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A stone 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.

Respuesta :

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
Lanuel

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

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

Given the following data:

  • Rate = 2.4 ft/secs
  • Radius A = 4 feet
  • Time = 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;

[tex]A = \pi r^{2}[/tex]

Where:

  • A is the area of a circle.
  • r is the radius of a circle.
  • [tex]\pi = 3.142[/tex]

Next, we would differentiate with respect to time;

[tex]\frac{dA}{dt} = 2\pi r (\frac{dr}{dt})[/tex]

Substituting the values, we have;

[tex]\frac{dA}{dt} =[/tex] [tex]2[/tex] × [tex]3.142[/tex] × [tex]4[/tex]  × [tex]2.4[/tex]

[tex]\frac{dA}{dt} = 60.33 \;ft^2/s[/tex]

Therefore, the area enclosed by the circular ripple is increasing at[tex]60.33 \;ft^2/s[/tex]  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;

[tex]Radius = \frac{dr}{dt}(t)\\\\Radius = 2.4(9)[/tex]

Radius, r = 21.6 feet.

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

[tex]\frac{dA}{dt} = 2\pi r (\frac{dr}{dt})[/tex]

Substituting the values, we have;

[tex]\frac{dA}{dt} =[/tex] [tex]2[/tex] × [tex]3.142[/tex] × [tex]21.6[/tex]  × [tex]2.4[/tex]

[tex]\frac{dA}{dt} = 325.76 \;ft^2/s[/tex]

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

Read more: https://brainly.com/question/14864445

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