the radius r of a sphere is increasing at a rate of 3cm/min. find the rate of change of the area when r=3cm
a = 4πr^2
da/dt = 8πr dr/dt
Now just plug in your numbers.
where did 8 come from
hello!?
the derivative of r^2 is 2r
4*2 = 8
?!?!?!??
ohhhh ya sorry I get it
To find the rate of change of the area of a sphere, we need to differentiate the formula for the area of a sphere with respect to time.
The formula for the surface area of a sphere is given by:
A = 4πr^2
Differentiating both sides of this equation with respect to time (t), we get:
dA/dt = d(4πr^2)/dt
Using the chain rule, we can compute:
dA/dt = 2(4πr) (dr/dt)
Now we know that the radius r of the sphere is increasing at a rate of 3 cm/min, which means dr/dt = 3 cm/min.
Substituting this value into the equation, we have:
dA/dt = 2(4πr) (3)
When r = 3 cm, we can substitute this value into the equation:
dA/dt = 2(4π(3)) (3)
Simplifying, we get:
dA/dt = 2(12π) (3)
dA/dt = 72π
Hence, when the radius of the sphere is 3 cm, the rate of change of the area is 72π square centimeters per minute.