Physics, 26.07.2019 02:10 kaseywright3418

Astone dropped into a still pond sends out a circular ripple whose radius increases at a constant rate of 2.4 ft/s. (a) how rapidly is the area enclosed by the ripple increasing when the radius is 4 feet? the area is increasing at ft2/sft2/s. (b) how rapidly is the area enclosed by the ripple increasing at the end of 9 seconds? the area is increasing at ft2/sft2/s.

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For the first question which asking how rapid is the area enclosed by the ripple increasing when the radius is 2 feet, we will use the formula which is Area A = pi*r^2 then apply differentiation dA/dt = pi * 2r * dr/dt
A) area enclosed increasing at speed dA/dt = pi * 2 * 2 feet * 2.4 feet/sec

For the second question B) After 9.5 seconds , radius = 9.5* 2.4 =22.8 feet area increasing at speed dA/dt = pi * 2 * 22.8feet * 2.4 feet/sec answer units = 343.82 feet^2 /sec

Answer from: madisonelizibet1

a. The area enclosed by the circular ripple is increasing at $60.33 \;ft^2/s$ when the radius is 4 feet.

b. The area enclosed by the circular ripple is increasing at $325.76 \;ft^2/s$ at the end of 9 seconds.

Given the following data:

Rate = 2.4 ft/secsRadius A = 4 feetTime = 9 seconds

a. To find how rapidly the area enclosed by the circular ripple is increasing when the radius is 4 feet:

Mathematically, we know that the area of a circle is given by the formula;

$A = \pi r^{2}$

Where:

A is the area of a circle.r is the radius of a circle. $\pi = 3.142$

Next, we would differentiate with respect to time;

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

Substituting the values, we have;

$\frac{dA}{dt} =$ × 3.142 × × 2.4

$\frac{dA}{dt} = 60.33 \;ft^2/s$

Therefore, the area enclosed by the circular ripple is increasing at $60.33 \;ft^2/s$ when the radius is 4 feet.

b. To find how rapidly the area enclosed by the circular ripple is increasing at the end of 9 seconds:

First of all, we would determine the radius after 9 seconds as follows;

$Radius = \frac{dr}{dt}(t)\\\\Radius = 2.4(9)$

Radius, r = 21.6 feet.

Now, we can determine the rate at which the area is increasing:

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

Substituting the values, we have;

$\frac{dA}{dt} =$ × 3.142 × 21.6 × 2.4

$\frac{dA}{dt} = 325.76 \;ft^2/s$

Therefore, the area enclosed by the circular ripple is increasing at $325.76 \;ft^2/s$ at the end of 9 seconds.

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Astone dropped into a still pond sends out a circular ripple whose radius increases at a constant ra...