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 \\ \\abla \equiv \hat{i} (\partial)/(\partial e) + \hat{j} (\partial)/(\partial n)`


`\hat{r} = (1)/( √(2) ) (-\hat{i}+\hat{j} )`

Then the rate of change of h with respect to the vector v is

`\\abla h . \hat{r} = (1)/(√(2))(50\hat{i} + 75\hat{j}).(-\hat{i}+\hat{j}) = (1)/(√(2)) (-50+75) =17.68`

Answer: 17.7 ft per mile