Respuesta :
Answer:
After 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec
Explanation:
Area of a circle = πr²
where;
r is the circle radius
Differentiate the area with respect to time.
[tex]\frac{dA}{dt} = 2\pi r\frac{dr}{dt}[/tex]
dr/dt = 4 ft/sec
after 12 seconds, the radius becomes = [tex]\frac{dr}{dt} X 12 = 4 \frac{ft}{sec} X 12 sec = 48 ft[/tex]
To obtain how rapidly is the area enclosed by the ripple increasing after 12 seconds, we calculate dA/dt
[tex]\frac{dA}{dt} = 2\pi r\frac{dr}{dt}[/tex]
[tex]\frac{dA}{dt} = 2\pi (48)(4)[/tex]
dA/dt = 1206.528 ft²/sec
Therefore, after 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec
Answer:
- [tex]\frac{dA}{dt} = 384\pi ft^2/sec[/tex]
Explanation:
Given,
[tex]\frac{dr}{dt} = 4ft/sec[/tex]
Area of ripple [tex]A = \pi r^2[/tex]
therefore,
[tex]\frac{dA}{dt} = 2\pi r\frac{dr}{dt}[/tex]
after 12 seconds radius becomes
[tex]\frac{dr}{dt} * 12 = 4*12 = 48ft[/tex]
therefore,
[tex]\frac{dA}{dt} = 2 * \pi * 48 * 4\\\\ = 384\pi ft^2/sec[/tex]
For more information on this visit
https://brainly.com/question/23379286