# Simple really.

Here is how the problem from class works out.

We can approximate $k!$ using $k^{k+\frac12}e^{-k}\sqrt{2\pi}$.

The probability that one voter makes the decisive vote is

$\frac{(2k)!}{k!k!2^{2k}}$

Plugging rack factorial in to the above we get:

$~\frac{(2k)^{2k+\frac12}e^{-2k}\sqrt{2\pi}}{k^{2k+1}e^{-2k}(2\pi)2^{2k}}$
$=\frac{(2k)^{2k+\frac12}\sqrt{2\pi}}{k^{2k+1}(2\pi)2^{2k}}$
$=\frac{(2k)^{2k+\frac12}}{k^{2k+1}\sqrt{2\pi}2^{2k}}$
$=\frac{(2k)^{2k}(2k)^{\frac12}}{k^{2k+1}\sqrt{2\pi}2^{2k}}$
$=\frac{k^{2k}\sqrt{2k}}{k^{2k+1}\sqrt{2\pi}}$
$=\frac{k^{2k}\sqrt{2k}}{k^{2k}k\sqrt{2\pi}}$
$=\frac{\sqrt{2k}}{k\sqrt{2\pi}}$
$=\frac{\sqrt{k}}{k\sqrt{\pi}}$
$=\frac{1}{\sqrt{k\pi}}$

At last.

# Simple really.

Here is how the problem from class works out.

We can approximate $k!$ using $k^{k+\frac12}e^{-k}\sqrt{2\pi}$.

The probability that one voter makes the decisive vote is

$\frac{(2k)!}{k!k!2^{2k}}$

Plugging rack factorial in to the above we get:

$~\frac{(2k)^{2k+\frac12}e^{-2k}\sqrt{2\pi}}{k^{2k+1}e^{-2k}(2\pi)2^{2k}}$
$=\frac{(2k)^{2k+\frac12}\sqrt{2\pi}}{k^{2k+1}(2\pi)2^{2k}}$
$=\frac{(2k)^{2k+\frac12}}{k^{2k+1}\sqrt{2\pi}2^{2k}}$
$=\frac{(2k)^{2k}(2k)^{\frac12}}{k^{2k+1}\sqrt{2\pi}2^{2k}}$
$=\frac{k^{2k}\sqrt{2k}}{k^{2k+1}\sqrt{2\pi}}$
$=\frac{k^{2k}\sqrt{2k}}{k^{2k}k\sqrt{2\pi}}$
$=\frac{\sqrt{2k}}{k\sqrt{2\pi}}$
$=\frac{\sqrt{k}}{k\sqrt{\pi}}$
$=\frac{1}{\sqrt{k\pi}}$

At last.

Dear 375 Students:

Here is a calendar showing test dates and other info, it will be updated as the semester rolls along. I have contacted the person in charge of rooms, with some luck we should be moving to more spacious digs soon enough!

-S. Donovan

# More Hints

Here is the problem:

Prove that the volume of a cone, (defined as the set of points on line segments joining a vertex, v, with a set of points in the same plane called the base) with height H, and base area B is $\frac{BH}{3}$.

What makes this tricky is at first it seems, with so many possible shapes for the base, there is no good way to express all of these 'different' areas with the same A(x). But, since the slices are similar we need only concern ourselves with the ratio of any side, radius, diagonal or width of a given slice to that same side, diagonal radius, or width on the base. With this we can find the area, independent of any specific formula (such as $s^2 = A$ or $\pi r^2 =A$)

To do this use the fact that the ratio of areas of similar plane figures is equal to the ratio of any one of their dimensions squared. And if we are looking for the area slice at x that ratio will be the same as the ratio of x to H, the total height.

So, this means that $\frac{A(x)}{B} = \frac {x^2}{H ^2}$. So, now you have a formula for A(x).

# Hints

Take home

#2 Hint:

#3 Hint:

You can use this trig identity
$\sin x \cos x = \frac12 \sin 2x$

#5 Hint:

Consider $f(t) = g(t) - h(t)$ , where g and h are the position functions of the two runners.

#6 Hint:

$\frac{d}{dx} |x| = sgn x = \frac{x}{|x|}$

#7 Hint:

Induction would be much much harder than L'Hopital.

