175 – Quiz 2

Here is quiz 2.

Problem 1 is (simplification of part of) exercise 6.6.16 from the book.

To solve the question, use coordinates as in the accompanying figure in the book, so the origin is at ground level, and y increases downwards. The units of y are feet. For a fix y with 0\le y\le 20, the thin slice of water in the tank at depth y and of tickness dy has volume dV=10\times 12\times dy and weighs dF=62.4\times dV=7488\,dy. This is a constant force, so the work required to remove it to ground level is just dW={\rm distance}\times dF, where {\rm distance} is the depth at which the slice is located, i.e., y. Hence, dW=7488 y\,dy. The total work is obtained by adding all these contributions, i.e., W=\int_0^{20} 7488 y\,dy=7488\times 200=1497600 ft-lb.

Problem 2 is exercise 9.2.16 from the book. 

The curve is r^2=-\cos\theta. Since \cos\theta=\cos(-\theta), the graph is symmetric about the x-axis (because whenever (r,\theta) is in the graph, then so is (r,-\theta)).

Since (-r)^2=r^2, the graph is symmetric about the origin (because whenever (r,\theta) is in the graph, then so is (-r,\theta)).

Since the graph is symmetric about both the origin and the x-axis, it is also symmetric about the y-axis.

To sketch the curve, look first at 0\le \theta< \pi/2. Here \cos\theta>0, so r^2<0, which is impossible, so there is nothing to graph here. Consider now what happens when \pi/2\le\theta\le \pi. As \theta increases, -\cos\theta increases, from 0 to 1. So the same occurs with r^2. This means that r increases from 0 to 1, and -r decreases from 0 to -1. The part with r gives us a curve in the second quadrant, and the part with -r gives us its reflection about the origin. This part of the curve is in the fourth quadrant. Their reflections on the x-axis complete the curve, which can be seen here.

Note that \tan(\pi/2) is undefined. This corresponds to the fact that at the origin the tangent to the curve is the y-axis, as can be seen from the graph.

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