Question

Consider the surface z=f(x,y)=2x2+5y2+1 and the curve C in the xy-plane given parametrically as x=cost and y=sint where 0≤t≤2π. a. Find z′(t).

b. Imagine that you are walking on the surface directly above the curve C in the direction of positive orientation. Find the values of t for which you are walking uphill (that is, z is increasing). a. Find the intermediate derivatives. ∂z∂x= (Type an expression using x and y as the variables.)

Answer 1

Let's go through each part of the problem step by step:

a. Find z′(t):

To find the derivative of z with respect to t, we'll use the chain rule. Given z = f(x, y) = 2x^2 + 5y^2 + 1, and the parametric equations for x and y are x = cos(t) and y = sin(t), we can express z as a function of t:

z(t) = 2x^2 + 5y^2 + 1

z(t) = 2(cos(t))^2 + 5(sin(t))^2 + 1

Now, let's find dz/dt using the chain rule:

dz/dt = ∂z/∂x * dx/dt + ∂z/∂y * dy/dt

First, find the partial derivatives of z with respect to x and y:

∂z/∂x = 4x

∂z/∂y = 10y

Now, plug in the values of x and y from the parametric equations:

x = cos(t)

y = sin(t)

∂z/∂x = 4cos(t)

∂z/∂y = 10sin(t)

Next, find dx/dt and dy/dt:

dx/dt = -sin(t)

dy/dt = cos(t)

Now, substitute these values into the chain rule formula:

dz/dt = (4cos(t)) * (-sin(t)) + (10sin(t)) * (cos(t))

Simplify:

dz/dt = -4cos(t)sin(t) + 10cos(t)sin(t)

Now, you have the derivative of z with respect to t:

z′(t) = -4cos(t)sin(t) + 10cos(t)sin(t)

b. Find the values of t for which you are walking uphill (z is increasing):

To find when you're walking uphill on the surface, you want to find the values of t for which z′(t) is positive. In other words, you want to find when the rate of change of z with respect to t is positive.

From part a, we already have the expression for z′(t):

z′(t) = -4cos(t)sin(t) + 10cos(t)sin(t)

Now, set z′(t) > 0 and solve for t:

-4cos(t)sin(t) + 10cos(t)sin(t) > 0

Factor out the common terms:

(10 - 4)cos(t)sin(t) > 0

6cos(t)sin(t) > 0

Now, consider the signs of cos(t) and sin(t) in different quadrants:

- In the first quadrant (0 ≤ t < π/2), both cos(t) and sin(t) are positive, so their product is positive.

- In the second quadrant (π/2 ≤ t < π), cos(t) is negative, and sin(t) is positive, so their product is negative.

- In the third quadrant (π ≤ t < 3π/2), both cos(t) and sin(t) are negative, so their product is positive.

- In the fourth quadrant (3π/2 ≤ t ≤ 2π), cos(t) is positive, and sin(t) is negative, so their product is negative.

So, the values of t for which z′(t) > 0 are in the first and third quadrants (0 ≤ t < π/2 and π ≤ t < 3π/2).

Therefore, you are walking uphill on the surface when 0 ≤ t < π/2 and π ≤ t < 3π/2.

Explanation: