Math 241 Test 2 explanations

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

1. Set the given function f(t) equal to the given point and solve for t. 2. The curvature equation is k=Area[v,a]/|v|^3, where v is velocity and a is acceleration (the first and second 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. Once we have the formula, k, we substitute the value for t to find the curvature at the given point.

Use a quadratic approximation to estimate each number a. (4.03)^4 b. (4)^4.03 c. (4.03)^4.03

1. To solve each problem we must create a function to best describe each scenario. 2. We can use a Taylor series approximation of degree 2 to find a quadratic approximation. 3. A Taylor series has the formula f(x)~f(a)+f'(a)(x-a)+f'(a)(x-a)^2/2 4. Where "a" is 4 (the number the series is centered about. And we evaluate 4.03 in the Series.

The length L, width W, and height H of a bos change with time. At a certain instant the dimension 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.

1. Equation for the surface area of the box is S=2*l*w+2*l*h+2*h*w 2. Differential for S(t): dS=Stdt 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. Subbing dl, dw, and dh into the second dS formula you get dS=Sl*lt*dt+Sw*wt*dt+Sh*ht*dt 8. Set dS equal to Stdt and put dt=1: St=Sl*lt+Sw*wt+Sh*ht 9. Evaluate this function for given l, w, and h values.

Find the minimal value of f(x,y)=7x^3+3x^2-4y^3+12y^2+3 on the curve (x/6)^2+(y/3)^2=1.

1. Finding the gradient of f(x,y) gives teh tangent line of the level curve of f(x,y) at a point (a,b). 2. We make the ellipse equal a level curve by setting the given ellipse equal to 0. 3. We know that the gradient of f and the gradient of the level curve are proportional at (a,b) B/C the tangent lines are the same at (a,b) since the tangent lines are at extremes. 4. (a,b) must also be on the curve g(x,y)=0. 5. We now find the x and y partial derivatives of f and g(x,y), setting them proportional to find the critical points. 6. With this we can form a system of equations with z being the proportionality constant and with the constraint of g(x,y)=0. 7. We now solve for x, y, and z, and substitute these values int f(x,y) to determine the min value.

At which point does the curve y=7e^(3x) have max curvature?

1. The curvature formula is K=Area[v,a]/|v|^3. 2. We must parameterize the curve by setting y=t and solving for x. 3. We solve the parameterized curve for the first and second derivative, which will be our velocity and acceleration. 4. B/C this is a 2-D problem, the area is equal to the determinate of velocity and acceleration. 5. Once we have the curvature equation, we take the derivative and set it equal to zero to solve for the critical points. 6. Evaluate all of the critical points to find the max curvature.

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=3/5t+2/5s,y=3/10ts.

1. The differential of f(x,y) is df=fx*dx+fy*dy 2. The differential for x(t,s) is dx=xt*dt+xs*ds 3. The differential for y(t,s) is dy=yt*dt+ys*ds 4. Subsitute dx and dy into the formula of df to get df=fx*(xt*dt+xs*ds)+fy*(yt*dt+ys*ds). 5. f is a function of x and y, x and y are a function of t and s. This makes f a function of t and s. 6. This makes the differential df=ft*dt+fs*ds 7. Equating the two formulas for df...ft*dt+fs*ds=fx*(xt*dt+xs*ds)+fy*(yt*dt+ys*ds) 8. Make dt=1 and ds=0 9. ft=fx*xt+fy*yt 10. Make ds=1 and dt=0 11. Fs=fx*xs+fy*ys

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

1. The directional derivative of a function f(x,y) at point (a,b) in the direction of a vector <p,q> is the rate of change of the function f in the direction of v. 2. We take the scalar component of gradf(a,b) in the direction of v. 3. This makes Dvf(a,b)=gradf(a,b).(unit vector v).

Use the Chair Rule to find zt at (x,y)=(2,3), where z=ln(3x^2+2y),x=2t^2,y=3t^-2, and t>0.

1. The function z is a function in terms of x and y. 2. x and y are individual function of t. 3. The differential for the function z is dz=zx*dx+zy*dy, where zx and zy are the x and y partial derivatives. 4. B/C x and y are in terms of t and z is in terms of x and y the differential can also be dz=zt*dt. 5. dx and dy are the differentials of x(t) and y(t) and can be written as dx=xt*dt and dy=yt*dt. 6. Substitute dx and dy into dz=zx*dx+zy*dy. 7. Set the two equations equal to one another zt*dt=zx*xt*dt+zy*yt*dt 8. Solve x and y for the t value 9. Evaluate zt by plugging in 1 for dt. The expression is now zt=zx*xt+zy+yt. 10. Plug in the t value into zt to get the final value.

Find the coeefficients c1,2 with the term (x-2)(y-1)^2 of the Taylor series for f(x,y)=cos(2xy) centered at the point (2,1).

