos[t],bsin[t]}, where a and b are constants from the ellipse and t is bounded by 0,2Pi. 3. We subdivide the time interval into smaller subintervals of length deltat. 4. Area swept by the position vector=the sum of all areas swept over smaller intervals. 5. Area over the shorter intervals~the area of the triangle determined by the position vector and the displacement vector. 6. Assume motion over short intervals has the same velocity so displacement can be represented by velocity(deltat). 7. Velocity=the first derivative of the position function. 8. Using the formula for the area of a triangle, total area over smaller intervals=(sigma)(1/2)Det[r(ti),v(ti)]*deltat. 9. As deltat gets smaller the approximation improves 10. This is the definition of the definite integral over the bounds 0,2Pi

Let V be a right triangular prism height 3 with the base PQR (and the top P'Q'R') being a right triagnle with the right angle at the vertex R (resp. R') and the sides |PR|=3 and |QR|=4. Set up and compute the triple integral where the function f is the square of the distance from the side PP'R'R. Test 3 #8 3.2.2 #2

1. Because f is the square of the distance from PP'R'R, f(x,y,z)=z^2. 2. We will integrate WRT x, then y, and then z. 3. We can see that x bounds are in terms of y and z, our y bounds will be in terms of z, and z bounds will be constants. 4. X is bounded from 0 to plane PP'QQ'. 5. We find the equation of the plane and solve for x to get the upper bound for x. 6. We then project the volume onto the yz plane to get the bounds for y. 7. This forms a rectangle making the bounds for y 0 to the height 3. 8. To find the bounds for z, we project the rectangle onto the z axis, making the bounds constants 0,4.

Let A=(triangle)PQR be a triangle with sides |PR|=9, |QR|=2, and with the right angle at the vertex R. Set up and compute the double integral where f is the squar of the distance from the side PR. Test 3 #6 3.2.1 #2

1. Because f is the square of the distance from the side PR, f=x^2 2. We integrate WRT x and then y. 3. We can tell x is bounded from 0 to line PQ. 4. So we find the equation of line PQ to find the bounds for x. 5. We project the triangle onto the y-axis, making the bounds 0 to 9.

Use the Chain Rule to find the partial derivatives ft and fs at the point (t,s)=(1,1), where f=(4y/x)^x, x=(2/5)t+(2/5)s,y=(3/10)ts. Test 2 #45 2.3.4 #6

1. Definition of the differential for F(x,y): dF=Fx(dx)+Fy(dy) 2. Differentials for x(t,s): dx=xt(dt)+xs(ds) 3. Differentials for y(t,s): dy=yt(dt)+ys(ds) 4. Substitue dx and dy into dF: dF=Fx(xt(dt)+xs(ds))+Fy(yt(dt)+ys(ds)) 5. Since F is a function of x and y, and x and y are functions of t and s, F is ultimately a function of t and s. 6. Because of step 5: dF=Ft(dt)+Fs(ds) 7. We equate the two formulas for dF. 8. Set dt=1 and ds=0: Ft=Fx(xt)+Fy(yt) 9. Set dt=0 and ds=1: Fs=Fx(xs)+Fy(ys).

Let f(x,y)=1+x+4y-3x^2-2xy+2y^2 be a function on a coordinate (x,y)-plane. Let A=(triangle)PQR be a triangle with vertices P(0,2), Q(7,0), and R(0,0). Use the Divergence Theorem to find the double integral of the function over the triangle. Test 3 #11 3.3.3 #1

1. Divergence theorem states: (integral)div(F)dA=(integral)F.nds 2. We first find a vector field F, such that the divergence of F=f. 3. We parameterize the boundary of the triangle counterclockwise. 4. We find the outward pointing unit normal vector to each side. 5. Note that ds=|v|dt. 6. We find the speed along each boundary. 7. We compute: (integral)fdA=(integral)div(F)dA=(integral)(F.n)*|v|dt.

