Day: April 24, 2021

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202104241713 Homework 1 (Q7)

The height of a mountain is given by h(x,y)=3000-2x^2-y^2, where the y-axis points east, the x-axis points north, and all distances are measured in meters. Suppose a mountain climber is at the point (30,\, -20,\, 800), will he ascend or descend if he moves in the southwest direction?

Solution.

(The solution below is based on the manuscript of 2015-2016 PHYS2155 Methods of Physics II Homework Solutions.)

The altitude h(x,y) is given by a function of x and y:

h(x,y)=3000-2x^2-y^2.

Now that the climber moves in the southwest direction

\begin{aligned} \mathbf{n} & =-1\,\hat{\mathbf{i}}-1\,\hat{\mathbf{j}} \\ \hat{\mathbf{n}} & = \frac{1}{\sqrt{2}}(-\hat{\mathbf{i}}-\hat{\mathbf{j}}) \end{aligned}

\begin{aligned} h_{\hat{\mathbf{n}}}'(x,y) & = \nabla h(x,y)\cdot \hat{\mathbf{n}} \\ & = (-4x\,\hat{\mathbf{i}}-2y\,\hat{\mathbf{j}}) \cdot \frac{1}{\sqrt{2}}(-\hat{\mathbf{i}}-\hat{\mathbf{j}}) \\ & = \frac{1}{\sqrt{2}} (4x+2y) \end{aligned}

At point (30,\, -20,\, 800),

\begin{aligned} h_{\hat{\mathbf{n}}}'(30,-20) & = \frac{1}{\sqrt{2}}\big( 4(30)+2(-20)\big) \\ & = \frac{1}{\sqrt{2}}\cdot 80 \qquad (>0) \end{aligned}

he will ascend southwesterly.