Class 11 - Review (9/28/2017, Thursday)
2nd Gateway
The second Gateway will start on Sept 26, Tuesday and you will earn extra 10% quiz points if you pass it by Sept 29, Friday. (Of course, 20% total if you passed 1st Gateway early.)
Webwork
- 8.2 due 9/28 Thur
- 8.4 due 10/2 Mon
Team HW 2
- Assignment 2 link
- Due 9/29 Fri
Here are hints for Team HW 2.
1a. To determine overestimate / underestimate, you need to know whether the function f(x) increases or decreases in the interval [1, 4]. This can be done by taking derivative:
f'(x) = \dfrac{(x^{2} + 2x) - 2x(x + 1)}{(x^{2} + 2x)^{2}}.
1b. Note that x^{2} + 2x = x(x + 2). Apply the partial fraction trick.
1c. Your computation will involve the natural log \ln.
1d. You should notice that you have some problem at x = 0.
2a. Note that each interval has equal length of 0.5.
2b. Sum 2^{-2} + 2^{-1} + 2^{0} + 2^{1} + 2^{2}.
2c. Suppose that the number c = 1/0 exists. Then 0 = 0 \cdot c = 0 \cdot (1/0) = 1.
2d. You don't need any hint for this.
3a. You are asked to compute f(3) - f(0).
3b. Think about g(t) separately for the intervals [0, 0.5], [0.5, 2], and [2, 3]. Recall that the line with slope m passing through (a, b) is given by y = m(x - a) + b. You should think about the rest of the problem on your own.
3c. Recall that average value of F(t) for a \leq t \leq b is equal to
\frac{1}{b - a}\int_{a}^{b}F(t)dt.
Think about the rest of the problem on your own.
3d. Possibly except at t = 0.5, 2, you are given H(t) = g'(t) - f'(t). For ii, the distance between Car 1 and Car 2 at time t = x is equal to
\left|\int_{0}^{x}g(t) - f(t)dt\right|.
Now, think about how to get rid of the absolute value by breaking the interval [0, 3] into pieces.
4a. You should start by writing
P(x) = -2 + \int_{1}^{x}p(t)dt.
4b. Write out p(x) explicitly by breaking up the interval [-3, 3]. Then use the hint for 4a to explicitly compute P(x).
Comments
Post a Comment