Test 2 will cover the material in sections 2.5 through 4.3. (Note, 4.4 on the fundamental theorem of calculus will be covered on the final.) I can’t emphasize enough that this test will be difficult. There is a lot of material covered here that you have learned in this class and not previously. For a review, here are the key topics and a sample question for each. We will have 1 hour on Monday 11/30 to review for the exam.
This allows you to find $dy/dx$ when x and y are related by an equation and not a function. Here is an example.
Find $dy/dx$ and $dy^2/dx^2$ when $y^2 - x^2 = 5$.
answer Okay. We have with $y'=dy/dx$
$$2y y' - 2x = 0 \rightarrow y' = x/y$$
As well, taking second derivatives gives:
$$2(y' y' + y y'') - 2 = 0 \rightarrow y'' = (2 - 2(y')^2)/y = (2 - 2(x/y)^2)/y = \frac{2y^2 - 2x^2}{y^3} (Thanks Allen).$$
Oh boy, word problems! Related rates boil down to the chain rule in practice. For example, if area and height are related, then the change in area and the change in height are related. Here is an example problem:
A funnel in the shape of cone has the ratio $h/r = 2$. The funnel is losing volume at the rate of 2 cm$^3$/minute. What is the rate of change of height?
answer: The funnel has volume given by $V = \pi/3 hr^2$. But $h/r=2$ so we can write $V = V(h) = \pi/3 h (h/2)^2 = \pi/12 h^3$ Takine $d/dt$ of $V(h)$ we get
$$dV/dt = \pi/12 \cdot 3 h^2 dh/dt.$$
Solve for dh/dt to get $dh/dt = (4/\pi)dV/dt h^{-2}$
The most important thing here is the followoing fact: A continuous function on a closed interval $[a,b]$ has a maximum value and a minumum value. It achieves these at a critical point or an endpoint. As well, we learned what an absolute extrema is and how it differs from a relative extrema.
Find the extrema over $[0,3]$ of the function
$$f(x) = x \sqrt{3-x}$$
answer: We know the answer lies in the critical points and the endpoints.
First, the critical points solve $f'(x) = 0$ or DNE. We have $f'(x) = \sqrt{3-x} + x (1/2) 1/\sqrt{3-x}(-1) = 1/\sqrt{3-x}(3 - x - x/2)$ From here we get x=3 and x=2 are critical points. The endpoints are 0 and 3. So we need to check 0,2 and 3: f(0)=0, f(3)=0, f(2)=2 which is the maximum.
This is an important theorem for proving things in mathematics. For example, we talked about how you can prove if $f'(x) =0$ we know $f(x)$ is a constant. I won’t ask a question on this on this exam, but you may have one on the common final.
We learned what it means to call $f(x)$ increasing or decreasing on an interval $I$. We learned the first derivative test which tells us when a critical point leads to a relative extrema.
Use the first derivative test to characterize the critical points of $f(x) = (x+3)/x^2$
answer: First the critical points solve $f'(x) =0 or
DNE$. We get
$$f'(x) = (x^2 - (x+3)(2x))/x^4$$
This DNE at x=0 but x=0 is not in the domain of $f(x)$ so is not a critical point – it will be a vertical asymptote. Note that $x^4$ is always positive, and so f’(x) =0 when $x^2 - 2x^2 - 6x = -(x^2+6x) = -(x(x+6)) = 0$. That is $x=0$ and $x=-6$ are critical points and $f'(x)$ is only positive in $(-6,0)$. This tells us that -6 is a relative min. (- to +) 0 as discussed before is not a critical point but a asymptote. It should however look like a relative max as the derivative changes from + to - there.
We learned what concavity is and what the second derivative test tells us. ($f''(x) > 0$ implies $f(x)$ is concave up).
Use the second derivative test to characterize the critical points of $f(x) = (x+3)/x^2$
answer: Again only 6 is a critical point. To apply the second derivative test we nee the derivative of $f'(x) = -(x^2 +
6)/x^4$ . That is
$$f''(x) = -(2x x^4 - (x^2+6)4x^3)/x^8 = - (2x^5- 4x^5 - 24x^3)/x^8.$$
This is positive for x=-6 and so we have a relative minimum.
Limits at infinity helped us investigate horizontal asymptotes. Remember, it took a new definition of a limit.
Find the horizontal asympotote(s) of
$$f(x) = \frac{x^2 + 2x}{\sqrt{x^4 + 2}}$$
answer: recall the square root trick: $\sqrt{x^4+2} =
\sqrt{x^4}\sqrt{1 + 2/x^4} = x^2 \sqrt{1 + 2/x^4}$. So the dominant term is $x^2$. This say we will have a horizontal asymptote of x=1 on both sides of the graph.
In calculus a sketched curve should contain the following information: intercepts, asymptotes, domain, range, symmetry, relative extrema, concavity, continuity and differntiablity. Your drawing should match the mathematics. Practice with the following
Sketch
$$f(x) = | x^2 + 5x + 6| + x$$
answer: The trick here is the derivative. Recall this is the same function as
$$f(x) = x^2 + 6x+6 if x^2+5x+6 > 0$$
and
$$f(x) = -x^2 - 4x - 6 if x^2 + 5x + 6 < 0 ( when x is in (-3,-2))$$
Thus, $f'(x) = 2x + 6$ when x is not in (-3,-2) and is $-2x -4$ when it is. While we are here, $f''(x) = 2$ when x is not in (-3,-2) and f”(x)=-2 when x is in (-3,-2).
