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KonstantinChe [14]
3 years ago
12

The radius of a circle is changing at the rate of 1/π inches per second. At what rate, in square inches per second, is the circl

e’s area changing when the radius is 5 inches?
Mathematics
1 answer:
MAXImum [283]3 years ago
4 0

Answer:

10 square inches per second.

Step-by-step explanation:

The radius of the circle is given by the equation:

r(t) = (1/π  in/s)*t

Where time in seconds.

Remember that the area of a circle of radius R is written as:

A = π*R^2

Then the area of our circle will be:

A(t) = π*( (1/π  in/s)*t)^2 = π*(1/π  in/s)^2*(t)^2

Now we want to find the rate of change (the first derivation of the area) when the radius is equal to 5 inches.

Then the first thing we need to do is find the value of t such that the radius is equal to 5 inches.

r(t) = 5 in =  (1/  in/s)*t

       5in*(π s/in) = t

        5*π s = t

So the radius will be equal to 5 inches after 5*π seconds, let's remember that.

Now let's find the first derivate of A(t)

dA(t)/dt = A'(t) = 2*(π*(1/π  in/s)^2*t = (2*π*t)*(1/π  in/s)^2

Now we need to evaluate this in the time such that the radius is equal to 5 inches, we will get:

A'(5*π s) = (2*π*5*π s)*((1/π  in/s)^2

              = (10*π^2  s)*(1/π^2  in^2/s^2) = 10 in^2/s

The rate of change is 10 square inches per second.

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How to solve this problem
rodikova [14]
\bf \textit{circumference of a circle}\\\\
C=2\pi r\quad 
\begin{cases}
r=radius\\
-----\\
r=10
\end{cases}\implies C=2\pi 10\implies \boxed{C=20\pi }\\\\
-------------------------------\\\\

\bf \textit{let's converte 1km to inches}
\\\\\\
km\cdot \cfrac{1000m}{km}\cdot \cfrac{100cm}{m}\cdot \cfrac{in}{2.54cm}\implies \cfrac{1000\cdot 100\cdot in}{2.54}\approx 39370.079\ in\\\\
-------------------------------\\\\
\textit{how many times does }20\pi \textit{ go into }39370.079?\implies \boxed{\cfrac{39370.079}{20\pi }}
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We're no strangers to love

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Never gonna run around and desert you

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Never gonna give you up, never gonna let you down

Never gonna run around and desert you

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