#8 (b)Hint:

Use $\int_a^b f(t)dt =-\int_b^a f(t)dt$

#8 (d)Hint:

Divide the square root of x in to each number in the numerator to make it easier to see what the anti derivative should be.

#9 Hint:

Draw a diagram look at the triangles.

# Answers to Routine Practice Problems.

1. $\lim_{x \rightarrow 0} \frac{\sin(x)}{x}$
This shows how powerful Hospital's rule really is. Recall, that we did a rather long geometric proof of this limit earlier in the semester. So, from that we know it's 1. This is the indeterminate form $\frac{0}{0}$, consider
$\lim_{x \rightarrow 0} \frac{\cos(x)}{1}$ and this is clearly 1.

2. $\lim_{x \rightarrow \infty} \left( \sqrt{x} - \sqrt{x+1} \right)$ Use the conjugate.
$= \lim_{x \rightarrow \infty} \frac{ \left( \sqrt{x} - \sqrt{x+1} \right)\left( \sqrt{x} + \sqrt{x+1} \right)}{\sqrt{x} + \sqrt{x+1} }$ $= \lim_{x \rightarrow \infty} \frac{ x - x+1}{\sqrt{x} + \sqrt{x+1} } = \lim_{x \rightarrow \infty} \frac{1}{\sqrt{x} + \sqrt{x+1} } = 0$

3. $\lim_{x \rightarrow \infty}\frac{x \ln(x)}{\sqrt{x}}$ We have $\frac{\infty}{\infty}$ So, consider:
$\lim_{x \rightarrow \infty}\frac{\frac{x}{x}+ \ln x}{\frac{1}{2\sqrt{x}}} = \lim_{x \rightarrow \infty}\frac{2\sqrt{x} + 2\sqrt{x}\ln x}{1} = \infty$

4. Find an equation of the tangent to the curve $y=x^2-x$ that is parallel to $y=3x+3$.
$y'=2x-1$, we want the derivative to be 3, the slope of $y=3x+3$. So, solve $3=2x-1$, and you get x=2. Now find the tangent line through the point (2, 2). $y=3x-4$

5. Find an equation of the tangent to the curve $y=\ln x$ that will pass through the origin.
We are looking for a line, so y=mx+b. It goes through the origin so we already know b=0. $y'=\frac{1}{x}$ so, if the point where this line is tangent is $( x_0, y_0)$ then $y_0 = \frac{1}{x_0}x_0 +b$ means $b = y_0 -1$. If b=0, then $y_0 =1$. This happens when $x_0 =e$. So, the slope, $m = \frac{1}{x_0} = \frac{1}{e}$. Our equation is: $y= e^{-1}x$

6. A man walks away from a bird perched on a 40-foot-high flag pole. The man walks at a rate of 5 feet per second. How fast is the distance (as the crow flies) between the man and the bird increasing when he is 30 feet away from the pole?

Find the situation is a right triangle. Call the base b, the height (the pole) a, and the hypotenuse c. We can use $a^2 + b^2 = c^2$ to find c when b=40... c=50. First a, the height of the pole is constant. so: $1600 + b^2 = c^2$ Now take the derivative of our equation w/ respect to time: $0 + 2b\frac{db}{dt} = 2c \frac{dc}{dt}$ since $\frac{db}{dt} = 5 \frac{feet}{second}$ and we know c=50 and b=30. Hence, $2* 30 * 5 \frac{feet}{second} = 2*50 * \frac{dc}{dt}$. We want to know $\frac{dc}{dt}$, solving we get $\frac{dc}{dt}= 3 \frac{feet}{second}$.

7. For $f(x) = 3x^4 - 12x^3+17$: State any interval(s) of increase. State any relative minimum(s) and relative maximum(s). State the (x,y) location of any point(s) of inflection.

Take the derivative: $f'(x) = 12x^3 - 36x^2 = 12x^2(x -3)$ our critical points are 0 and 3. Test values in each interval:

$f'(-1) = -12 - 36 = neg$ So, $(-\infty, 0)$ is an interval of decrease.
$f'(2) = 12 \dot 8 - 36 \dot 4 = neg$ So, $(0, 3)$ is also an interval of decrease
$f'(4) = 12 \dot 64 - 36 \dot 16 = pos$ So, $(0, 3)$ is an interval of increase.

