Question

Consider the surface z = x 2+4y 2+1. Suppose you are walking on this surface directly above a curve C in the xy-plane, where the parameterized curve C is given by C : x(t) = cost, y(t) = sin t. Find the values of t for which you are walking uphill (increasing z).

Answer 1

With $x(t)=\cos t$ and $y(t)=\sin t$, we have

$z(t)=x(t)^2+4y(t)^2+1$

$z(t)=\cos^2t+4\sin^2t+1$

$z(t)=3\sin^2t+2$

Then $z(t)$ has critical points where

$\dfrac{\mathrm dz}{\mathrm dt}=6\sin t\cos t=3\sin2t=0$

$\implies\sin2t=0\implies2t=n\pi\implies t=\dfrac{n\pi}2$

where $n$ is any integer.

$z(t)$ is increasing wherever $\sin2t>0$, which happens for

$\dfrac{2n\pi}2<t<\dfrac{(2n+1)\pi}2$

$\boxed{n\pi<t<\dfrac{(2n+1)\pi}2}$

Answer 2

The value of t in the curve when one is you are walking uphill (increasing z) will be t < (2n + 1)π/2.

How to compute the value?

From the information, x(t) = cos t and y(t) = sin t. Also z(t) has critical points.

In this case, dz/dt = 6sintcost = 3sin2t = 0. Therefore, the value of t in the curve when one is you are walking uphill (increasing z) will be t < (2n + 1)π/2.

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