The length L, width W, and height H of a bos change with time. At a certain instant the dimensions are L=7m, W=3m and H=6m, and L and W are decreasing at a rate of 9m/s while H is increasing at a rate of 5m/s. At that instant find the rate At of change of the surface area of the box. Test 2 #47 2.3.4 #8

1. Equation for surface area of a box: S=2lw+2lh+2hw 2. Differential for S(t): dS=St(dt) 3. Differential for S(l,w,h): dS=Sl(dl)+Sw(dw)+Sh(dh) 4. Differential for l(t): dl=lt(dt) 5. Differential for w(t): dw=wt(dt) 6. Differential for h(t): dh=ht(dt) 7. Substituting dl, dw and dh into the second dS formula: dS=Sl(lt)(dt)+Sw(wt)(dt)+Sh(ht)(dt) 8. Set dS equal to St(dt) and put dt=1: St=Sl(lt)+Sw(wt)+Sh(ht) 9. Evaluate the function for given l, w, h values.

Find the angle, theta, between two line y=4x+7 and y=5x+9. Test 1 #6 1.1.3 #1

1. First we find points on each given line. 2. We use the points to compute a vector for each line. 3. Find the cosine of the angle between the vectors, which is the dot product of vectors over magnitudes of vectors). 4. Find the angle by taking the inverse cosine.

Let f(x,y)=4+8x+3y+9x^2+7xy+y^2 be a function on a coordinate (x,y)-plane. Let A=(triangle)PQR be a triangle with vertices P(0,3), Q(4,0), and R(0,0). Use the Green's Theorem to find the double integral of the function over the triangle. Test 3 #12 3.3.3 #2

1. Green's Theorem states: (integral)curl(F)dA=(integral)F.Tds 2. We first find a vector field F such that curl of F=f. 3. We parameterize the boundary of the triangle counterclockwise. 4. We find the unit tangent vector to each side. 5. Note that ds=|v|dt. 6. We find speed along each boundary. 7. We compute: (integral)fdA=(integral)curl(F)dA=(integral)(F.T)*|v|dt.

Find the area of the triangle PQR where P(3,1), Q(-3,0), R(2,6). Test 1 #36 1.3.1 #1

1. Knowing the formula for the area of a triangle using [(1/2)Det[u,v]], we define u and v to be vectors of the triangle. 2. We use the determinate because it calculates the area of a parallelogram that the triangle is apart of. 3. Because of the area of a triangle, we multiply the determinate by 1/2. 4. This gives the area of the parallelogram which is the area of the triangle.

Find the coefficient c1,2 with the term (x-2)(y-1)^2 of the Taylor series f(x,y)=cos(2xy) centered at the point (2,1). Test 2 #30 2.3.2 #1

1. Taylor series has the form: f(x,y)=c0,0+c1,0(x-2)+c1,1(x-2)(y-1)+c1,2(x-2)(y-1)^2+c0,1(y-1)+c0,2(y-1)^2+ other terms of (x-2) higher than degree 1 and (y-1) higher than degree 2. 2. Then fx(x,y)=c1,0+c1,1(y-1)+c1,2(y-1)^2+other terms of (x-2) higher than degree 0 and (y-1) higher than degree 2 fxy(x,y)=c1,1+2c1,2(y-1)+ other terms of (x-2) higher than degree 0 and (y-1) higher than degree 1 and fxyy(x,y)=2c1,2+other terms of (x-2) higher than degree 0 and (y-1) higher than degree 0. 3. Evaluating x=2 and y=1: fxyy(2,1)=2c1,2 so c1,2=fxyy(2,1)/2

Find the directional derivative of the function f(x,y)=(2y/x)e^(xy) at the given point (2,3) in the direction of vector v=<2,5>. Test 2 #37 2.3.3 #5

1. The directional derivative of f(x,y) at pt (a,b) in the direction of v is the rate of change of the function f in the direction of v. 2. To find the directional derivative take the scalar comp of (upside down triangle)f(a,b) in the direction of v. 3. This makes Dvf(a,b)=(upside down triangle)f(a,b).(unit vector v).

Find the extreme values of f(x,y)=3x^3+2x^2+2y^3+8y^2+1 on the region D={(x,y):(x/4)^2+(y/2)^2<1}. Test 2 #51 2.3.5 #4

1. To find the extreme values, we find the critical pts of f(x,y) in the interior of the region and on the boundary given. 2. To find the crit pts in the interior we find the gradient of f and set it equal to 0 and limit the pts by region D. 3. To find the crit pts on the boundary of region D we parameterize the ellipse, and take the first derivative of the ellipse and set it equal to 0 and solve for t restricted from 0 to 2Pi. 4. Lastly we evaluate all the critical points we found into f(x,y) to find the extreme values.

