Chapter 6: Q14P (page 295)

(a) Suppose that a hill (as in Fig. 5.1) has the equation $32 - x^{2} - 4 y^{2}$ , where $z = height measured from some refrence level$ (in hundreds of feet). Sketch acontour map (that is, draw on one graph a set of curves $z =$ const.); use the contours $z = 32, 19, 12, 7, 0$ (b) If you start at the point $(3, 2)$ and in the direction $i + j$ , are you going up hillor downhill, and how fast?

Short Answer

Expert verified

a) Contour is sketched below

b) $- 11 \sqrt{2}$ $- 11 \sqrt{2}$

Step by step solution

Given Information.

A hill has the equation $32 - x^{2} - 4 y^{2}$

Definition of gradient.

Gradient is defined by the equation mentioned below

$\nabla z = i \frac{\hat{\partial z}}{\partial x} + j \frac{\hat{\partial z}}{\partial y}$

Sketch the graph.

(a)

The graph looks like as presented below

Calculate the vector u.

$u = \frac{i + j}{\sqrt{1^{2} + 1^{2}}} = \frac{1 + j}{\sqrt{2}}$

Compute the partial derivates.

Take the partial derivatives.

$\frac{\partial z}{\partial x} = - 2 x \frac{\partial z}{\partial y} = - 8 y$

Substitute the partial derivates in the gradient.

The values of the partial derivatives calculated is substituted in the gradient.

$\nabla z = i (2 x) + j (- 8 y) = - 2 x i - 8 y j$

Substitute the partial derivates in the gradient.

(b)

Calculate the gradient at (3,2).

$\nabla z . (3, 2) = - 2 (3) i - 8 (2) j = - 6 i - 16 j$

Find the directional derivate.

Calculate the dot produce of u and the gradient to find the directional derivative.

$\nabla z . u = (- 6 i - 16 j) . (\frac{i + j}{\sqrt{2}}) = - 6 \times \frac{1}{\sqrt{2}} - 16 \times \frac{1}{\sqrt{2}} = - 3 \sqrt{2} - 8 \sqrt{2} = - 11 \sqrt{2}$

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Short Answer

Step by step solution

Given Information.

Definition of gradient.

Sketch the graph.

Calculate the vector u.

Compute the partial derivates.

Substitute the partial derivates in the gradient.

Substitute the partial derivates in the gradient.

Find the directional derivate.

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