The radius of sphere at therate of 4cm per second .at what ratis its surface area chang when theradius is 6cm
You seem to have left out some words in your question, such as:
The radius of sphere IS INCREASING at the rate of 4cm per second.
Surface area is
A= 4*pi*R^2
The rate of increase of surface area is
dA/dt = 8*pi*R*(dR/dt)
Plug in 6 cm for R, and dR/dt = 4 cm/s, and compute
R=4t (cm)
6=4t
t=1.5 s.
A=4πR²=4π(4t)²=
=64πt²=64π(1.5)²=452.4 cm²
dA/dt = 64πt²=128 πt=128 π•1.5=603.2 cm²/s
To determine the rate at which the surface area of a sphere changes with respect to its radius, we can use the formula for the surface area of a sphere:
Surface Area = 4πr^2,
where r is the radius of the sphere.
We are given that the radius is changing at a rate of 4 cm per second, which can be written as dr/dt = 4 cm/s.
We need to find the rate at which the surface area is changing with respect to time, dA/dt, when the radius is 6 cm.
To find this derivative, we can apply the chain rule.
Derivative of the surface area with respect to time:
dA/dt = dA/dr * dr/dt,
where dA/dr is the derivative of the surface area with respect to the radius.
Taking the derivative of the surface area formula with respect to the radius:
dA/dr = d/dt(4πr^2) = 8πr.
Plugging in the given radius of 6 cm:
dA/dr = 8π(6) = 48π cm^2.
Finally, multiplying the derivative by the rate of change in the radius, which is 4 cm/s:
dA/dt = dA/dr * dr/dt = 48π cm^2 * 4 cm/s = 192π cm^2/s.
Therefore, the rate at which the surface area of the sphere is changing when the radius is 6 cm is 192π cm^2/s.