Question

A hiker is hiking in a valley. The height of the valley is h(x,y)=4x2+y2 where x and y are the east-west and north-south distances from the valley floor respectively. The hiker follows an elliptical path x(t)=2 cos(t), y(t)=4 sin(t) around the valley. Compute the time derivative of the height in two ways. First expand the composite function h(x(t),y(t)) explicitly in terms of t. Second use the chain rule.

Answer 1

Answer:

A. $\frac{\partial{h}}{\partial{t}}=0$

Step-by-step explanation:

A. The problems asked for 2 ways to solve it, expanding the equation with the substitution  x(t)=2 cos(t) and y(t)=4 sin(t) to differentiate it . The other way is by chain rule.

Expanding and differentiating:

We start by substituting x(t)=2 cos(t) and y(t)=4 sin(t) in h(x,y)=4x2+y2:

$h(x,y)=4x^{2}+y^{2}= 4(2cos(t))^{2}+(4sin(t))^{2}\\h(x,y)=4(4cos^{2}(t))+(16sen^{2}(t))\\h(x,y)=16cos^{2}(t)+16sen^{2}(t)=16(sen^{2}(t)+cos^{2}(t))\\h(x,y)=16$

So, in the path that the hiker chose:

$\frac{\partial{h}}{\partial{t}}=0$

Chain rule:

We start differentiating h(x,y) using chain rule as follows:

$\frac{\partial{h}}{\partial{t}}= \frac{\partial{h}}{\partial{x}}\frac{\partial{x}}{\partial{t}}+\frac{\partial{h}}{\partial{y}}\frac{\partial{y}}{\partial{t}}$

Now, it´s easy to find all these derivatives:

$\frac{\partial{h}}{\partial{x}}=8x\\\frac{\partial{x}}{\partial{t}}=-2sin(t)\\\frac{\partial{h}}{\partial{y}}=2y\\\frac{\partial{y}}{\partial{t}}=4cos(t)$

Now we replace them in the chain rule, with the replacement x=2cos(t) and y=4sin(t) in the x,y that are left and we operate everything:

$\frac{\partial{h}}{\partial{t}}= 8x(-2sin(t))+2y(4cos(t)$

$\frac{\partial{h}}{\partial{t}}= 8(2cos(t))(-2sin(t))+2(4sin(t))(4cos(t)$

$\frac{\partial{h}}{\partial{t}}= -32cos(t)sin(t)+32sin(t)cos(t)$

$\frac{\partial{h}}{\partial{t}}= 0$

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