The 4- inch radius of a circle is increasing at a rate of 2 inches per second. How fast is the area increasing?
A = πr^2
dA/dt = 2πr dr/dt
Now plug in your numbers
So im plugging in 2pie(4)? If so wjat do I do with the 2 inches per second ?
c'mon, man. In the first place, its pi, not pie.
They told you the radius is changing at 2in/s ... that is dr/dt
dA/dt = 2πr dr/dt = 2π*4*2 = 16π in/s
lol sorry im hungry. Thank you.
To find how fast the area of the circle is increasing, we need to use the formula for the area of a circle, A = πr², where A is the area and r is the radius.
Given that the radius is increasing at a rate of 2 inches per second, we can differentiate the area function with respect to time to find the rate of change of the area:
dA/dt = d/dt (πr²)
To find dA/dt, we need to differentiate the function πr² with respect to t. Applying the chain rule, we get:
dA/dt = 2πr(dr/dt)
Now, we have all the information we need:
- The radius, r, is 4 inches.
- The rate at which the radius is changing, dr/dt, is 2 inches per second.
Plugging these values into the equation, we get:
dA/dt = 2π(4)(2)
Simplifying:
dA/dt = 16π
So, the area of the circle is increasing at a rate of 16π square inches per second.