If a snowball melts so that its surface area decreases at a rate of 10 cm^2/min, find the rate at which the diameter decreases when the diameter is 11 cm.
i got -5/11pi, but it says i'm wrong. where did i mess up?
Given:
d(SA)/dt = 10cm^2/min
d(d)/dt = ?
d = 11
SA = 4πr^2 = 4π(d/2)^2 = π*d^2
Calculation:
SA = π*d^2
d(SA)/dt = 2πd[d(d)/dt]
10 = 2π(11)[d(d)/dt]
10 = 22π[d(d)/dt]
5/(11π) = d(d)/dt
Looks like your mistake was the negative there somehow lol
Remember that 5/(11π) is the rate that the diameter is decreasing! If it decreased at a rate of -5/(11π), then it's increasing at a rate of 5/(11π)!
Be very careful
To find the rate at which the diameter decreases, you need to relate the rate of change of the surface area to the rate of change of the diameter. The formula for the surface area of a sphere is given by:
Surface Area = 4πr^2
where r is the radius of the sphere.
To relate the radius to the diameter, we use the formula:
diameter = 2 * radius.
Let's differentiate the surface area formula with respect to time (t) using the chain rule:
d(Surface Area)/dt = d(4πr^2)/dt = 8πr * (dr/dt)
where dr/dt represents the rate of change of the radius with respect to time.
We are given that d(Surface Area)/dt = -10 cm^2/min (negative because the surface area is decreasing). And we need to find the value of dr/dt when the diameter is 11 cm.
From the given information, we can deduce that when diameter = 11 cm, radius = diameter/2 = 11/2 = 5.5 cm.
Let's plug in the values we have into the equation:
-10 = 8π(5.5) * (dr/dt)
Now, we can solve for dr/dt:
dr/dt = -10 / (8π * 5.5)
= -5 / (4π * 5.5)
= -5 / (22π)
= -5 / (22 * 3.14)
= -0.072 cm/min
So, the rate at which the diameter decreases when the diameter is 11 cm is approximately -0.072 cm/min.
It seems like you made an error in your calculation. Please double-check your calculations to find the mistake.