Given four points P(1,5,2), Q(-4,3,0), R(5,9,6), S(7,8,4) in space, find the distance d from the point P to the plane QRS. Test 1 #38 1.3.4 #4

1. To solve, we need to find the height of the parallelepiped spanned by vectors QP, QS, and QR. 2. We find this value by finding the volume of the parallelepiped by dividing it by the area of the face QRS. 3. The area of the face is found by taking the cross product of QR and QS. 4. The volume of the parallelepiped can be found by taking the determinate of the set of vectors that make up the sides of the parallelepiped.

Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane 2x+5y+3z=4. Test 2 #49 2.3.5 #2

1. Volume of a rectangular box: V=lwh=xyz 2. To put the vertex on the given plane, solve the plane equation for one variable and substitute that into the volume equation. 3. This also makes the volume equation a function of two variables. 4. We then find the critical pts of the volume function. 5. Find crit pts by setting the partial derivatives of volume WRT the remaining variables equal to 0. 6. Crit pts=(dV/dx=0 and dV/dy=0) 7. Evaluate the critical points into the volume function to find the max value.

Find the coefficient c3 with the term (x-4)^3 of the Taylor series for f(x)=sinx centered at the point 4. Test 2 #1 2.1 #1

1. We are looking for the coefficient c3 for the polynomial of the Taylor Series f(x)=sin(x) centered at given pt x=a. 2. General formula: f(x)=c0*(x-a)^0+c1*(x-a)^1+c2*(x-a)^2+c3*(x-a)^3+c4*(x-a)^4+... 3. To determine c3 we differentiate f(x) 3 times WRT x and sub x=a. 4. f'(x)=cos(x)=c1+2*c2*(x-4)+3*c3*(x-4)^2+ other terms higher than degree 2. 5. f''(x)=-sin(x)=2*1*c2+3*2*c3*(x-4)+ other terms higher than degree 1. 6. f'''(x)=-cos(x)=3*2*1*c1+other terms higher than degree 0. 7. Evaluate f'''(4): c3=-cos(4)/(3*2*1)

Find the minimal value of f(x,y)=6x^3+2x^2-4y^3+18y^2+3 on the curve (x/9)^2+(y/3)^2=1. Test 2 #29 2.3.1 #4

1. We consider the level curves f(x,y)=V, which is given, and g(x,y)=0 where g(x,y) is the ellipse given set equal to zero. 2. The minimum value of f on the ellipse occurs at (x0,y0) such that the level curves touch. 3. Level curves only touch if their tangent lines to the level curves are proportional, where z is the proportionality constant. 4. Tangent lines of the level curves: fx(x0,y0)(x-x0)+fy(x0,y0)(y-y0)=0 and gx(x0,y0)(x-x0)+gy(x0,y0)(y-y0)=0 5. Because the tangent lines are proportional: fx(x0,y0)=z*gx(x0,y0) and fy(x0,y0)=z*gy(x0,y0). 6. To find the critical pts we solve the system of equations of set up by the tangent lines with the addition of g(x0,y0)=0. 7. We evaluate the function f at each of the critical pts to find the min value.

Determine whether the given vectors are orthogonal. u=<-2,4>, v=<4,2>. Test 1 #5 1.1.2 #2

1. We create given vectors u and v. 2. We know that 2 vectors are orthogonal if and only if their dot product is equal to zero. 3. We compute the dot product of u and v.

The force F=<5,9,-8> acts on the point P(2,3,5). How much (NUMERICALLY) does the force pull toward the origin? Test 1 #12 1.1.3 #7

1. We define given point, P, and the origin. 2. We define vectors F and PO. 3. Derivation of the formula. 4. Suppose F=Fpa+Fpe is teh Parallel-Perpendicular Decomposition of F along PO. 5. Since Fpa is parallel to PO, we have the Direction-Magnitude Decomposition. 6. Fpa=(PO/|Po|)*compPO(F), where we need to find the number compPO(F). 7. Dot product the equality F=Fpa+Fpe with PO (both sides):PO.F=PO.Fpa+PO.Fpe 8. Since Fpe is perpendicular to PO, the dot product PO.Fpe=0. 9. New formula PO.F=(PO.PO/|PO|)*compPO(F) 10. PO.PO=|PO|^2 11. Because of step 10, compPO(F)=PO.F/|PO|

