Question

A car is driving northwest at v mph across a sloping plain whose height, in feet above sea level, at a point n miles north and e miles east of a city is given by h(n,e)=1500+75n+50e. (a) at what rate is the height above sea level changing with respect to distance in the direction the car is driving?

Answer 1

Refer to the diagram shown.

Given:
$h(n,e) = 1500 + 75n + 50e$

Define 
$\hat{r} = unit \, vector \, along \, \vec{v} \\ \hat{i} = unit \, vector \, east \\ \hat{j} = unit \, vector \, north \\ \nabla \equiv \hat{i} \frac{\partial}{\partial e} + \hat{j} \frac{\partial}{\partial n}$

$\hat{r} = \frac{1}{ \sqrt{2} } (-\hat{i}+\hat{j} )$

Then the rate of change of h with respect to the vector v is
$\nabla h . \hat{r} = \frac{1}{\sqrt{2}}(50\hat{i} + 75\hat{j}).(-\hat{i}+\hat{j}) = \frac{1}{\sqrt{2}} (-50+75) =17.68$

Answer: 17.7 ft per mile