Review Sheet for Test 3
Math 231 – Calculus I
The test is on Friday 11/20. (By the way the 229 exam was moved from 11/19 to 12/3). The test will cover the material in chapter 3 – applications of the derivative.
Here are some sample problems to think about:
Find the extrema (absolute maxs and mins) of $g(x) = 2x + 5\cos(x)$ on the interval $[0,2\pi]$.
Solve for $g'(x) = 2 - 5 \sin(x)$. This has 2 solutions in $[0,2\pi]$ namely $\sin^{-1}(2/5)$ and $\pi - \sin^{-1}(2/5)$. One needs to check the 4 points: $,0,2\pi,\sin^{-1}(2/5), \pi - \sin^{-1}(2/5)$. Here is the Matlab code:
>> x = [0, asin(2/5), pi-asin(2/5), 2*pi]
>> 2*x + 5*cos(x)
ans =
5.00000 5.40561 0.87758 17.56637Inspect to see the absolute minimum is 0.87758 and the absolute maximum is 17.56637.
Let $f(x) = (x-1)^2 (x-3)$.
find $f'(x)$.
what are the critical points of $f(x)$?
Where is $f(x)$ increasing?
Use the first derivative test to classify the critical points.
First $f(x) = x^3 - 5x^2 + 7x -3$ so $f'(x) = 3x^2 -10x + 7$ a nice parabola that opens upward. To find the critical points solve $f'(x) = 0$ to get $x = 1 or 7/3$. We know from the shape of $f'(x)$ that it is negative in the interval $(0,7/3)$ and otherwise non negative. So $f(x)$ is decreasing only on this same interval. The first derivative test tells us at $x=1$ we will have a relative maximum (the derivative changes sign + to -) and at $x=7/3$ we have a relative minimum (the derivative changes sign - to +).
Let $f(x) = x+\cos(x)$.
Find $f'(x)$, $f''(x)$.
Does the first derivative change sign + to - or vice versa? What does this say about the existence of relative extrema?
Does the second derivative change sign + to - or vice versa? What does this imply about the concavity, and what do we call the points where this happens?
First
$$f'(x) = 1 - sin(x),\quad f''(x) = -cos(x).$$
We see that $f'(x) \geq 0$. We know then there are no relative extrema from the first derivative test. (the derivative never changes sign!).
The second derivative does change sign. From - to + at $\pi/2$ and from + to - at $3\pi/2$. These two points are inflection points. They tell us when the concavity is changing.
Notice that this example has a critical point $\pi/2$ that is not a relative extrema, but instead an inflection point.
Find the limits
$$\lim_{x\rightarrow\infty}\frac{2x^2 - 1/x}{3x^2 + 5x},\quad \lim_{x\rightarrow-\infty}\frac{3x }{\sqrt{x^2 + 2}},\quad \lim_{x\rightarrow\infty}\frac{3}{x} - \frac{2}{\sqrt{x+2}}$$
The first one will be the same as
$$\lim_{x\rightarrow\infty} \frac{2x^2}{3x^2} = 2/3$$
This is simply the ratio of the leading terms.
For the second, factor out the $x^2$ in the bottom and remember that $\sqrt{x^2} = |x|$ not just $x$ to see that the limit will be the same as
$$\lim_{x\rightarrow-\infty} \frac{3x}{|x|} = -3$$
Finally we get an easy one. Plug in $\infty$ to get 0 - 0 which is not indeterminate its just 0. Notice plugging in in the first two simply says “More Work!”