Find the equation of the plane containing the points P(1,1,12), Q(2,6,4), ad R(-8,4,8). (Let x, y, and z denote the standard coordinate variables). Test 1 #20 1.2.3 #1

1. We define given points P, Q, and R. 2. We form 3 equations by subbing in the given x, y, and z into the equation Ax+By+Cz=D. 3. We solve for A, B, C, and D in the system of equations. 4. Answer can vary proportionally, so we sub 1 for A and find the rest of the values, B, C, and D. 5. We the sub A, B, C, and D back into Ax+By+Cz=D to find the solution.

Find the distance d from the point P(2,3,1) to the plane 2x+8y+3z=5. Test 1 #32 1.2.3 #5

1. We define given pt P. 2. Find a pt Q on the plane, and define vector QP. 3. Define another vector as the coefficients for the plane equation because it is the gradient as well as orthogonal to the plane. 4. The distance is equal to the magnitude of the projection of QP on to the gradient.

Find the shortest distance dmin from the point P(0,2,4) to the line through the points Q(-1,4,1) and R(3,-5,2). Test 1 #23 1.2.1 #7

1. We define given pts P, Q, and R 2. We define vector QR 3. We use QR and the points to define the parametric presentation of our line f(t)=Q+QR*t. 4. We need to find the distance from our point P to our line, f(t). 5. Because of step 4, we define vector, V as V=P-f and take its magnitude, F. 6. We use Calc 1 techniques to find critical pts to determine the min distance. 7. So we find the 1st derivative of F, set it equal to 0, and solve for t. 8. We evaluate the value for t in F to find the shortest distance.

The force F=<2,1,3> acts at the point P(-4,4,4). How much does the force pull toward the origin? Test 1 #13 1.1.3 #8

1. We define points P and the origin and vectors F and PO. 2. We use the projection formula and plug in the vectors using vector F with respect to vector PO.

Find the vector U=<ux,uy,uz> orthogonal to the plane through the points P(1,3,5), Q(5,0,2), and R(3,4,1). Test 1 #31 1.2.3 #4

1. We define pts P, Q, and R. 2. Plug P, Q, and R into the equation Ax+By+Cz=D. 3. Solve the system of equations for A, B, C, and D because the plane passes through each of the points. 4. Since we are solving for 4 variables with only 3 equations, one of the variables will be arbitrary (A=1).

Find an equation of the tangent line to the curve x^2/9^2+y^2/2^2=1 at the point (9Sqrt[3]/2,1). Test 2 #15 2.3.1 #2

1. We define the function f(x,y)=x^2/9^2+y^2/2^2-1 2. We find the partial derivatives of f WRT x and y. 3. Evaluate the partial derivatives at the point P. 4. Tangent line EQ: fx(x0,y0)(x-x0)+fy(x0,y0)(y-y0)=0.

Find an equation of the tangent plane to the graph of the function at the point (-0.2,-0.6) and use the equation to find the z-intercept of the plane, f(x,y)=9-3x+9y+2x^2-4xy+5y^2+2x^3-3x^2y+6xy^2+2y^3 Test 2 #33 2.3.3 #1

1. We define the given function and point f and P respectively as well as an arbitrary point Q(x,y). 2. We find the gradient of the function and evaluate the gradient at the given point P. 3. Equation of the tangent line: z=f(P)+gradf(P).(Q-P), which is the same as finding the first degree Taylor Series (finding the tangent line from the derivative). 4. We evaluate x=0 and y=0 into the tangent eq to find the z-intercept.

Find the area of the triangle QRS where Q(-1,3,2), R(8,3,6), and S(5,4,7). Test 1 #39 1.3.4 #1

1. We define the pts that will for the 3 corners of the triangle. 2. We pick one of the 3 pts to define the vectors that make up the triangle (QR and QS). 3. We use the Area of a parallelogram in 3D space as the magnitude of the cross product of two of its adjacent vectors. 4. Area of a triangle= (1/2) the area of the parallelogram 5. We use this to find the final answer.

