Name: Math 241: Calculus & Analytic Geometry A
Exam 2 Solution Guide
November 1 2002

There is always one moment in childhood when the door opens and lets the future in.
- Graham Greene, The Power and the Glory.

Instructions: Show all work to receive full or partial credit. All University rules and guidelines for student conduct are applicable.

Beneath each problem is percentage of students who earned 80% or more of the points for that question.

1. [15 pts] Find an equation of the tangent line to the curve
   $$y = \tan\left(\frac{\pi x^2}{4}\right)$$
   at the point $(1,1)$. Simplify your answer as much as possible.

   
   

   Section 016: 64.3% Section 017: 87.5% Section 018: 60% All sections: 71.2%

   Very similar to problem 3.6: 47a.

   This problem is a direct application of the chain rule:

   $$y' = \frac{\pi}{2} x \sec^2\left(\frac{\pi x^2}{4}\right).$$
   Thus, the slope of the tangent line at $(1,1)$ is
   $$m = y'(1) = \pi.$$
   Therefore, the equation of the line tangent to the curve at $(1,1)$ is
   $$(y-1) = \pi (x-1).$$

2. [15 pts] Find the equation to the tangent line to the curve
   $$3 x^2 y^2 = (y+1)^2 (4-y^2)$$
   at $(2,1)$. Simplify your answer as much as possible.

   
   Section 016: 50% Section 017: 75% Section 018: 83.3% All sections: 69.5%

   Very similar to 3.7: 30.

   This is a direct application of implicit differentiation. Differentiating, we obtain:

   $$6(xy^2 + x^2 y y') = 2\left[(y+1)y'(4-y^2) + (y+1)^2 (-y y')\right]$$
   Solving for $y'$, one gets
   $$y' = \frac{3 x y^2}{-3x^2 y + (y+1)(4-y^2) -y(y+1)^2}.$$
   If we evaluate $y'$ at $(2,1)$, we find that $y' = -\frac{3}{5}$. Thus, the equation of the tangent line is
   $$y-1 = -\frac{3}{5}(y-2).$$

   
   

3. [20 pts] A vertical post (located at the equator during Spring Equinox) stands 10 meters out of the ground so that the sun passes directly overhead at noon local time. How fast is the tip of the post's shadow moving at 3:00 pm? Simplify your answer as much as possible.

   
   

   Section 016: 14.2% Section 017: 37.5% Section 018: 20% All sections: 23.2%

   
   Similar to 3.9: Example 5.

   Very similar to 3.9: 32.

   The sun goes around once every 24 hours. Alternatively, one knows that the sun traverses the sky overhead in 12 hours. Thus, $\frac{d \theta}{dt}$ = $\frac{\pi}{12}$ radians per hour. To relate $x$ and $\theta$ together, we see that

   $$\tan \theta = \frac{x}{10}.$$
   Differentiating both sides,
   $$\sec^2(\theta) \frac{d \theta}{dt} = \frac{1}{10} \frac{d x}{dt}$$
   At 3:00, the sun makes an angle of $\theta = \frac{\pi}{4}$, so that
   $$\frac{dx}{dt} = \frac{5 \pi}{3} \frac{\rm m}{\rm hr}.$$

4. [10 pts] Find $f^{(50)}(x)$ where
   $$f(x) = (x+3)^{25} (x-2)^{25}.$$
   Justify your answer.

   
   

   Section 016: 35.7% Section 017: 45.8% Section 018: 50% All sections: 43.9%

   Very similar to example discussed in lecture.

   Similar to 3.8: Example 3.

   From our knowledge of Pascal's triangle and multiplication of polynomials, we know that

   $$f(x) = (x^{25} + ...)(x^{25} + ...)$$
   where the ``...'' represents the powers of $x$ that are less than 25. Thus, the highest power of $x$ in $f(x)$ is $x^50$. Taking 50 derivatives will eliminate every power of $x$ that is 49 or less. In other words
   $$f(x) = x^{50} + ...$$
   and all the ``...'' stuff gets eliminated after we differentiate 50 times. Thus,
   $$f^{(50)} (x) = 50!.$$

   
   

5. [10 pts] Find a general expression for $f^{(n)}(x)$ where
   $$f(x) = \frac{1}{(1-x)^2}.$$

   
   Section 016: 64.2% Section 017: 87.5% Section 018: 80% All sections: 76.8%

   Similar to 3.8: Example 4.