Sketch a graph of these functions
$$f(x) = \frac{4}{1 + 3x^2},\quad g(x) = \frac{1}{\pi}(2 \sin(\pi x) - \sin(2\pi x))$$
First. $f(x)$ is even so there is symmetry, $f(x) > 0$ so no $x$-intercepts, $f(0) = 4$ which is the $y$ intercept and $y=0$ is a horizontal asymptote. Now for the calculus:
$$f'(x) = -24 \frac{x}{(1+3x^2)^2},\quad f''(x) = 24 \frac{9x^2 -1}{(1+3x^2)^3}.$$
We see that $f'(x)$ is always defined and is 0 when $x = 0$. Since $f''(0)$ is negative this says this is a relative maximum. Notice $f''(x) = 0$ at 3 and -3. These will be inflection points because we can easily verify that $f''(x)$ changes sign at these three points. Since $f''(x)<0$ only in the interval $(-3,3)$ $f(x)$ is concave down just there, and otherwise concave up.
Second. $g(x)$ is not symmetric. It has no asymptotes as the sine function doesn’t. However this function is periodic. The period of one of the sine functions 2, and the other is 1 ($T = 2\pi/b$!) so this repeats itself every 2 units. So concentrate on the interval $[0,2]$.
Now
$$g'(x) = 1/\pi (2\pi \cos(\pi x) - 2\pi \cos(2\pi x)) = 2\cos(\pi x) - 2\cos(2\pi x),\quad g''(x) = -2\pi\sin(\pi x) + 4\pi \sin(2\pi x).$$
We have that $g'(x)$ is always defined, and $g'(x) = 0$ when $\cos (\pi x) - (2\cos^2(\pi x) -1) = 0$ (double angle) This we can solve with: $\cos(\pi x) = 1$ or $\cos(\pi x) = -1/2$, or $x = 0,2,4/6,8/6$. (Good luck!). Plug these numbers into the second derivative to check if they correspond to positive or negative numbers. Here is the MATLAB
>> x = [0,4/6,8/6,2]
x =
0.00000 0.66667 1.33333 2.00000
>> -2*pi*sin(pi*x) + 4*pi*sin(2*pi*x)
ans =
0.00000 -16.32419 16.32419 -0.00000So the second derivative test is inconclusive at $0$ and $2$, but tells us we have a relative max at $x = 4/6$ ($f''(4/6) < 0$) nad a relative min at $x = 8/6$ ($f''(8/6) > 0$). What about at $0$ and $2$? Since the second derivative here is 0, we can’t conclude that the points are relative maxs or mins, or inflection points. To check this, we can see if they are inflection points by making sure $f''(x)$ changes sign at these points. Lets plug in values to the left and the right, that are small.
octave> x = [-.1 .1];
octave> -2*pi*sin(pi*x) + 4*pi*sin(2*pi*x)
ans =
-5.4447 5.4447This indicates that the second derivative is changing from - to + at 0. Aha! an inflection point. We don’t need to check at 2 it will be also. Why? Because the function is periodic. Here is a graph of the function on the interval $[-.5 ,2.5]$ along with the derivatives:
Show that the greatest area of any rectangle inscribed in an equilateral triangle is one-half that of the triangle.
Look at the picture. The length is labeled $l$ and the point of intersection is $a$. Based on this picture there are two triangles that tell us what $y$ and $x$ are. One is the 30-60-90 triangle with hypotenuse $l-a$ and height $y$. The other is a similar triangle with hypotenuse $a$ and width $x$. Solve the triangles to get the following ($\theta = 60^\circ$)
$$x = a \cos(\theta),\quad y = (l-a)\sin(\theta),\quad A = (2x)*y = 2 \sin\theta \cos\theta a(l-a) = \sin(2\theta) a(l-a).$$
Notice this is a nice function of $a$ (a parabola). Finding the critical points we take a derivative with respect to $a$ and see that we need to solve
$$A'(a) = \sin(2\theta) (1 - 2a) = 0, \text{or} a = l/2$$
This is clearly a max. (or take the second derivative, it is always negative) So the maximum area is when $a=l/2$ this yields an area of $\sin(120^\circ) (l/2)(l - l/2) = \sqrt{3}/2 l^2/4 = l^2 \sqrt{3}/8$. The area of the triangle is $(l\cos(\theta) l\sin(\theta) = l^2 \sqrt{3}/4$ or twice as big.
