Class 8 - More volumes and shell method (9/21/2017, Thursday)
Here are some logistics to begin with.
You may notice that the integral in the absolute value is non-negative for any $0 \leq t \leq 3,$ so you may eliminate writing the absolute value signs to have
\[d(t) = \int_{0}^{t}f(x) - g(x)dx\]
By the second Fundamental Theorem, we have $d'(t) = f(t) - g(t),$ so now the critical points $t$ making $d'(t) = 0$ is the same as the ones where $f(t) = g(t).$ Hence, you can maximize $d(t)$ using Math 115 skills.
Many people have plugged in a bunch of $t$ to compute the integral, but this won't give you a complete verification of your answer.
Now, when $-3 < x \leq -2$, you have $p(x) = -x + 1,$ so $P(x) = - 2 + (-x^{2}/2 + x).$ When $-2 < x < 0,$ you have $p(x) = -1,$ so $P(x) = -2 - x$. When $x = 0,$ you have $P(x) = P(0) = -2,$ so you can see that the value they have given you would be the unique value that makes $P(x)$ continuous.
Remark: optional reading. This should look a bit weird that you are forced to choose one point to make $P(x)$ continuous. There is something cool happening here: the functions that are given by integral can "ignore" finitely many points. Roughly this is our vague idea that says: "For area, a few points do not matter." A more rigorous way to understand this phenomenon is available by formalizing the concept of area. This leads to the theory of measures.
In any case, let's continue. For $0 < x \leq 2$, we have $p(x) = 1,$ so $P(x) = -2 + x$. For $2 < x < 3,$ we have $p(x) = -2x + 5,$ so $P(x) = -2 -x^{2} + 5x.$ Note that $\lim_{x \rightarrow 0+}P(x) = -2 = P(0),$ as we have expected.
Deadline to drop
Sept 25, Monday: Keep in mind that Math 116 will require you a significant amount of work outside the class! It won't be possible, even if you had good mathematical background, to just sit in the courses without work.
1st Gateway
The first Gateway will close on Sept 25, Monday. Please take it because otherwise there is a huge penalty on your final letter grade.
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
- 7.2 due 9/21 Thurs
- Partial Fractions due 9/21 Thurs
- 7.5 due 9/25 Mon
Remarks on Team HW 1
Let's discuss briefly about some points that people have missed on Team HW 1.- For 3a, when you wrote piece-wise function form of $f'(t),$ you should not have included $t = 1, 2, 2.5$ because at these points $f(t)$ is not differentiable.
- For 3e, the distance between Car 1 and Car 2 at time $t$ is can be written as
\[d(t) = \left|\int_{0}^{t}f(x) - g(x) dx\right|\]
You may notice that the integral in the absolute value is non-negative for any $0 \leq t \leq 3,$ so you may eliminate writing the absolute value signs to have
\[d(t) = \int_{0}^{t}f(x) - g(x)dx\]
By the second Fundamental Theorem, we have $d'(t) = f(t) - g(t),$ so now the critical points $t$ making $d'(t) = 0$ is the same as the ones where $f(t) = g(t).$ Hence, you can maximize $d(t)$ using Math 115 skills.
Many people have plugged in a bunch of $t$ to compute the integral, but this won't give you a complete verification of your answer.
- For 4, you can explicitly write $P(x)$ as follows:
\[P(x) = -2 + \int_{0}^{x}p(t)dt.\]
Now, when $-3 < x \leq -2$, you have $p(x) = -x + 1,$ so $P(x) = - 2 + (-x^{2}/2 + x).$ When $-2 < x < 0,$ you have $p(x) = -1,$ so $P(x) = -2 - x$. When $x = 0,$ you have $P(x) = P(0) = -2,$ so you can see that the value they have given you would be the unique value that makes $P(x)$ continuous.
