Isabelle takes the bus to work. The bus ride to work costs her $2.50 each time. She could buy a bus pass for a one-time fee of $
2 answers:
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I think that is the answer because it says 2/5 where two rows are shaded and three more rows remain unshaded so, when you add that three and two it's altogether 5, so I think that is the answer, please check out the document I have uploaded.
Answer:
The center of mass is located at coordinates
Step-by-step explanation:
Consider the disk and its' portion located at the first quadrant. We are being said that the density
is proportional to the distance of the point to the x-axis. Given a point of coordinates (x,y), its' distance to the x-axis is y. Therefore, we know that the density function
for some constant k, whose value is not relevant. Let (x_0,y_0) be the center of mass and let D be the region occupied by the lamina. The center of mass's coordinates fulfill the following equations:
where M is the mass of the region, which in this case is given by .
Let us calculate the mass of the lamina. For this, we will use the polar coordinates. Recall that they are given by the change of coordinates , where r and theta are the new parameters. Given a point (x,y) in the plane, r is the distance from (x,y) to the origin and theta is the angle formed between the line that joins the origin and the point, and the x-axis. We want to describe the region D in terms of the new parameters. Replacing the values of x,y in the given inequality, we get that
. Since
and r>0 we get that r<=5. On the other side, in order to describe the whole region of the first quadrant, we need to sweep the angle theta from 0 to
. With that, we can calculate the mass of the lamina as follows
M =
In here, the extra r appears as the jacobian of the change of coordinates (the explanation of why this factor occurs is beyond the scope of this answer. Please refer to the internet for further explanation).
Then, we need to calculate the following integrals.
(recall
).
Then,
On the other hand:
(recall
Then,
To check that the answer makes sense, the center of mass must lie in the disk, that is, it should satisfy the equation. We can easily check that