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anastassius [24]
3 years ago
10

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?

Physics
1 answer:
coldgirl [10]3 years ago
3 0
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

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