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lutik1710 [3]
3 years ago
9

You’re baking a circular cake for a party. Your cake can be any size you want. What will the radius of your cake be and how many

slices will you be able to cut?

Mathematics
1 answer:
REY [17]3 years ago
7 0
What will the radius of your cake be?

This is a problem of geometry. Given that the cake is circular, the greater the cake the greater the radius of it. So, as shown in the figure 1, the radius will be the distance from the center to any extreme point of the circle.

How many slices will you be able to cut?

The total area of a circle is given by:

A= \pi  r^{2}

We need to fin how many slices will be cut, so let's calculate the area of the circular sector which can be obtained simply applying rule of three, so:

A_{cs}= \pi r^{2}( \frac{\theta}{2\pi})= \frac{r^{2}\theta}{2}

Let's name n the number of slices, if we divide the total area by n this result, each area  must equal, then:

\frac{\pi r^{2}}{n}=\frac{r^{2}\theta}{n}

Finally, we will be able to cut:

n= \frac{2\pi}{\theta} Slices

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By inspecting the integrand, the "obvious" choice for substitution would be

<em>u</em> = <em>y</em> + <em>x</em>

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<em />

Solving for <em>x</em> and <em>y</em>, we would have

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<em>y</em> = (<em>u</em> + <em>v</em>)/2

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J=\begin{bmatrix}x_u&x_v\\y_u&y_v\end{bmatrix}=\dfrac12\begin{bmatrix}1&-1\\1&1\end{bmatrix}\implies|\det J|=\left|\dfrac12\right|=\dfrac12

The trapezoid <em>R</em> has two of its edges on the lines <em>x</em> + <em>y</em> = 8 and <em>x</em> + <em>y</em> = 9, so right away, we have 8 ≤ <em>u</em> ≤ 9.

Then for <em>v</em>, we observe that when <em>x</em> = 0 (the lowest edge of <em>R</em>), <em>v</em> = <em>y</em> ; similarly, when <em>y</em> = 0 (the leftmost edge of <em>R</em>), <em>v</em> = -<em>x</em>. So

-<em>x</em> ≤ <em>v</em> ≤ <em>y</em>

-(<em>u</em> - <em>v</em>)/2 ≤ <em>v</em> ≤ (<em>u</em> + <em>v</em>)/2

-<em>u</em> + <em>v</em> ≤ 2<em>v</em> ≤ <em>u</em> + <em>v</em>

-<em>u</em> ≤ <em>v</em> ≤ <em>u</em>

<em />

So, the integral becomes

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=\displaystyle\frac52\int_8^9\frac u7(\sin7-\sin(-7))\,\mathrm du

=\displaystyle\frac57\sin7\int_8^9u\,\mathrm du

=\displaystyle\frac5{14}\sin7(9^2-8^2)=\boxed{\frac{85}{14}\sin7}

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