Question

related rates prob. cylinder fills us with water at a rate of 314ft^3 per min. the radius is 10 ft. how fast is the depth of oil changing (dh/dt)?

Answer 1

fills up**

Answer 2

so i guess the volume of the cylinder is
\[100\pi h\] right?

Answer 3

that is
\[V=100\pi h\] and so
\[V'=100\pi h'\] you are told that
\[V'=314\] so solve
\[314 = 100 \pi h'\] for
\[h'\]

Answer 4

the volume is not 314, its increasing at a rate of 314 ft^3 per min. which is dv/dt=314

Answer 5

right. it is
\[V'=314\]

Answer 6

the derivative is 314 that is what you need. the volume is changing

Answer 7

right, but the way you explained above is finding the height when the volume is changing that fast. i am looking for "how fast" the depth of oil is changing with reference to volume

Answer 8

in other words, im looking for dh/dt

Answer 9

yes, that is what i found.

Answer 10

maybe my notation is confusing. i wrote
\[V= 100 \pi h\] and then
\[V'= 100\pi h'\]

Answer 11

you can write
\[V= 100 \pi h\] and so
\[\frac{dV}{dt}=100 \pi\frac{dh}{dt}\]

Answer 12

you are given the rate of change of the volume, and asked for the rate of change of the height. i wrote the volume as a function of the height, then differentiated both sides wrt t

Answer 13

ohhh, okay. thanks a lot. got it

Answer 14

so your answer should be
\[h'=\frac{314}{100 \pi}\] or if you prefer
\[\frac{dh}{dt}=\frac{314}{100 \pi}\]

Answer 15

yw (find it easier to write y' instead of dy/dt, but i realize that the other is most frequently used in related rate problems)