The surface area of a melting snowball decreases at a rate of
3.4cm^2/min. Find the rate at which its diameter decreases when the diameter is 24cm . (Round your answer to three decimal places if required)
cm/min
SA = 4π r^2
dSA/dt = 8π r dr/dt
when diameter = 24, r = 12
so just plug in your values.
Could you explain more? Your answer is very confusing to me.
SA = 4π r^2 <---------- You must know that formula for the surface area
dSA/dt = 8π r dr/dt <----- I assumed you take Calculus
when diameter = 24, r = 12 <---- The question gave us that
so just plug in your values. <----- You should at least do some of the work.
at the end, double dr/dt, since the diameter d = 2r
or, start with A = πd^2
dA/dt = 2πd dd/dt
To find the rate at which the diameter of the snowball decreases, we first need to relate the surface area and the diameter.
The surface area of a sphere can be expressed as:
A = 4πr^2
where A is the surface area and r is the radius.
Since the diameter (d) of the snowball is twice the radius, we can express the surface area in terms of the diameter:
A = 4π (d/2)^2
A = πd^2/4
Now, we differentiate both sides of this equation with respect to time (t) to find the rate of change of the surface area with respect to time:
dA/dt = (π/4) * 2d * dd/dt
dA/dt = (π/2) * d * dd/dt
We know that the rate at which the surface area is decreasing is given as 3.4 cm^2/min. So, we can substitute this value into the above expression:
3.4 = (π/2) * 24 * dd/dt
Now, we can solve this equation for dd/dt, the rate at which the diameter is decreasing when the diameter is 24 cm:
dd/dt = (3.4 * 2) / (π * 24)
dd/dt = 0.283 (rounded to three decimal places)
Therefore, the rate at which the diameter is decreasing when the diameter is 24 cm is approximately 0.283 cm/min.