Respuesta :
Answer:
[tex]\frac{dA}{dt} = 7200\pi t[/tex]
a) [tex]\frac{dA}{dt} = 7200\pi\ cm^2/s[/tex]
b) [tex]\frac{dA}{dt} = 21600\pi\ cm^2/s[/tex]
c) [tex]\frac{dA}{dt} = 36000\pi\ cm^2/s[/tex]
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:
[tex]A = \pi*r^2[/tex]
The rate that the radius of the circle increases is 60 cm/s, so we have:
[tex]\frac{dr}{dt} = 60[/tex]
[tex]dr = 60dt \rightarrow r = 60t[/tex]
To find the rate that the area increases, let's take the derivative of the equation of the area in relation to time:
[tex]\frac{dA}{dt} = \pi*\frac{d}{dt} r^2[/tex]
[tex]\frac{dA}{dt} = \pi *\frac{dr^2}{dr} \frac{dr}{dt}[/tex]
[tex]\frac{dA}{dt} = \pi *2r *\frac{dr}{dt}[/tex]
[tex]\frac{dA}{dt} = 2\pi *(60t) *60[/tex]
[tex]\frac{dA}{dt} = 7200\pi t[/tex]
a)
Using t = 1, we have:
[tex]\frac{dA}{dt} = 7200\pi *1 = 7200\pi\ cm^2/s[/tex]
b)
Using t = 3, we have:
[tex]\frac{dA}{dt} = 7200\pi *3 = 21600\pi\ cm^2/s[/tex]
c)
Using t = 5, we have:
[tex]\frac{dA}{dt} = 7200\pi *5= 36000\pi\ cm^2/s[/tex]
We can conclude that the area of the circle increases faster when the time increases.
