170- Quiz 7

Quiz 7 is here.

Solutions follow.

Problem 1 asks for the derivative of \arctan(e^x).

Recall that \displaystyle (\arctan x)'=\frac1{1+x^2}.

If you do not remember this, it can be found easily: If y=\arctan x, then \tan y=x, so (\sec^2 y)y'=1 or \displaystyle y'=\frac1{\sec^2y}. But \sin^2y+\cos^2y=1, so (dividing this identity by \cos^2y) \tan^2y+1=\sec^2y. Now, since \tan y=x, we get \tan^2y=x^2 and \sec^2y=1+x^2. Finally, \displaystyle y'=\frac1{1+x^2}.

Using the chain rule, we obtain

\displaystyle (\arctan(e^x))'=\frac{e^x}{1+(e^x)^2}=\frac{e^x}{1+e^{2x}}.

Problem 2 defines f(\theta)=\cos^2(\theta)-2\sin(\theta), and asks for the intervals where f is increasing and where f is decreasing within {}[0,2\pi]. Then, it asks to use this information to find the local maxima and minima of f.

We know that is f is differentiable, then f increases where f' is positive, and decreases where f is negative. Since our function f(\theta) is certainly differentiable, we may start by finding the points where f' is 0:

f'(\theta)=2\cos(\theta)(-\sin(\theta))-2\cos(\theta)=-2\cos(\theta)(\sin(\theta)+1), so f'(\theta)=0 iff \cos(\theta)=0 or \sin(\theta)+1=0. In the first case, \displaystyle\theta=\frac{\pi}2 or \displaystyle \frac{3\pi}2 (remember that we only consider values of \theta in {}[0,2\pi]). In the second case, \sin(\theta)=-1 or \displaystyle \theta=-\frac{3\pi}2. Below I indicate the distinguished points within the interval {}[0,2\pi]:

We have identified 3 intervals in {}[0,2\pi]: The interval where \theta<\pi/2, the one with \pi/2<\theta<3\pi/2, and the one where 3\pi/2<\theta. In each of these intervals, the sign of f'(\theta) does not change, so to determine it, it suffices to evaluate f'(\theta_0) for specific points \theta_0 chosen in each interval. In the first one, we can actually take \theta_0=0 because 0 is not a critical point of f even though it is a distinguished point of our analysis, being an endpoint. We have

f'(0)=-2\cdot1\cdot(0+1)=-2<0.

It follows that

f is decreasing in (0,\pi/2).

Since f(2\pi)=f(0), we can take \theta_0=2\pi in the last interval, and it follows that

f is decreasing in (3\pi/2,2\pi).

Finally, we can take \theta_0=\pi in the middle interval, and we see that f'(\pi)=-2(-1)(0=1)=2>0, and therefore

f is increasing in (\pi/2,3\pi/2).

We have f(0)=1, f(\pi/2)=-2, f(3\pi/2)=2 and f(2\pi)=1. From our analysis, we conclude that:

  • (0,1) is a local maximum.
  • (\pi/2,-2) is a local minimum. (Actually, this is the global minimum of f.)
  • (3\pi/2,2) is a local maximum. (Actually, this is the global maximum of f.)
  • (2\pi,1) is a local minimum.

The graph of f, shown below, confirms our findings.

Click the image above to enlarge.

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This entry was posted on Monday, October 18th, 2010 at 2:37 pm and is filed under 170: Calculus I. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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