As of October, 2016 the world`s largest pumpkin weighed 1190 kilograms. In order to grow a

sjeikdom0

sjeikdom0

Answered question

2021-08-16

As of October, 2016 the world`s largest pumpkin weighed 1190 kilograms. In order to grow a record-setting pumpkin, the fruits mass must increase by an average of 1.27ft3/day. If we assume that the dinsity of raw pumpkin is approximately 0.50g/cm3 we can convert this to an increase in volume of 1.27ft3/day. Question: If the volume of a spherical pumpkinn is increasing at 127ft3/day, then how fast is the surface area of the pumpkin changing when the diameter of the pumpkin is b ft?
Give your answer with two decimal places, and ensure that you include unots.
Note: Equations for the volume and surface area of a sphere are on D2L in several places.

Answer & Explanation

Sadie Eaton

Sadie Eaton

Skilled2021-08-17Added 104 answers

Let V be the volume an S be the surface area of the pumpkin at any time t. Let r be its radius.
Volume is increasing at 1.27ft3/day
dvdt=1.27
Rate of change of surface area is to be calculated
dsdt=? when diameter is b
By formula for volume of sphere,
V=43πR3
dvdt=43π3R2drdt
dvdt=4πr2drdt
1.27=4πr2drdt|:dvdt=1.27
drdt=1.274πr2(1)
By formula for surface area of sphere,
S=4πr2
dsdt=(4πr2)(drdt)
dsdt=8πr(1.274πr2)
dsdt=2.54r
=12x=01x[(1+x1)2(x2+11)2]dx
=12x=01x[xx4]dx
=12x=01(x2x5)dx
dsdt=2.54b2
dsdt=5.08b
Thus,surface area is increasing at 5.08bft3/day when diameter is b units.

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