# Related rates help :/

• joe32725

## Homework Statement

2. Coffee is draining from a conical filter into a cylindrical coffee pot at the rate of 10 in3/min.

a) How fast is the level in the pot rising? ____________

b) How fast is the level in the cone falling when the level in the cone is 5 in.? _________

## Homework Equations

i know i need to use the volume formulas for a cylinder and a cone, but i don't know how to differentiate them. here's the wrong answers i got:

V= pi(r^2)h --> d/dt[V]= pi(2r)(dh/dt)--> dV/dt= 2(pi)(r)*dh/dt
dh/dt= (1/(2(pi)(r)))*dV/dt

dV/dt= 2/3(pi)(r)(dh/dt)
3/(2(pi)(r))*(dV/dt)

## The Attempt at a Solution

when i differentiate both sides with respect to time i get:

d/dt[V]= 2/3(Pi)(r)*(dh/dt) ---> dV/dt= 2/3(pi)(r)(dh/dt)

dh/dt= 3/(2(pi)(r))*(dV/dt)

dh/dt= 3/(2(pi)(r))*(10in^3/min)

30/2(pi)r = 15/(pi)(r) in^3/min

p.s. this is really getting frustrating. if calc 1 is throwing me curveballs like this that i can't solve, my quest to be a mechanical engineer seams very bleak.

## Homework Statement

2. Coffee is draining from a conical filter into a cylindrical coffee pot at the rate of 10 in3/min.

a) How fast is the level in the pot rising? ____________

b) How fast is the level in the cone falling when the level in the cone is 5 in.? _________

## Homework Equations

i know i need to use the volume formulas for a cylinder and a cone, but i don't know how to differentiate them. here's the wrong answers i got:

V= pi(r^2)h --> d/dt[V]= pi(2r)(dh/dt)--> dV/dt= 2(pi)(r)*dh/dt
dh/dt= (1/(2(pi)(r)))*dV/dt
Okay, that's how fast the height of the coffee in the pot is increasing.

dV/dt= 2/3(pi)(r)(dh/dt)
3/(2(pi)(r))*(dV/dt)
?? $V= (1/3)\pi r^2 h$. You seem to have differentiated r2 with respect to r but then multiplied by dh/dt rather than dr/dt. Using the product rule
[tex]\frac{dV}{dt}= (2/3)\pi r h \frac{dr}{dt}+ (1/3)\pi r^2 \frac{dh}{dt}[/itex]

## The Attempt at a Solution

when i differentiate both sides with respect to time i get:

d/dt[V]= 2/3(Pi)(r)*(dh/dt) ---> dV/dt= 2/3(pi)(r)(dh/dt)
No, as I said above, you you have to use the product rule. Remember that the cone formed by the water always has the ratio of r to h that the filter has. Are you given the height and radius of the filter cone in the problem?

dh/dt= 3/(2(pi)(r))*(dV/dt)

dh/dt= 3/(2(pi)(r))*(10in^3/min)

30/2(pi)r = 15/(pi)(r) in^3/min

p.s. this is really getting frustrating. if calc 1 is throwing me curveballs like this that i can't solve, my quest to be a mechanical engineer seams very bleak.

You need to know the ratio of radius to height of the original filter cone and you haven't given that here.

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