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kvv77 [185]
3 years ago
6

As part of his work for NASA, Dr. Murdock was asked to find out what percentage of people in the continental united States saw H

aley's Comet when it was last visible. He randomly selected three major cities, Seattle, Cleveland, and Boston, and polled 1000 randomly selected people from these cities. He finds that fewer than 5% of the people he interviewed saw the comet, so he concludes that fewer than 5% of people in the continental United States saw the comet.
Required:
Discuss whether Murdock is using a generalization or an analogy, name the sample and the target, and discuss whether there are any fallacies present in the argument (if so, why; if not, why not?).
Physics
1 answer:
mario62 [17]3 years ago
5 0

Answer:

Please, in the Explanation section you will find the explanation of the answer.

Explanation:

The exercise shows the continental United States and 3 cities used in the study carried out by Murdock. It can be said that the sample taken is part of the objective. There are several inconsistencies in Murdock's argument: the first has to do with the fact that the sample that was taken cannot represent the entire American population. A much larger, scientifically calculated sample would be required. The second is that their study did not take into account small cities or people living in the interior of the United States.

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A torus is formed when a circle of radius 3 centered at (5 comma 0 )is revolved about the​ y-axis. a. Use the shell method to wr
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Answer:

a) V=4\pi\int\limits^8_2 {x\sqrt{9-(x-5)^{2}}} \, dx

b) V=20\pi\int\limits^3_{-3} {\sqrt{9-y^{2}}} \, dy

c) V=90\pi ^{2}

Explanation

In order to solve these problems, we must start by sketching a drawing of what the graph of the problem looks like, this will help us analyze the drawing better and take have a better understanding of the problem (see attached pictures).

a)

On part A we must build an integral for the volume of the torus by using the shell method. The shell method formula looks like this:

V=\int\limits^a_b {2\pi r y } \, dr

Where r is the radius of the shell, y is the height of the shell and dr is the width of the wall of the shell.

So in this case, r=x so dr=dx.

y is given by the equation of the circle of radius 3 centered at (5,0) which is:

(x-5)^{2}+y^{2}=9

when solving for y we get that:

y=\sqrt{9-(x-5)^{2}}

we can now plug all these values into the shell method formula, so we get:

V=\int\limits^8_2 {2\pi x \sqrt{9-(x-5)^{2}} } \, dx

now there is a twist to this problem since that will be the formula for half a torus.Luckily for us the circle is symmetric about the x-axis, so we can just multiply this integral by 2 to get the whole volume of the torus, so the whole integral is:

V=\int\limits^8_2 {4\pi x \sqrt{9-(x-5)^{2}} } \, dx

we can take the constants out of the integral sign so we get the final answer to be:

V=4\pi\int\limits^8_2 {x\sqrt{9-(x-5)^{2}}} \, dx

b)

Now we need to build an integral equation of the torus by using the washer method. In this case the formula for the washer method looks like this:

V=\int\limits^b_a{\pi(R^{2}-r^{2})} \, dy

where R is the outer radius of the washer and r is the inner radius of the washer and dy is the width of the washer.

In this case both R and r are given by the x-equation of the circle. We start with the equation of the circle:

(x-5)^{2}+y^{2}=9

when solving for x we get that:

x=\sqrt{9-y^{2}}+5

the same thing happens here, the square root can either give you a positive or a negative value, so that will determine the difference between R and r, so we get that:

R=\sqrt{9-y^{2}}+5

and

r=-\sqrt{9-y^{2}}+5

we can now plug these into the volume formula:

V=\pi \int\limits^3_{-3}{(5+\sqrt{9-y^{2}})^{2}-(5-\sqrt{9-y^{2}})^{2}} \, dy

This can be simplified by expanding the perfect squares and when eliminating like terms we end up with:

V=20\pi\int\limits^3_{-3} {\sqrt{9-y^{2}}} \, dy

c) We are going to solve the integral we got by using the washer method for it to be easier for us to solve, so let's take the integral:

V=20\pi\int\limits^3_{-3} {\sqrt{9-y^{2}}} \, dy

This integral can be solved by using trigonometric substitution so first we set:

y=3 sin \theta

which means that:

dy=3 cos \theta d\theta

from this, we also know that:

\theta=sin^{-1}(\frac{y}{3})

so we can set the new limits of integration to be:

\theta_{1}=sin^{-1}(\frac{-3}{3})

\theta_{1}=-\frac{\pi}{2}

and

\theta_{2}=sin^{-1}(\frac{3}{3})

\theta_{2}=\frac{\pi}{2}

so we can rewrite our integral:

V=20\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {\sqrt{9-(3 sin \theta)^{2}}} \, 3 cos \theta d\theta

which simplifies to:

V=60\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {(\sqrt{9-(3 sin \theta)^{2}}} \, cos \theta d\theta

we can further simplify this integral like this:

V=60\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {(\sqrt{9-9 sin^{2} \theta}}} \, cos \theta d\theta

V=60\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {3(\sqrt{1- sin^{2} \theta})}} \, cos \theta d\theta

V=180\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {(\sqrt{1- sin^{2} \theta})}} \, cos \theta d\theta

V=180\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {(\sqrt{cos^{2} \theta})}} \, cos \theta d\theta

V=180\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {(cos \theta})} \, cos \theta d\theta

V=180\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {cos^{2} \theta}} \, d\theta

We can use trigonometric identities to simplify this so we get:

V=180\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {\frac{1+cos 2\theta}{2}}} \, d\theta

V=90\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {1+cos 2\theta}}} \, d\theta

we can solve this by using u-substitution so we get:

u=2\theta

du=2d\theta

and:

u_{1}=2(-\frac{\pi}{2})=-\pi

u_{2}=2(\frac{\pi}{2})=\pi

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V=45\pi\int\limits^{\pi}_{-\pi} {1+cos u}} \, du

when integrating we get that:

V=45\pi(u+sin u)\limit^{\pi}_{-\pi}

when evaluating we get that:

V=45\pi[(\pi+0)-(-\pi+0)]

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