Find the orthogonal component of the vector F=<3,5,2> with respect to the vector v=<-3,4,1>. Test 1 #14 1.1.3 #9

1. We define the vectors F and v. 2. We derive the formula for the orthogonal component. 3. Orthogonal component of F WRT v is deonted as orthv(F)=F-projv(F) or Fpe=F-Fpa, where Fpe reps the orthogonal component and is perpendicular to v and Fpa is parallel to v. 4. Using Parallel-Perpendicular Decomposition of the vectors we derive a formula for the orth component of F WRT v. 5. F=Fpa+Fpe, we using Direction-Magnitude Decomposition to find Fpa. 6. Fpa=(v/|v|)*compv(F), where compv(F) is the unknown comp of F WRT v. 7. Dot product the equation F=Fpa+Fpe with v to get v.F=v.Fpa+Fpe. 8. Fpe was define as being perpendicular to v, so the dot product v.Fpe=0. 9. Now, sub Direction-Magnitude Decomposition for Fpa and 0 for v.Fpe in the equation v.F=v.Fpa+v.Fpe. 10. Now the equation is v.F=(v.v/|v|)*compv(F). 11. v.v=|v|^2 12. Equation is now compv(F)=(v.F/|v|). 13. Fpa=(v/|v|)*compv(F) 14. This makes the equation Fpa=(v.F/|v|^2)*v. 15. Sub step 14 into the original equation Fpe=F-Fpa, so Fpe=F=(v.F/v.v)*v. 16. Now we find the orth comp, Fpe.

Use the Second Derivative Test to find all points of local maximum, points of local minimum, and saddle points of the function. (Order your answers from smallest to largest x, then from smallest to largest y.) Test 2 #54 2.3.6 #2

1. We find the critical pts of f by solve the system of equation from setting the gradient of f=0. 2. Once we find the critical points, we use the Second Derivative Test to determine if they are a local max, local min, or saddle point. 3. To do this, we take the second degree Taylor approximation centered at (x0,y0). 4. Taylor approximation: T2(x,y)=f(x0,y0)+fx(x0,y0)(x-x0)+fy(x0,y0)(y-y0)+fxx(x0,y0)/2(x-x0)^2+fxy(x0,y0)(x-x0)(y-y0)+fyy(x0,y0)/2(y-y0)^2. 5. If the approximation at the critical point has a stable max, f has a local max at this point. 6. If the approximation at the critical point has a stable min then f has a local minimum at this point. 7. If the approximation at the critical point has saddle then f has a saddle point at this point.

Find the maximum rate of change Dmax of f(x,y)=(5x^2+3y^2)^-1 at the point (1,1). Test 2 #38 2.3.3 #6

1. We know that the max rate of change of f(x,y) at pt (a,b) is the mag of the gradient of f at (a,b). 2. We take the x and y partial derivatives of f(x,y) to find the gradient. 3. We take the magnitude of the gradient after given pt has been evaluated in the gradient.

At what point does the curve y=7e^(3x) have maximum curvature? Test 2 #23 2.2.3 #2

1. We know the formula for curvature is k=Area[v,a]/|v|^3. 2. We parameterize the curve by setting y=t and solving for x. 3. We find the first and second derivative of the step 2 to get velocity and acceleration. 4. Once we have the curvature equation we take the first derivative, set it equal to 0, and solve for the critical points. 5. Evaluate the critical points in the curvature eq to find the max curvature.

Find the angle theta between the planes 9x+5y+7z=4 and 4x+6y+8z=6. Test 1 #33 1.2.3 #6

1. We need to find the gradient vector of the two planes (the coefficients from their plane equations). 2. We know that the angle between the gradient vectors is equal to the angle between their corresponding planes. 3. This is because the gradient is orthogonal to its respective plane. 4. We find the angle with the Law of Cosines.

Find a point P(p1,p2,0) of intersection of the plane z=0 and the tangent line to the curve f(t)=(5t^2,4t^2-6t+3,6*2^t) at the point (125,73,192). Test 2 #19 2.2.2 #5

1. We need to find the tangent line to the curve f(t) at the given pt Q. 2. The tangent line is given by Q+v*t, where v is the velocity vector, which is the first derivative of f(t) at Q. 3. We need to find the time t0 at which f(t0)=Q. 4. Using the tangent line eq, we solve for the time when the z component of the tangent line=0. 5. Once we have that value, we evaluate it in the eq for the tangent line to find the pt of intersection.