   Very similar to 3.8: 34.

   Just differentiate and look for a pattern.
   

   $$f(x) = \frac{1}{(1-x)^2}$$

   $$f'(x) = \frac{2}{(1-x)^3}$$

   $$f''(x) = \frac{2 \cdot 3}{(1-x)^4}$$

   $$f'''(x) = \frac{2 \cdot 3 \cdot 4}{(1-x)^5}$$

   
   
   By inspection, we see that
   $$f^{(n)}(x) = \frac{2 \cdot 3 \cdot 4 \ldots (n+1)}{(1-x)^{n+2}}.$$

6. [10 pts] Classify all the critical points of
   $$y = \sin (x) - \frac{1}{2} \sin^2 (x) + 3, \ \ \ \ -2 \pi \leq x \leq 2 \pi.$$

   
   

   Section 016: 10.7% Section 017: 37.5% Section 018: 23.3% All sections: 23.2%

   Similar to 4.1: Example 9, problem 57, 58.

   We need to find the roots of $y'$.

   $$y' = \cos x - \sin x \cos x = \cos x ( 1-\sin x).$$
   The roots of $\cos x$ on the given interval are $\pm \frac{\pi}{2}$ and $\pm \frac{3 \pi}{2}$. The roots of $1-\sin x$ on the given interval are simply $\frac{\pi}{2}$ and $- \frac{3 \pi}{2}$. Both of the roots are already roots of the first term. Now, we must classify the roots.

   Interval	Incr/Decr
   $\left(-2 \pi,-\frac{3 \pi}{2}\right)$	Increasing
   $\left(-\frac{3 \pi}{2},-\frac{\pi}{2}\right)$	Decreasing
   $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$	Increasing
   $\left(\frac{\pi}{2},\frac{3 \pi}{2}\right)$	Decreasing
   $\left(\frac{3 \pi}{2},2 \pi \right)$	Increasing

   Thus, $y$ has local maxima at $\frac{\pi}{2}$ and $- \frac{3 \pi}{2}$ and local minima at $-\frac{\pi}{2}$ and $\frac{3 \pi}{2}$.

   
   

7. [5 pts] Find the inflection point(s) for the function and interval in problem 6.

   
   

   Section 016: 7.1% Section 017: 8.3% Section 018: 3.3% All sections: 6.1%

   We must find the roots of $y''$ that correspond to changes in the sign of $y''$.

   $$y'' = - \sin x - \cos^2 x + \sin^2 x = -\sin x -1 + 2\sin^2 x.$$
   We can factor this expression:
   $$y'' = (2 \sin x +1)(\sin x -1).$$
   Notice that the second term has roots at $\frac{\pi}{2}$ and $- \frac{3 \pi}{2}$, but we already know that these are local maximums, so they cannot be inflection points. Therefore, we simply inspect the roots of $2 \sin x +1$. In other words, for what angles is
   $$\sin x = - \frac{1}{2}.$$
   If you inspect the unit circle, you will find that $x = -\frac{5 \pi}{6}, -\frac{\pi}{6}, \frac{7 \pi}{6}, \frac{11 \pi}{6}$. Since, the second term, $\sin x -1$, is always negative except at its roots, we know that this term is negative at each of the roots of the first term. Thus, the roots of the roots of the first term indicate sign changes and are inflection points.

8. [15 pts] Find the absolute minimum and maximum values of
   $$y = 2 x^3 - 3 x^2 -12 x +8$$
   on the interval $-2 \leq x \leq 2$.

   
   

   Section 016: 67.9% Section 017: 95.8% Section 018: 83.3% All sections: 81.7%

   Similar to 4.1: 47-52.

   We must check all the critical points and the endpoints of the interval.
   

   $$y' = 6 x^2 - 6 x - 12$$

   $$= 6(x^2 - x -2 )$$

   $$= 6(x-2)(x+1)$$

   
   

   Checking each critical point and endpoint we find that
   

   $$y(-2) = 4$$

   $$y(-1) = 15$$

   $$y(2) = -12$$

   
   
   So, the absolute maximum value is 15, and the absolute minimum value is -12.