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Mekhanik [1.2K]

2 years ago

9

Coach barron bought 66 pounds of ice cream for the end of the year little league social. there are 8 teams, each with 12 people.

if the ice cream is to be divided equally amongst each person, how many ounces of ice cream will each person get?

1 answer:

frozen [14]2 years ago

4 0

Each person gets about 9.75 ounces of ice cream

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1 year ago

Determine the answer to 3 + (-5) and explain the steps using a number line

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2 years ago

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw

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First, complete the square in the equation for the second circle to determine its center and radius:

<em>x</em> ² - 10<em>x</em> + <em>y</em> ² = 0

<em>x</em> ² - 10<em>x</em> + 25 + <em>y </em>² = 25

(<em>x</em> - 5)² + <em>y</em> ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

<em>x</em> = <em>r</em> cos(<em>θ</em>)

<em>y</em> = <em>r</em> sin(<em>θ</em>)

Then

<em>x</em> ² + <em>y</em> ² = 100 → <em>r </em>² = 100 → <em>r</em> = 10

<em>x</em> ² - 10<em>x</em> + <em>y</em> ² = 0 → <em>r </em>² - 10 <em>r</em> cos(<em>θ</em>) = 0 → <em>r</em> = 10 cos(<em>θ</em>)

<em>y</em> = 5 → <em>r</em> sin(<em>θ</em>) = 5 → <em>r</em> = 5 csc(<em>θ</em>)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to <em>y</em> = 5 at the point (0, 5).

Split up the region at 3 angles <em>θ</em>₁, <em>θ</em>₂, and <em>θ</em>₃, which denote the angles <em>θ</em> at which the curves intersect. They are

<em>θ</em>₁ = 0 … … … by solving 10 = 10 cos(<em>θ</em>)

<em>θ</em>₂ = <em>π</em>/6 … … by solving 10 = 5 csc(<em>θ</em>)

<em>θ</em>₃ = 5<em>π</em>/6 … the second solution to 10 = 5 csc(<em>θ</em>)

Then the area of the region is given by a sum of integrals:

$\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)$

$=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta$

To compute the integrals, use the following identities:

sin²(<em>θ</em>) = (1 - cos(2<em>θ</em>)) / 2

cos²(<em>θ</em>) = (1 + cos(2<em>θ</em>)) / 2

and recall that

d(cot(<em>θ</em>))/d<em>θ</em> = -csc²(<em>θ</em>)

You should end up with an area of

$=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta$

$=\boxed{25\sqrt3+\dfrac{125\pi}3}$

We can verify this geometrically:

• the area of the larger circle is 100<em>π</em>

• the area of the smaller circle is 25<em>π</em>

• the area of the circular segment, i.e. the part of the larger circle that is bounded below by the line <em>y</em> = 5, has area 100<em>π</em>/3 - 25√3

Hence the area of the region of interest is

100<em>π</em> - 25<em>π</em> - (100<em>π</em>/3 - 25√3) = 125<em>π</em>/3 + 25√3

as expected.

3 0

2 years ago