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PLEASE HELP with this related rates problem.

PLEASE HELP with this related rates problem.-example-1
User FerDensetsu
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1 Answer

23 votes
23 votes

Answer:

160/13 ≈ 12.3 units per second

Explanation:

The product of pressure and volume is proportional to temperature, and temperature is 100 when pressure and volume are 10 and 13, respectively. You want to know the rate of change of temperature when volume decreases at 1 per second and pressure increases at 2 per second.

Relation

The given proportion can be written with constant of proportionality k as ...

PV = kT

The value of k is ...

k = PV/T = (10)(13)/100 = 1.3

Solving the relation for T, we have ...

T = PV/k = PV/1.3

Rate of change

The rate of change is found by differentiating this product:

T' = (P'V +PV')/k

T' = ((+2)(13) +(10)(-1))/1.3 = (26-10)/1.3 = 16/1.3

T' = 160/13 ≈ 12.3

Temperature is increasing at about 12.3 units per second.

User Chirag Solanki
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