We use f’(x) to find the critical points. Notice 2x+6 is 0 when x = -3 and -2x-4 is 0 when x = -2. However, this does not say f’(x) = 0 when x = -3 or -2. In fact, since -3 and -2 are on the boundary, f’(x) does not exist at these points. (The left and right derivatives are not equal). S0 the critical points are x = -3 and x = -2 not because f’(x) is 0 there but instead f’(x) does not exist. Notice the 2nd derivative is never 0.
Let’s look at tables of f’(x) and f”(x):
| $(-\infty,-3)$ | -3 | (-3,-2) | -2 | (-2,$\infty$) | |
|---|---|---|---|---|---|
| f’(x) | - | DNE | + | DNE | + |
| f”(x) | + | DNE | - | DNE | + |
From the first derivative test we see that x = -3 is a relative minumum and x=-2 is not a relative extrema. Since f”(x) changes sign at both -3 and -2 these are both inflection points.
The zeores of f(x) are -3 - $\sqrt{3}$ and -3 + $\sqrt{3}$ as you can check with the quadratic equation. This gives us the following points to connect:
$$(-3 - \sqrt{3},0), (-3,-3), (-2,-2), (-3+\sqrt{3},0),(0,6)$$
The graph should look like two parabolas. one pointing upward (concave up) for most of the graph, with a bite taken out of it between -3 and -2 which points downward (concave down)
More word problems. (Expect atleast two on the exam!) Optimization problems apply the material learned in section 3.1 to finding the largest of smallest possible values.
Let two positive numbers add to 100. What is their largest product? What is their smallest product?
(Expect a drawing on the exam. Know the area of a triangle, rectangle, square and sphere. I’ll give formulas if I use a cone, or a pyramid or some other shape.)
answer: Okay, we know x+y = 100, and we wish to maximize xy = x(100-x). The interval must be [0,100] as we assume x and y are positive (okay, secretly it should be (0,100) for the purists.). To maximize over a closed interval we recall that we find the endpoints, (0 and 100) and the critical points and then take the largest function value for these. Note that (x(100-x))’ is 0 at x = 50 so we need to check 0,50 and 100. A quick investigation shows 0 and 100 have answers which are 0 and 50 has one which is 2500, obviously the maximum.
We learned what $dy \approx \Delta y$ and $f(x)
\approx f(c) + f'(c)(x-c)$ mean to use. Essentially, remember the tangent line approximates the function nearby. Then remember the formula for the tangent line.
Find the tangent line approximation for $f(x)$ near $x =0$ ($c$ is a constant).
$$f(x) = \frac{1}{\sqrt{1 - (x/c)^2}}$$
Use the tangent line approximation to estimate $f(.1c)$.
Next, this formula comes from the theory of relativity. What is the approximate value of $f(x)$ as $x$ approaches $c$?
answer: Okay, we have the tangent line approximation
$$f(x) \approx f(0) = f'(0) (x - 0)$$
Now figure out f(0) = 1 and f’(0). TO do the latter, first find f’(x) and then plug in:
$$f'(x) = [(1-(x/c)^2)^{-1/2}] = (-1/2) (1-(x/c)^2)^{-3/2} (-2(x/c) (1/c))$$
At x=0 we see f’(0) = 0.
So the tangent line approximation is the flat line $y = 1$. So, we should say that f(.1c) $\approx$ 1.
The second question is about
$$\lim_{x\rightarrow c -} f(x)$$
That is, a question about vertical asymptotes. Clearly the left limit above is $\infty$.
We learned how to find answers to differential equations, what the indefinite integral and why we don’t forget to use the constant of integration.
Suppose $dv/dt = t$. Find a formula for $x(t)$. (In the notes we derived the formula when $dv/dt$ is a constant – constant acceleration. This has you find a formula when the assumption is increased acceleration.)
answer:
From $dv/dt = t$ we know $v(t) = = t^2/2 + c$. Find c by looking at v(0). We see $v(0) = 0^2/2 + c$, or$ c = v_0$ and
$$v(t) = t^2/2 + v_0$$
Now find x(t) from the differential equation $dx/dt = v(t)$. That is $x(t)$ is an antiderivative of $v(t)$ so it must look like
$$x(t) = t^3 /(3(2)) + v_0t + x_0$$
We learned about subscript notation, summation notation and the meaning of
$$\Sigma_1^n f(x_i) \Delta x_i$$
As well we learned the the limit of above when $n \rightarrow \infty$ is the area (if we mind the assumptions). We also defined
$$\int_a^b f(x) dx$$
to be the area under $f(x)$ between $a < b$ provided $f(x) \geq 0$.
Find the area under $f(x) = x^3$ between $0$ and $1$. Use a Riemann sum to give the answer.
answer: We break the interval [a,b] into n equal sized pieces of size $\Delta = (b-a)/n$ Label the points $x_0, x_1,
... x_n$ and see that we must have
$$x_i = x_0 + i \Delta = 0 + i (1/n) = i/n$$
The $c_i$ is any point in the interval $[x_{i-1},x_i]$ for simplicity take $c_i = x_i=i/n$ and then our approximate sum becomes
$$\Sigma_{i=1}^n f(c_i) \Delta = \Sigma_{i=1}^n (i/n)^3 (1/n)= (1/n)^4 \Sigma_{i=1}^n i^3 = (1/n^4) (n(n+1)/2)^2$$
The area is the limit as n goes to $\infty$ which can be seen to be 1/4.