Mathematics: A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of 60 cm/s. Find the rate which the area within the circle is increasing after a) 1 second, b) 3 seconds, and c) 5 seconds. What can you conclude?

wt.f...Step-by-step explanation:...

A stone is dropped into a lake, creating a circular

soliskb

30.01.2022

$\frac{dA}{dt} = 7200\pi t$

a) $\frac{dA}{dt} = 7200\pi\ cm^2/s$

b) $\frac{dA}{dt} = 21600\pi\ cm^2/s$

c) $\frac{dA}{dt} = 36000\pi\ cm^2/s$

We can conclude that the area of the circle increases faster when the time increases.

Step-by-step explanation:

First let's write the equation for the area of the circle:

$A = \pi*r^2$

The rate that the radius of the circle increases is 60 cm/s, so we have:

$\frac{dr}{dt} = 60$

$dr = 60dt \rightarrow r = 60t$

To find the rate that the area increases, let's take the derivative of the equation of the area in relation to time:

$\frac{dA}{dt} = \pi*\frac{d}{dt} r^2$

$\frac{dA}{dt} = \pi *\frac{dr^2}{dr} \frac{dr}{dt}$

$\frac{dA}{dt} = \pi *2r *\frac{dr}{dt}$

$\frac{dA}{dt} = 2\pi *(60t) *60$

$\frac{dA}{dt} = 7200\pi t$

Using t = 1, we have:

$\frac{dA}{dt} = 7200\pi *1 = 7200\pi\ cm^2/s$

Using t = 3, we have:

$\frac{dA}{dt} = 7200\pi *3 = 21600\pi\ cm^2/s$

Using t = 5, we have:

$\frac{dA}{dt} = 7200\pi *5= 36000\pi\ cm^2/s$

We can conclude that the area of the circle increases faster when the time increases.

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