Word problem involving differentiation..

• Dec 1st 2008, 10:17 AM
struck
Word problem involving differentiation..
A water tank has a rectangular base 1.5m by 1.2m. The sides are vertical and water is being added to the tank at a constant rate of 0.45m^3. At what rate is the depth of water in the tank increasing?

So far, I can say that dV/dt = 0.45, i.e. the rate of change in volume with respect to time is 0.45m^3 / minute.

I do know that the volume would be 1.5 x 1.2 x height.

But I cannot put things together..! (Headbang)

p.s. I have to use the chain rule.
• Dec 1st 2008, 10:21 AM
Moo
Hello !
Quote:

Originally Posted by struck
A water tank has a rectangular base 1.5m by 1.2m. The sides are vertical and water is being added to the tank at a constant rate of 0.45m^3. At what rate is the depth of water in the tank increasing?

So far, I can say that dV/dt = 0.45, i.e. the rate of change in volume with respect to time is 0.45m^3 / minute.

I do know that the volume would be 1.5 x 1.2 x height.

But I cannot put things together..! (Headbang)

p.s. I have to use the chain rule.

Let $h(t)$ denote the depth of the water at time t.

Like you said, the volume would be : $V(t)=1.5 \times 1.2 \times \text{height}=1.8 \times h(t)$

Now if you differentiate both sides, not by using the chain rule but by using : $(a u(x))'=au'(x)$, where a is a constant, it'll give :

$\frac{dV}{dt}(t)=1.8 h'(t)$

$0.45=1.8 h'(t)$

and h'(t) is the rate of change of the depth !
• Dec 1st 2008, 10:23 AM
andreas
Quote:

Originally Posted by struck
A water tank has a rectangular base 1.5m by 1.2m. The sides are vertical and water is being added to the tank at a constant rate of 0.45m^3. At what rate is the depth of water in the tank increasing?

So far, I can say that dV/dt = 0.45, i.e. the rate of change in volume with respect to time is 0.45m^3 / minute.

I do know that the volume would be 1.5 x 1.2 x height.

But I cannot put things together..! (Headbang)

p.s. I have to use the chain rule.

What is $0.45m^3$/minute it is the volume that gets in one minute inside the tank, right? By how much would this volume increase the depth ? Volume/(Area of the base) or $0.45/(1.5*1.2)$ meters/minute