If a snowball melts so that its surface area decreases at a rate of 5 cm2/min, find the rate at which the diameter decreases when the diameter is 11 cm.
volume=4pI*r^3 /3
dV/dt=4Pi*r^2 dr/dt
(note, diameter=2r) so ratediamter=2 radius rate.
dV/dt= 3PI(diamter/2)^2 * dDiamter/dt * 1/4
if I did the algebra in my head right, then
rate diatmeter=16 dV/dt * 1/(3PI*diameter^2)
My typo:
S = 4 π R²
S = 4 π ∙ ( D / 2 )² = 4 π ∙ D² / 4
S = π ∙ D²
dS / dD = 2 π ∙ D
To find the rate at which the diameter decreases, we need to use the chain rule in calculus. The chain rule is a formula used to find the derivative of a composition of functions.
Let's begin by finding an equation that relates the surface area of the snowball to its diameter.
The surface area, A, of a sphere with diameter d is given by the formula:
A = 4πr^2,
where r is the radius of the sphere. Since the diameter is twice the radius, we can rewrite the equation as:
A = 4π(d/2)^2,
A = πd^2/4.
Now, we can differentiate both sides of the equation with respect to time, t, to find the derivative of the surface area:
dA/dt = (d/dt)(πd^2/4).
Let's simplify this expression:
dA/dt = π/4 * (d^2/dt).
Since we are given that the surface area decreases at a rate of 5 cm^2/min, we can substitute this value into the equation:
-5 = π/4 * (d^2/dt).
Now, we need to solve for (d^2/dt) to find the rate at which the diameter decreases. Rearranging the equation, we get:
(d^2/dt) = -5 * 4/π.
Substituting this value into the equation, we find:
(d^2/dt) = -20/π cm^2/min.
So, the rate at which the diameter decreases when the diameter is 11 cm is -20/π cm^2/min.
R = Radius
D = Diameter
The surface area of a sphere:
S = 4 π R²
S = 4 π ∙ ( R / 2 )² = 4 π ∙ D² / 4
S = π ∙ D²
dS / dD = 2 π ∙ D
You know:
dS / dt = - 5 cm² / min
d = 11 cm
The derivative by the chain rule:
dS / dD = dS / dt ∙ dt / dD
2 π ∙ D = ( - 5 ) ∙ dt / dD
2 π ∙ 11 = ( - 5 ) ∙ dt / dD
22 π = ( - 5 ) ∙ dt / dD
22 π / ( - 5 ) = dt / dD
- 22 π / 5 = dt / dD
Take reciprocal value:
- 5 / ( 22 π ) = dD / dt
dD / dt = - 5 / ( 22 π ) cm² / min