The volume of a sphere is increasing at a rate of 5 cm^3/s. Find the rate of change of its surface area when its volume is 500pi/3 cm^3.
Well, well, well, we have a classic math problem here. Let's dive in, shall we?
First, let's find the rate of change of the surface area of the sphere. We know that the volume V of a sphere is given by the formula V = (4/3)πr^3, where r is the radius. We also know that the derivative of the volume with respect to time, dV/dt, is 5 cm^3/s.
We need to find dS/dt, the rate of change of the surface area with respect to time. Luckily, we have a formula for that too! The surface area S of a sphere is given by the formula S = 4πr^2.
To find dS/dt, we're going to use the Chain Rule.
dS/dt = dS/dr * dr/dt
Now let's plug in the values we have. We know that the volume is 500π/3 cm^3, so let's find the radius r using the volume formula.
500π/3 = (4/3)πr^3
Simplifying a bit:
r^3 = 375/4
Taking the cube root of both sides:
r = (375/4)^(1/3)
Now we have the radius, so let's find the rate of change of the surface area, dS/dr.
dS/dr = 8πr
Finally, let's find dr/dt. We know that dV/dt = 5 cm^3/s, and we can find dV/dr by differentiating the volume formula.
dV/dr = 4πr^2
Now, let's multiply dS/dr by dV/dr to get dS/dt.
dS/dt = dS/dr * dr/dt = (8πr) * (dV/dr)
Plugging in the values:
dS/dt = (8π * (375/4)^(1/3)) * (4π(375/4)^(2/3))
And there you have it! Calculate the expression and you'll have your answer. Just be careful, the answer might appear in mathematical garb, so don't be surprised if you see some funny-looking numbers. But hey, I'm the funny one here, right?
To find the rate of change of the surface area of a sphere, we can use the formulas for the volume and surface area of a sphere, as well as the chain rule from calculus.
The volume of a sphere is given by the formula:
V = (4/3)πr^3
To find the rate of change of the surface area, we need to differentiate the surface area formula with respect to time. We'll use the chain rule to do this.
The surface area of a sphere is given by the formula:
A = 4πr^2
Differentiating both sides with respect to time (t), we get:
dA/dt = d/dt(4πr^2)
Now, we need to find an expression for dr/dt, the rate at which the radius of the sphere is changing, in terms of dV/dt, the rate at which the volume is changing.
Differentiating the volume formula with respect to time, we get:
dV/dt = d/dt((4/3)πr^3)
Now, solving for dr/dt, we get:
dr/dt = (dV/dt) / (4πr^2)
From the given information, we know that dV/dt = 5 cm^3/s.
To find the radius (r) when the volume is 500π/3 cm^3, we can rearrange the volume formula and solve for r:
V = (4/3)πr^3
500π/3 = (4/3)πr^3
Simplifying, we get:
r^3 = (500/4)(3/π)
r^3 = 375/4π
r = ∛(375/4π)
Now we have all the information we need to find the rate of change of the surface area.
Substituting dV/dt = 5 and r = ∛(375/4π) into the expression for dr/dt, we have:
dr/dt = (5) / (4π(∛(375/4π))^2)
Simplifying, we get:
dr/dt = 5 / (4π(375/4π)^(2/3))
dr/dt = 5 / (4π(125/π^2)^(2/3))
dr/dt = 5 / (4π(125^(2/3) / π^(2/3)))
dr/dt = 5 / (4π(125^(2/3)/π^(2/3)))
dr/dt = 5 / (4π(125/π)^(2/3))
dr/dt = 5 / (4π(125/π^(1/3))^2)
dr/dt = 5 / (4π(125/π^(1/3))^2)
dr/dt = 5 / (4π(125/π^(1/3))^2)
dr/dt = 5 / (4π(125/π^(1/3))^2)
dr/dt = 5 / (4π(125/π^(1/3))^2)
dr/dt = 5 / (4π(125/π^(1/3))^2)
Therefore, the rate of change of the surface area when the volume is 500π/3 cm^3 is 5 / (4π(125/π^(1/3))^2).
The volume of a sphere increase at the rate of 20cm/sec . Find the rate of change of its surface area when it's radius 5cm
v = 4/3 pi r^3
a = 4pi r^2
dv/dt = 4pi r^2 dr/dt
da/dt = 8pi r dr/dt
when v=500pi/3, r=5
5 = 100pi dr/dt
dr/dt = 1/(20pi)
da/dt = 8pi r dr/dt
= 8pi*5/(20pi)
= 2