Calculus problem?

Question

A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of 70 cm/s. Find the rate at which the area within the circle is increasing after each of the following. Evaluate your answer numerically.
A. 3 seconds
B. 5 seconds
C. 6 seconds

Answer 1

The radius of the circle after x seconds is 70x

area A = pie(70x)^2

A = 15386x^2
dA/dx = 30772x

plug in your values and you will get the rate of change

eg after 3 seconds - rate of change is

30772 x 3 = 92316 cm^2/s or 9.23 m^2/s

Answer 2

Looks like a related rates problem.

Area of a circle is: a = πr²

You need to find how quickly the area changes....this is (in calculus terms) da/dt.

==> da/dt = 2πr(dr/dt)

--> dr/dt = 70 cm/s (given in the question)

==> da/dt = 2π(70 cm/s)r

So....technically speaking the area is changing at the rate of 140πr [cm/s].



Now ----- for now the sake of clarity and ease, let's set 140π [cm/s] = Q

==> da/dt = Qr

Notice that the question states that the circle increases OUTWARDLY at the rate of blah blah blah blah blah. What this means is...the RADIUS is increasing.

So...imagine any single point on the circumference of the circle that is drawing a straight line outwardly at a constant rate --- which is given --- 70 cm/s.

So we can, in essence say that this point is traveling according to the distance/time/rate equation......d = vt. Where "v" = 70 cm/s.

So the radius "r" can be replaced by "d":

Therefore: r = d = vt

Then: da/dt = Qr = Q(vt)

Well..........v = 70 cm/s. Remember...Q = 140π [cm/s], so Q(vt) = (140π [cm/s])(70 cm/s)t = 9800πt [cm²/s²]....but let's keep this "Q" notation for a moment.

What we have now is this: da/dt = Qt



Where t will take the values: 3 s, 5 s, 6 s....so...let's put these in along with the actual value of "Q".

The case of t = 3 s
------------------------

da/dt = 9800π [cm²/s²]t = 9800π [cm²/s²](3 s) = 29400π [cm²/s]




The case of t = 5 s
------------------------

da/dt = 9800π [cm²/s²]t = 9800π [cm²/s²](5 s) = 49000π [cm²/s]




The case of t = 6 s
------------------------

da/dt = 9800π [cm²/s²]t = 9800π [cm²/s²](6 s) = 58800π [cm²/s]


NOTE! Look at the units on each result. They are in "cm²/s". Those units are correct. You are dealing with an AREA that changes per unit time! And you have square centimeters per second. This is what I claim is the result of this question. Check this over.....seems strange to me they'd give you a question like this. It does not seem to result in a nice 'clean' answer, but the results seem reasonable to me.

Answer 3

well this question cant be answered. the acceleration cant be calculated without the speed at several intervals. if your saying that its goes 70 cm/s continuously then at 3, 5, and 6 seconds the rate of increase will remain constant (70 cm/s)

Answer 4

I am going to assume that by 70cm/s you are referring to the diameter of the circle. Therefore the rate of change of the radius would be 35cm/s and the rate of the change of the area would be (35)²pi cm/s. So just multiply the seconds by the rate to get the area.
A. 35²pi * 3 =
B. 35²pi * 5 =
C. 35²pi * 6 =

Answer 5

A = pi r^2
dA/dt = (dA/dr)(dr/dt)
dA/dt = 2pir*70 = 439.82r
= 1319.5 cm^/sec after 3 seconds
= 2199,1 cm^2/second after 5 seconds
= 2638.9 cm^2/sec after 6 seconds