We only have a relative min. and it's at x=3. Next for the inflection point we look at the 2nd derivative:

$f''(x) = 36x^2 - 72x = 36x(x -2)$ Possible inflection points at 0 and 2.

$f''(-1) = 36 + 72 = pos$ concave up.
$f''(1) = 36 -72 = neg$ concave down.
$f''(10) = 3600 - 720 = pos$ concave up.

The concavity changes at both points so there are two inflection points: at 0 and 2. What could such a function look like?

8. Find $\frac{dy}{dx}$ for $\sin^3(x+y) =1$.

Using the chain rule: $3 \sin^2 (x+y) \cos(x+y) (1 + \frac{dy}{dx}) =0$ then solve for $\frac{dy}{dx}$. You get -1.
This may seem like too simple of a solution but, if $sin^3 (x+y) = 1$ then take the cube root of both sides and:
$sin (x+y) = 1$

The sine is 1 at $\frac{\pi}{2}, \frac{\pi}{2} + 2\pi, \frac{\pi}{2} + 4\pi, ... , \frac{\pi}{2} + n 2\pi$... so:

• $x+y = \frac{\pi}{2}$
• $x+y = \frac{\pi}{2} + 2\pi$

• $x+y = \frac{\pi}{2} + 4\pi$

• $\vdots$

Then subtract x in each equation:

• $y = -x + \frac{\pi}{2}$
• $y = -x + \frac{\pi}{2} + 2\pi$

• $y = -x + \frac{\pi}{2} + 4\pi$

• $\vdots$

These are linear equations. If we graph all of them at once then that is the set of (x,y) pairs that solve our equation:

(imagine more evenly spaced parallel lines that fill the whole graph...)

The slope of any tangent to these lines in -1 everywhere. So, our solution makes sense.

9. Find $\frac{dy}{dx}$ for $\sqrt{xy} = 2 + x^2 y^2$. You will get $\frac{4xy^2 \sqrt{xy} -y}{x- 4yx^2 \sqrt{xy} }= \frac{y(4xy \sqrt{xy} -1)}{x(1- 4yx \sqrt{xy}) } = \frac{-y}{x}$

10. Find an antiderivative of $f(x) = 3x^2 - 2x -1$.
$F(x)= x^3 - x^2 - x + 5$ (The constant can be anything including 0.)

11. Find an antiderivative of $f(x) = \frac {2x \sin (x) - x^2 \cos (x)}{\sin^2 x}$
$F(x) = \frac{x^2}{\sin x} + 9$ (The constant can be anything including 0.)

12. Find the linearization of $y=x^3 -2x +1$ centered at 0.
This is the same as finding the tangent to the curve at (0, 1). The slope is given by $y'=3x^2 -2$ if x=0 then the slope is -2. Then find the equation $y = -2 x +1$

1. Find an equation of the tangent to the curve $y = e^x$ that is parallel to x−4y = 1.

$Y = \frac{1}{4}x + \frac{1-\ln (1/4)}{4}$

(It is better to leave the natural log in the equation, don't find a decimal value... Though you may use such value for graphing.)

2. Find an equation of the tangent to the curve $y = e^x$ that will pass through the origin.

The the tangent to the curve at $x_0$ will have the form $y = e^{x_0}(x+1-x_0)$ ... Plug in (0,0) and you find you are seeking the x-value where the natural log is zero. So the equation is $y= ex$.

# Extending L' Hopital. (For my students. )

How can one deal with $\lim_{x \to 0^+} x^{x^2}$? It appears to be the indeterminate form $0^0$. The graph shows that it has a nice limit:

(Notice, how this function produces imaginary vales when  it is between -1 and 0. )

But, this is the wrong kind of indeterminate form for L'Hopital.  If we could somehow show that it was, in fact, the same as $\frac{0}{0}$ we could use the derivative to help us... Here is an extension of L'Hopital that will do just that:

$\lim_{x \to c} f(x)^{g(x)} = \exp \lim_{x \to c} \frac{\ln f(x)}{1/g(x)}$

See if you can apply it. We will talk more about this in class. If you simply said "we can't use L'Hopital" that is correct for this problem based on what you knew up until now. If you applied it anyway and tried to say that:  $\lim_{x \to c} f(x)^{g(x)} =\lim_{x \to c} f'(x)^{g'(x)}$ ... then that is incorrect! (And I believe it leads to the wrong answer here.)