• Jan 17th 2008, 01:11 PM
noselord
Can anyone help me out with these questions? I'm so lost in this unit... lol

I have 3 questions left and I really need help on them.

Water runs into a tank that is in the shape of an inverted cone at the rate of 9cubic feet/min. The tank has a height of 10 ft and a base radius of 5 ft. How fast is the water level rising when the water is 4 ft deep?

The length of the base of a right triangle is increasing at the rate of 12in/min. At the same time, the height of the triangle is decreasing in such a way that the length of the hypotenuse remains 10 inches. How quickly is the height of the triangle changing when the length of the base is 6 inches?

A spherical balloon is being inflated so that its volume is increasing at the rate of 5 cubic meters/min. At what rate is the diameter increasing when the diameter is 12m? (V = 4/3(pi)r^3)

• Jan 17th 2008, 03:21 PM
Edit: my answer to the first question is wrong. See Jhevon's post for the correct method.

Quote:

Water runs into a tank that is in the shape of an inverted cone at the rate of 9cubic feet/min. The tank has a height of 10 ft and a base radius of 5 ft. How fast is the water level rising when the water is 4 ft deep?
This question is confusing because you are given more information than you need. All you have to do is follow the normal procedure and ignore some of it.

$\frac {dh}{dt} = \frac {dv}{dt} \bullet \frac{1} {\frac {dv}{dh}}$
$
\frac {dv}{dt} = 9$

$
\frac {dv}{dh} = 4^2 \pi$

etc

Quote:

The length of the base of a right triangle is increasing at the rate of 12in/min. At the same time, the height of the triangle is decreasing in such a way that the length of the hypotenuse remains 10 inches. How quickly is the height of the triangle changing when the length of the base is 6 inches?
$b^2+h^2 = 10^2$. You want $\frac {dh}{dt}$ and have $\frac {db}{dt}$. Implicit differentiation looks easiest for finding $\frac {dh}{db}$, but if you haven't done that yet you can just rearrange.

Quote:

A spherical balloon is being inflated so that its volume is increasing at the rate of 5 cubic meters/min. At what rate is the diameter increasing when the diameter is 12m? (V = 4/3(pi)r^3)
$
\frac {dD}{dt} = \frac {dD}{dr} \frac {dr}{dV} \frac {dV}{dt}
$
• Jan 17th 2008, 03:23 PM
Jhevon
Quote:

Originally Posted by noselord
Can anyone help me out with these questions? I'm so lost in this unit... lol

I have 3 questions left and I really need help on them.

Water runs into a tank that is in the shape of an inverted cone at the rate of 9cubic feet/min. The tank has a height of 10 ft and a base radius of 5 ft. How fast is the water level rising when the water is 4 ft deep?

i will do this question. the principles i use here is similar to the principles you must use in the others.

TIPS:
- when doing related rates problems, ALWAYS draw diagrams if applicable
- label anything that changes in your problem with variables. anything that stays constant, label it by its value, not a variable! in short, label what you know and don't know. and state what you want to find, to keep an objective in your head
- find an appropriate formula that relates all the knowns and unknowns
- differentiate implicitly and solve for what you want. often you have to go back to the formula in the previous step, or use some other technique to solve for any unknown besides the one the problem asked you to find

now on to your question. see the diagram below.

the radius and height of the cone are constant, so i write in their values, 5 ft and 10 ft respectively. the radius of the water in the cone, i label r. the height of the water in the cone, i label the volume of water in the cone V.

as i wrote in the diagram, we want the rate of change of the height when the height is 4 ft. we know dV/dt, the rate of change of volume, so i filled that in as well.

now, i will use the equation for the volume of a cone to relate all my variables, as this does the job

so we will work with the formula:

$V = \frac 13 \pi r^2 h$

now the problem is, i know nothing about r. the question said nothing about the radius. so i need to get rid of that. PAY ATTENTION, WHAT I AM ABOUT TO DO IS DONE VERY OFTEN WITH CONE PROBLEMS. i will use similar triangles to solve for r in terms of h. the diagram of the right triangle beside the cone, is a cross-section of one-half of the cone. by similar triangles:

$\frac {\mbox{Base of small triangle}}{\mbox{Height of small tirangle}} = \frac {\mbox{Base of large triangle}}{\mbox{Height of large triangle}} \implies \frac rh = \frac 5{10} \implies r = \frac h2$

now we can plug that expression in for r. we get:

$\Rightarrow V = \frac 1{12} \pi h^3$

now we're in business. differentiate implicitly, we get:

$\Rightarrow \frac {dV}{dt} = \frac {\pi}4h^2 ~\frac {dh}{dt}$

now, we are concerned with the instant where h = 4, and we know that dV/dt = 9. so plug in our knowns:

$\Rightarrow 9 = 4 \pi ~ \frac {dh}{dt}$

now solve for dh/dt, we get:

$\boxed{\frac {dh}{dt} = \frac 9{4 \pi}}$

now try the others using the same procedure (as outlined in "TIPS")
• Jan 17th 2008, 04:30 PM
noselord
i think i got the answers

In order

1) 9/4(pi)
2) -3/4
3) 5(pi) / 72
• Jan 17th 2008, 04:39 PM