1. The 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. 2. fx(x,y)=c1,0+c1,1*(y-1)+c1,2*(y-1)^2 3. fxy(x,y)=c1,1+2*c1,2*(y-1) 4. fxyy(x,y)=2*c1,2 5. We evaluate these at x=2 and y=1 to get fxyy(2,1)=2*c1,2 6. Which takes the final form c1,2=fxyy(2,1)/2.

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.

1. The volume of a rectangular box is V=l*w*h=x*y*z 2. To place the vertex on the given plane, solve the plane equation for one of the variable and plug it back into the volume equation. 3. This also makes the volume equation an function of two variables. 4. Find the critical point of the volume by taking the partial derivatives of the remaining two variables and setting them equal to zero. 5. Plug the critical points into the volume function to find the max value.

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}.

1. To find the extreme values of the function (the min and max), we find the critical point of the interior and on the boundary of the given region. 2. To find the critical points in the interior, take the gradient of the function f and set it equal to 0, gradf=0. 3. We don't need all of the critical points of f so we limit the point to those inside the region D. 4. We define the boundary of D as {acost,bsint} where a and b are the constants given for x and y. 5. The critical points of the boundary are restricted to the interval 0 to 2Pi. 6. To find the critical points of the boundary take the derivative of f with respect to t and set it equal to 0. 7. Evaluate each critical point to determine the extreme values.

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.

1. We consider the level curves f(x,y)=V, which is given, and g(x,y)=0 2. we define g(x,y) as the given ellipse set equal to zero. 3. the minimal value of f on the ellipse occurs at a point (x0,y0) such that the level curves touch. 4. This makes the tangent lines to the level curves the same. 5 the tangent lines of the level curves are fx(x0,y0)*(x-x0)+fy(x0,y0)*(y-y0)=0 and gx(x0,y0)*(x-x0)+gy(x0,y0)*(y-y0)=0. 6. If the tangent lines are the same, there must be a scalar, z, such that fx(x0,y0)=z*gx(x0,y0) and fy(x0,y0)=z*gy(x0,y0) 7. we now have a system of equations for us to solve for the min value.

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.

1. We define a function, f, as the distance formula to measure the distance from point P to a point on the ellipse. 2. The shortest distance occurs at a point {x,y} such that the level curves g(x,y)=0 and f(x,y)=C touch. 3. For the level curves to touch, they must share a tangent line. 4. The general equation for the tangent line is fx(x0,y0)*(x-x0)+fy(x0,y0)*(y-y0)=0. 5. The equation for the ellipse g(x,y)=0 at the same point is gx(x0,y0)*(x-x0)+gy(x0,y0)*(y-y0)=0. 6. We take the x and why partial derivatives of each function to find the gradient vector of each tangent line. 7. If the tangent lines are the same, the gradient vectors must be proportional with the condition that g=0. 8. This sets up a system of equations for us to solve for the shortest distance.

Find the distance, d, from the point P(3,4,0) to the plane 9x+5y+7z=2.

1. We define an arbitrary point Q on the plane with coordinates (x,y,z). 2. This allows for formation of a the distance equation from the point P and the plane. We'll use this to find the minimal distance. 3. Find the critical point of the distance formula by finding the x and y partial derivatives and setting them equal to zero. 4. The critical points will be the x and y coordinates for point Q. 5. Evaluate the critical points to find the min value.

The demensions of a closed rectangular bos are measured as 24cm, 35cm, and 15cm, respectively, with a possible error of .3cm in each dimension. Use differentials to estimate the max error in calculating the surface area of the box.

1. We define dx, dy, and dz as the possible error for each side because the problem statement has .3 as the max error for each side. 2. The surface area formula is f=2*x*y+2*x*z+2*y*z 3. Take the partial x and y derivatives, and plug in the given measurements. 4. The definition of the differential is df=fx(x0,y0,z0)*dx+fy(x0,y0,z0)*dy+fz(x0,y0,z0)*dz. 5. Plug in all of the given values into the differential definition to find the max error in the surface area.

Find the distance, d, between the skew lines {x=5+t,y=4+3t,z=2t} and {x=1+4t,y=3+5t,z=-2+6t}.

1. We define the first skew line in terms of t and the second skew line in terms of s. 2. We make a function from the distance formula f(t,s)=|PQ| 3. Find the critical points by taking the t and s partial derivatives of f(t,s) and set each derivative equal to zero. 4. Evaluate all critical points to find the min value.