(Harder one) A ladder is being moved around a corner. One hallway has width 4 feet the other width 6 feet. Find the length of the longest ladder that can be moved around this corner.
Look at the figure. FOr each $\theta$, the maximum length ladder possible is drawn. If we find the $\theta$ which makes this the smallest, then this length is the largest possible ladder.
Now $L = l1 + l2$ where we figure out each $l1$ and $l2$ from a triangle with an angle $\theta$. From the trigonometry we have
$$\frac{4}{l1} = \sin\theta,\quad \frac{6}{l2}=\cos\theta$$
or
$$L(\theta) = l1 + l2 = \frac{4}{\sin\theta} + \frac{6}{\cos\theta}$$
Now we are in business. We need to minimize $L(\theta$ over the interval $[0,\pi/2$.
$$L'(\theta) = -4 \frac{\cos\theta}{\sin^2\theta} + 6\frac{\sin\theta}{\cos^2\theta} = \frac{6\sin^3\theta - 4\cos^3\theta}{\sin^2\theta\cos^2\theta}$$
Now the denominator is alway positive except at the obvious maximum $x=0$ and $x=\pi/2$, so we need only concentrate on the numerator. When is this 0? Unfortunately, this isn’t easy to solve. Lets look at a graph:
We see only one time where the derivative is 0, so we try Newton’s method to solve this non-linear equation. Here is the MATLAB for
$$f(x) = 6\sin^3(x) - 4\cos^3(x),\quad f'(x) = 6(3)\sin^2(x)\cos(x) - 4(3)\cos^2(x)(-\sin(x)) = 18*\sin^2(x) \cos(x) + 12 \cos^2(x) \sin(x)$$
>> x = .8; %% the initial guess
>> x = x - (6*sin(x)^3 - 4*cos(x)^3)/(18*sin(x)^2*cos(x) + 12*cos(x)^2*sin(x))
x = 0.71891
>> x = x - (6*sin(x)^3 - 4*cos(x)^3)/(18*sin(x)^2*cos(x) + 12*cos(x)^2*sin(x))
x = 0.71803
>> x = x - (6*sin(x)^3 - 4*cos(x)^3)/(18*sin(x)^2*cos(x) + 12*cos(x)^2*sin(x))
x = 0.71803
>> 6*sin(x)^3 - 4*cos(x)^3
ans = 1.2212e-13Whew! the answer is x = .71803. We need to plug this into the original formula to get the length. Here is the MATLAB
>>4/sin(x) + 6/cos(x)
ans = 14.047So we can slide a 14 foot ladder through this hallway, but not a 15 foot one.
Wow, that’s quite abit harder than anything you’ll see on the test, (Hmm, it might make a nice MATLAB problem) but if you can do it who knows you might be able to get a job at Home Depot!
Use Newton’s method to find a root of $x^3 - 3x -1$.
Here is the MATLAB. First you need to know $f'(x) = 3x^2 - 3$.
>> x= 0; %%% the initial guess
>> x = x - (x^3 - 3*x - 1)/(3*x^2 -3)
x = -0.34722
>> x = x - (x^3 - 3*x - 1)/(3*x^2 -3)
x = -0.34730
>> x = x - (x^3 - 3*x - 1)/(3*x^2 -3)
x = -0.34730That’s a root. To double check you might try the roots command
>> roots([1 0 -3 -1])'
ans = 1.87939 -1.53209 -0.34730Which shows we got -0.34. If you started elsewhere, you might get a different answer.
Use the tangent line approximation to find an approximate value for
$$\frac{1}{1 - .17}$$
without a calculator.
We know that
$$\frac{1}{1-x} \approx 1 + x$$
This is the tangent line approximation with $f(x) = (1-x)^{-1}$ and $c=0$. So the answer is simply 1.17. Notice we have the errors are
>> error = 1/(1-.17) - 1.17
error = 0.034819 % the difference from the correct answer
>> relative_error = error/1.17
relative_error = 0.029760 % the percentage difference