Griffiths 1.12 The height of a certain hill (in feet) is given by

Griffiths 1.12 The height of a certain hill (in feet) is given by

P
945 points

The height of a certain hill (in feet) is given by,

$$h(x,y)=10(2xy-3x^2-4y^2-18x+28y+12)$$

where y is the distance (in miles) north, x the distance east of South Hadley.

(a) Where is the top of the hill located?

(b) How high is the hill?

(c) How steep is the slope (in feet per mile) at a point 1 mile north and one mile east of South Hadley? In what direction is the slope steepest, at that point?

Griffiths

P
945 points

The function $h(x,y)$ looks like

part a

The top of the hill is located where the gradient is zero,

$$\nabla h=0$$

$$\rightarrow 10(2y-6x-18) \hat{x}+10(2x-8y+28)\hat{y}=0$$

$$\left\{ \begin{array}{c} 2y-6x-18=0 \\ 2x-8y+28=0 \\ \end{array} \right.$$

$$\rightarrow x=-2,\,\,y=3$$

and so the top of the hill is 3 miles north and 2 miles west of South Hadley

part b

plug the values from part a into the function to gets its height

$$h(-2,3)=720\text{ feet}$$

part c

Use $x=1, \, y=1$ and the gradient you found in part a

$$\nabla h(1,1) = 10[(2-6-18) \hat{x} + (2-8+28) \hat{y}] = -220 \hat{x} + 220 \hat{y}$$

is the direction and

$$\left\lvert \nabla h \right\rvert = \sqrt{(-220)^2+220^2} = 220\sqrt{2} \approx 311$$

in ft per mile

So the steepness of the slope is 311 ft/mile in the northwest direction

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