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

1. We define the function f(x,y) as the given ellipse set equal to zero. 2. Next we take the partial derivatives of f with respect to x and y. 3. We then evaluate the partial derivatives at the point P. 4. The equation of the tangent line is 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 (-.2,-.6) and use the equation to find the z-intercept of the plane. f(x,y)=9-3x+9y+2x^2-4xy+5y^2+5y^2+2x^2-3x^2y+6xy^2+2y^3

1. We define the function f, the give point P0, and an arbitrary point P(x,y) on the plane. 2. Next we find the gradient of the function and evaluate the gradient at the given point P0. 3. The equation of the tangent plane is z=f(P0)+gradf(P0).(P-P0). 4. This is the same as the first degree Taylor series (like finding the tangent line from taking the derivative. 5. Lastly, evaluate x=0 and y=0 in the equation to find the z-intercept.

Find the shortest distance, dmin, from the point P(0,3.5) on the y-axis to the ellipse x^2/6^2+y^2/5^2=1.

1. We define the given point P and an arbitrary point Q on the xy-plane. 2. We use these points to form a function f(x,y) from the distance formula. 3. We make the given ellipse a level curve by setting the given ellipse equal to 0. 4. Using the Lagrange Multiplier method, a level curve of f touches the level curve g(x,y)=0 at a minimum of t. 5. This means that the gradients of f and g are perpendicular to their level curves making them proportional B/C they are colinear. 6. Solve a system of equations with the constraint of g=0 and that f and g gradients are proportional. 7. Evaluate these points in f to determine the min value.

Find the critical numbers of the function. Then apply (if possible) the Second Derivative Test (Using a Taylor approximation of degree 2) to each critical number to conclude whether they correspond to local min or max of the function, or state that the test fails.

1. We find the critical pts of f(x) by taking the derivative and setting it equal to zero. 2. The Taylor series follows the formula f(x)~f(a)+f'(a)*(x-a)+f''(a)*(x-a)/2. This is the same as a Second Derivative Test.

Find the maximum rate of change, Dmax, of f(x,y)=(5x^2+3y^2)^-1 at the point (1,1)

1. We know the max rate of change of a function f(x,y) at point (a,b) is the magnitude of the gradient of f at (a,b). 2. Gradient is found by taking the x and y partial derivatives of f(x,y). 3. To find the max rate of change we need teh magnitude of the gradient 4. You do this by finding the norm of the gradient after point (1,1) has been evaluated in the gradient.

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

1. We make the ellipse into a level curve by setting the ellipse equal to 0. 2. The gradient at (a,b) is normal to the level curve (making it normal to the ellipse). 3. We can use the gradient to find the normal line at the given point by solving gradf(x0,y0) where (x0,y0) is the given point. 4. We parameterize the normal line as g(t)=(x0,y0)+gradf(x0,y0)*t 5. Finally we solve one of the components for t and plug it into the other component to get the equation.

Find an equation of the tangent line to the curve x^2/8^2+y^2/3^2=1 at the point (8Sqrt[5]/3,2).

1. We make the given ellipse a level curve by setting the ellipse equal to 0. 2. Next we find the gradient of the level curve, f, at point P. 3. The gradient vector of f is perpendicular to the level set at the point P. Making the gradient vector perpendicular to the tangent line. 4. An equation of the tangent line is found by using the orthogonality condition gradf(P).(Q-P)=0 where Q is an arbitrary point on the tangent line.

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

1. We must find the tangent line to the curve f(t) at the point Q. f(t)=Q+v*t where v is the velocity vector. 2. The velocity vector is the derivative of f(t) and the point Q. 3. We must find time t0 at which f(t0)=Q. 4. Once we have the tangent line equation we can solve for the time when the z component of the tangent line is 0. 5. We plug the now found z component into the tangent line equation to find the point of intersection.

Find and equation of the tangent line to the curve x^2/7^2+y^2/9^2=1 at the point (7Sqrt[17]/9,8).

1. We parameterize the ellipse as f(t)={acost,bsint} where a and b are the constants for the x and y on the ellipse given. 2. Next we find t0 where f(t0)=P. 3. Then we find f'(t0) at point P. 4. Now we form a line equation for the tangent line with g(s)-f(t0)+f'(t0)*s. 5. We then solve the x component of g(s) and plug the result into the y component of g(s).

Find the shortest distance, dmin, from the point P(0,2.6) on the y-axis to the ellipse x^2/5^2+y^2/4^2=1.

1. We parameterize the ellipse as f(t)={acost,bsint}, where a and b are the constants for the x and y on the given ellipse. 2. We will use the distance formula to find the distance from P to f. 3. We take the derivative of the distance formula and set it equal to 0 to find the critical points. 4. Once we find the critical points we evaluate each point into the distance function to find the min value.

Find the shortest distance, dmin, from the point P(0,1,3) to the line through the point Q(-2,3,4) and R(5,-4,0)

1. We write a function f(t)=Q+QR*t to define the line QR. 2. We use the distance formula to find the distance from point P to f(t). 3. We then find the first derivative of the distance formula and set it equal to zero to find the critical points. 4. Once we have the critical points we will evaluate them in the distance formula to find the min value.