i'm trying to confirm my answer?

Question

if Snow balls melts so that it's surface area decreases at a rate of 1cm^2/min,find the rate at which the diameter decreases when the diameter is 29cm. answer in cm/min

Answer 1

The formula for surface area is
S = 4 pi r^2
The diameter is equal to twice the radius.
D = 2r, so r = D/2
Replacing r with D/2 in the surface area formula we have
S = 4 pi (D/2)^2
S = 4 pi D^2/4
S = pi D^2
Now take the derivative with respect to time t.
dS/dt = (2 pi D) dD/dt
You given that dS/dt = 1 and D = 29
You can solve for dD/dt
1 = (2 pi)(29) dD/dt
1/(58pi) = dD/dt
dD/dt = .005488101486

Answer 2

Surface area = S(r) = 4 (pi) r^2

S'(r) = the rate of change = 8 (pi) r * r' ==>formula "1"

since r = 29/2 and S'(r) = S'(29/2) = 1cm^2/min,

then ==> formula "1" ==> 1 = 8 (pi) * (29/2) * r'

find r'

the rate of diameter decrease = 2 * the rate of radious decrease = 2 * r' (the answer is in cm/min

Answer 3

known: rate of decrease in surface area (dA/dt = -1 cm^2/min)

find: rate at which diameter decreases when diameter is 29 cm
format: answer in cm/min

problem: relate rate of surface area to rate of diameter

Formula for area is: Area = 4 * pi * r ^2
or
Area = pi * D ^2.

take derivative of both sides of equation
d(Area)/dt = pi * d(D^2)/dt
or
d(Area)/dt = pi * d(D^2)/dD * dD/dt
or
d(Area)/dt = pi * (2 * D) * dD/Dt

you are given d(Area)/dt as a value. This is algebra. Isolate the time rate of change of the diameter, then substitute in the numeric values.

Now ask yourself:
What is it about this answer, besides that I got it online, that makes it correct? What about the method is correct? How can I check it?

(I recommend a unit analysis to determine if its correct)

Answer 4

SA of sphere = 4pi (D/2)^2 = 4pi (D^2/4) = pi D^2
D = diameter
T = time (in this case min)

dSA/dT = 1cm^2/min
dSA/dD = 2pi D

dD/dT = (dSA/dT) / (dSA/dD)
= 1 / (2pi D)

So dD/dT when D = 29:
1 / (2pi x 29) = 0.00549cm/min

Hope that this is clear enough. Did you get this answer too?

Answer 5

Diameter,D=29cm
Area of snow ball A=pi*D^2/4

dA/dt=1
d(pi*D^2/4)/dt=1
d(D^2)/dt=4/pi
2DdD/dt=4/pi
dD/dt=2/(pi*D)=2/(3.1416*29)
=0.021952355cm/min(ans)

Answer 6

A = 4πr² = 4π(d/2)² = πd², and
dA/dt = dA/dd • dd/dt
2 = 2πd • dd/dt
1/π = 29 • dd/dt

so dd/dt = 1/(29π) cm/min

Answer 7

what r u talkin about?

Answer 8

-0.005488101