Let V be a triangular pyramid PQRS with the three edges SP, SQ, and SR being mutually orthogonal. Suppose the |SP|=4, |SQ|=3, and |SR|=4. Set up and compute the triple integral where function f is the distance from the side SPQ. Test 3 #9 3.2.2 #3

1. We set S at the origin, make R the z intercept, Q the y intercept, and P the x intercept. 2. Because f is the distance from the side SPQ f(x,y,z)=z. 3. We integrate WRT x, then y, then z. 4. We can see that for any choice of y and z, x is bounded from 0 to plane PQR. 5. So we find the equation of the plane PQR and solve for x to get x's upper bounds. 6. We then project the volume onto the yz plane, which is triangle SQR. 7. We can see that y is bounded from 0 to line RQ. 8. So we get the equation of line RQ and solve for y to get y's upper bound. 9. We then project triangle SQR onto the z axis, making the z bounds 0,4.

Find the curvature k of f(t)=<2t+4,4t^2-2t+3,2t^2+3t+4> at the point (10,33,31). Test 2 #25 2.2.3 #4

1. We set f(t) equal to the given pt and solve for t. 2. To find curvature we use k=Area[v,a]/|v|^3, where v, velocity, is the first derivative and a, acceleration, is the 2nd derivative of f(t). 3. To find the area, we dot the cross product of a and v with itself and take the square root. 4. This is the mag of the vector that is the cross product of a and v. 5. Divide step 4 with the magnitude of v cubed, which is the cube of speed. 6. Once we have k, we plug in the value for t to find the curvature at the given point.

Find the equation of the normal line to the curve x^2/8^2+y^2/9^2=1 at the point (8Sqrt[17]/9,8). Test 2 #39 2.3.3 #8

1. We set the curve equal to 0 to get f(x,y), so f(x,y) is now a level curve. 2. The gradient at a given pt is normal to the level curve (making the gradient also normal to the ellipse). 3. We use the gradient to find the normal line at the given pt by solving for (upside down triangle)f(x0,y0) where (x0,y0) is given. 4. We parameterize the normal line as g(t)=(x0,y0)+(upside down triangle)f(x0,y0)*t. 5. Once we have g(t) we solve one of the components for t and plug it into the other component.

Find the parametric presentation for the line through the points P(1,-3,0) and Q(5,-2,2). (Use t as your parameterized variable). Test 1 #19 1.2.1 #6

1. We use parametric presentation to describe a line. 2. We define give pts P and Q. 3. We use the points to define the vector PQ. 4. The equation of the line f(t)=P+PQ*t 5. This gives us the new parametric points WRT t.

Let F=<-3x+2y,2x-3y> be a field of forces in a coordinate (x,y)-plane. Consider the motion of a particle in the plane given by position function f(t)=(-3t^2+3t-1,2t^2+t+2), where t |0,3|. 1. Find the work W1 done by the field of forces on the particle during this motion. 2. Find the work W2 done by the field of forces on the particle if the particle was moving linearly from the point f(0) to the point f(3). Test 3 #1 3.1.1 #1

1. Work=Area under the curve created by the dot product of the vector field of forces and the displacement vector. 2. To find W1 during motion f(t), we divide the interval into smaller subintervals of length deltat. 3. It can be assumed that motion of the particle is approximately linear over each subinterval. 4. It can be assumed force acted upon the particle is constant 5. This makes Fi=F(f(ti)) and Di=v(ti)*deltat, where Di is displacement and v is the velocity of the particle. 6. Work over the entire motion is approximately the sum of the work done over the smaller intervals. 7. W=(sigma)Wi=(sigma)(Fi.Di). 8. The approximation improves as the limit of delta t approaches 0. 9. This is the definition of the definite integral F(f(t).v(t))dt over the entire time interval. 10. To find W2, first we parameterize the position function. 11. When t=0 the particle is at point P 12. When t=3 the particle is at point Q. 13. We use these points to find vector PQ. 14. Parameterize the linear motion as g(t)=P+PQ*t, where t is bounded 0,1. 15. Use this equation of g(t) to find the new values of x and y in terms of t. 16. Plug these values into the given force EQ. 17. Integrate this new value of force dotted with v(t)=PQ from 0 to 1.