Calculus 3 help. This is a take home, open book, quiz, but I need help on it! I

Question

Calculus 3 help. This is a take home, open book, quiz, but I need help on it! I am mainly stuck on the last 4. If I could just get steps on how to approach the answers, that would help me a lot! Thank you in advance!

A baseball is hit from 3 feet above home plate (the location of the origin of a rectangular coordinate system), with an initial velocity of feet/sec. Neglect all forces except for gravity. Find the position and velocity vectors of the ball as a function of time. Find the length of the curve whose position vector is r rightarrow (t) = (cos3 t)j rightarrow + (sin3 t)j rightarrow, between the points (0, 1) and (1.0). Consider the curve r rightarrow (t) = (2 cost)I rightarrow + (2sin t)j rightarrow + 4tk rightarrow. Reparameterize this curve using arc-length. Image that you are walking of the group of f (x, y) = x2 + 4y2 + 1 directly above the curve x = cost, y = sin t. Find the values of t for which you are walking uphill. Find dw/dt, where w = xyz, x = 2t4, y = 3t-1, z = 4t-3 Find the equation of the tangent plane to the graph of f (x, y) = x2 - xy + y2 at the point (2, -1, 9). Find the equation of the line normal to the contour of f (x, y) = x2 - xy + y2 that passes through the point (1, -1) Find the absolute maximum of f (x, y) = x2 + y2 - 2x - 2y + 5 on the {(x, y)|x2 + y2 4} Use differentials to approximate the value of f (x, y) = 5 / x2 + y2 at the point (-0.93,1.94). Hint: use the value of f at (-1, 2) in your estimate. What is the rate of change in f (x, y) = x2 + 4y2 + 1 at the point (2,3) in the point (2,3) in the direction of the vector ?

Explanation / Answer

10. grad(f) = <2x,8y> = <4,24> and <4,24>*<3/5,4/5> = 108/5

9. f(-0.93,1.94) = f(-1,2) +0.07*fx -.06*fy = 1 + 0.07(-10(-1)/5^2) - 0.06(-10*2/5^2) = 1.076

8. grad(f) = <2x-2,2y-2> so only critical point is at (1,1). Parametrize boundary by x = 2cos(t), y = 2sin(t) to get g(t) = 9-4cos(t)-4sin(t) so g'(t) = -4sin(t)+4cos(t)=0 means tan(t) =1 means t = pi/4 or 5pi/4 so the other critival points are at (sqrt(2),sqrt(2)) and (-sqrt(2),-sqrt(2)).

Now check f(1,1) = 3, f(sqrt(2),sqrt(2)) = 9-2sqrt(2) = 6.17 and f(-sqrt(2),-sqrt(2)) = 9+2sqrt(2) = 11.82 so 3 is abs. min and 11.82 is abs. max.

7. controus is x^2-xy+y^2 = 1, so 2x - y -xy' + 2yy' = 0 and y'(2y-1) = y-2x ans y' = (y-2x)/(2y-1) = -1/1 = -1 so slope of normal is 1 and passes through (1,1): y = x

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