Remark: optional reading. This should look a bit weird that you are forced to choose one point to make $P(x)$ continuous. There is something cool happening here: the functions that are given by integral can "ignore" finitely many points. Roughly this is our vague idea that says: "For area, a few points do not matter." A more rigorous way to understand this phenomenon is available by formalizing the concept of area. This leads to the theory of measures.
In any case, let's continue. For $0 < x \leq 2$, we have $p(x) = 1,$ so $P(x) = -2 + x$. For $2 < x < 3,$ we have $p(x) = -2x + 5,$ so $P(x) = -2 -x^{2} + 5x.$ Note that $\lim_{x \rightarrow 0+}P(x) = -2 = P(0),$ as we have expected.
Today we will cover a part of Section 8.2 of your book.
Computing the area of a volume: shell method
There is another method to compute the volume, when we do not want to cut up the volume when we try to use disk method. Here is an example:
Exercise. Consider the graph $y = x(x-1)(x-2)$ and the area between the graph and the $x$-axis from $x = 0$ to $x = 2$. Rotate the area with respect to the $y$-axis. Compute the volume obtained by this rotation.
Optional reading: deriving shell method. With disk method, you will have to write a bunch of integrals. However, with shell method this becomes much simpler. First, find any position $0 \leq x \leq 2$ and consider a thin cylinder (i.e., "shell") whose height is given by $f(x) = |x(x-1)(x-2)|.$ Then the volume of the thin cylinder must be given by
$\pi(x + \Delta x)^{2}f(x + \Delta x) - \pi x^{2}f(x)$.
Note that we can write the above as
$\pi((x + \Delta x)^{2}(f(x + \Delta x) - f(x)) + ((x + \Delta x)^{2} - x^{2})f(x)),$
which can again be written as
$\pi \left( \Delta x(x + \Delta x)^{2}\left(\dfrac{f(x + \Delta x) - f(x)}{\Delta x}\right) + (2x + \Delta x)f(x)\Delta x\right) \approx 2\pi xf(x)\Delta x,$
where the last approximation is valid when $\Delta x$ is small and $f(x)$ is differentiable except at finitely many places.
Hence, if $f(x)$ is differentiable except at finitely many places, letting $\Delta x$ go to $0$ will lead us to conclude that the area is given by
\[\int_{0}^{2}2 \pi xf(x) dx = \int_{0}^{2}2 \pi |x(x-1)(x-2)| dx = 2\pi \left(\int_{0}^{1}x(x-1)(x-2)dx - \int_{1}^{2}x(x-1)(x-2)dx\right).\]
It is important to remember the following (which need to be modified in different settings):
The volume of a shell: $2\pi x f(x) \Delta x,$ where $x$ is radius and $f(x)$ is height of the shell. That is, you can memorize this as the area of side of a cylinder times some thickness!
Let's tackle the problem that was somewhat difficult before again:
Exercise* Solve 6b in Exam 1, 2017 Winter. (Hint: the volume of each shell must be $2\pi (3 - x)f(x) \Delta x.$ Figure out what $f(x)$ should be and where $x$ varies.)
Remark. It is surprising that using shell method for the problem above is much simpler. This means that applying shell method in the right places will save a ton of time on your exam!
More problems. Here are a bunch more problems. Hints to any problem in this posting would be
Exercise*. Solve 7b in Exam 1, 2016 Winter.
Exercise* Solve 6b in Exam 1, 2017 Winter. (Hint: the volume of each shell must be $2\pi (3 - x)f(x) \Delta x.$ Figure out what $f(x)$ should be and where $x$ varies.)
Remark. It is surprising that using shell method for the problem above is much simpler. This means that applying shell method in the right places will save a ton of time on your exam!
More problems. Here are a bunch more problems. Hints to any problem in this posting would be
- Draw a picture;
- Find thickness;
- Find where such thickness ranges;
- Compute volume of a slice;
- Sum up (i.e., integrate).
Exercise* Solve 3a in Exam 1, 2017 Winter.
Exercise* Solve 7a in Exam 1, 2016 Winter.
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