2 Related Rates Problems!!

• Oct 17th 2012, 04:08 PM
illicitkush
2 Related Rates Problems!!
This 2 questions on my review are killing me and I don't know what to do. Please any help on how to solve these two? Instructions would be helpful.

1. Water is drained out of a tank in the shape of an inverted circular cone with height 9 feet and base radius 6 feet. The water level is decreasing at 1 ft/min. At what rate is the volume decreasing when the water is 6 feet deep?

2. An observer stands 200 meters from the launch site of a hot-air balloon. The balloon rises vertically at a constant rate of 4 m/s. At what rate is the angle between the ground and the observers line of sight to the balloon changing 25 s after the launch pad.

• Oct 17th 2012, 05:40 PM
skeeter
Re: 2 Related Rates Problems!!
Quote:

Originally Posted by illicitkush
This 2 questions on my review are killing me and I don't know what to do. Please any help on how to solve these two? Instructions would be helpful.

1. Water is drained out of a tank in the shape of an inverted circular cone with height 9 feet and base radius 6 feet. The water level is decreasing at 1 ft/min. At what rate is the volume decreasing when the water is 6 feet deep?

note that r/h = 6/9 ... manipulate the volume formula for a cone to get V strictly in terms of h, then take the time derivative, sub in your given values and determine dV/dt.

2. An observer stands 200 meters from the launch site of a hot-air balloon. The balloon rises vertically at a constant rate of 4 m/s. At what rate is the angle between the ground and the observers line of sight to the balloon changing 25 s after the launch pad.

right triangle trig ... balloon height is the opposite side, 200 m is the adjacent side. Assign a variable for the height and set up a trig equation in terms of $\displaystyle \theta$ and the height. Take the time derivative, sub in your given/calculated values and determine the angle's rate of change w/r to time.

...
• Oct 17th 2012, 07:14 PM
illicitkush
Re: 2 Related Rates Problems!!
Quote:

Originally Posted by skeeter
...

I can't get the equation for the 2nd problem. How do I create the equation?
• Oct 17th 2012, 07:35 PM
MarkFL
Re: 2 Related Rates Problems!!
You know the adjacent and opposite sides of the triangle...which trig. function relates the angle to these two sides?
• Oct 17th 2012, 07:40 PM
illicitkush
Re: 2 Related Rates Problems!!
Well sin? I just don't know how to build the function itself.
• Oct 17th 2012, 08:08 PM
MarkFL
Re: 2 Related Rates Problems!!
No, it is the tangent function which is opposite over adjacent (although you can use the Pythagorean theorem to use any trig function you want, I always try to use the given values, rather than compute other values to use if I can help it). Let x be the height of the balloon in meters at time t, and so you may state:

$\displaystyle \tan(\theta)=\frac{x}{200}$

Now, differentiate with respect to $\displaystyle t$. You are given $\displaystyle \frac{dx}{dt}$. Then solve for $\displaystyle \frac{d\theta}{dt}$, and now use the Pythagorean theorem to find the hypotenuse in terms of $\displaystyle x$. Once you have done this, then use $\displaystyle x=4t$, and plug in the given value of $\displaystyle t$.
• Oct 17th 2012, 08:24 PM
illicitkush
Re: 2 Related Rates Problems!!
When i differentiate both sides i get: Sec^2(dtheta/dt)=(1/200)*(dx/dt). I think I differentiated wrong?
• Oct 17th 2012, 08:32 PM
MarkFL
Re: 2 Related Rates Problems!!
On the left side you should get:

$\displaystyle \sec^2(\theta)\cdot\frac{d\theta}{dt}$

and on the right, you simply have a constant times $\displaystyle x$...think of it as $\displaystyle \frac{1}{200}\cdot x$ and so when you differentiate, the right side will be:

$\displaystyle \frac{1}{200}\cdot\frac{dx}{dt}$

and so you have:

$\displaystyle \sec^2(\theta)\cdot\frac{d\theta}{dt}=\frac{1}{200 }\cdot\frac{dx}{dt}$
• Oct 17th 2012, 08:36 PM
illicitkush
Re: 2 Related Rates Problems!!
Okay thats exactly what i got. Now You said solve for dtheta, so i divided the sec on both sides and got.
dtheta=((1/sec^2)*(1/200)*(dx/dt). How do I solve this? Idk how to input (1/sec^2) into my calculator.
I know (dx/dt=4 m/s)
• Oct 17th 2012, 08:47 PM
MarkFL
Re: 2 Related Rates Problems!!
I would use $\displaystyle \frac{1}{\sec^2(\theta)}=\cos^2(\theta)$

Now, cosine is adjacent over hypotenuse, and so by Pythagoras, we may state:

$\displaystyle \cos^2(\theta)=\left(\frac{200}{\sqrt{200^2+x^2}} \right)^2=\frac{200^2}{200^2+x^2}$

So now we have:

$\displaystyle \frac{d\theta}{dt}=\frac{200^2}{200^2+x^2}\cdot \frac{1}{200}\cdot\frac{dx}{dt}$

Now, as you observed, use $\displaystyle \frac{dx}{dt}=4$ and let $\displaystyle x=4t$ to get...
• Oct 17th 2012, 09:00 PM
illicitkush
Re: 2 Related Rates Problems!!
Okay. Phew. So for the x=4t, I got 100? and after plugging in 100 into x and solving for dtheta, I got .016? It doesn't seem right to me? Anything I missed?
• Oct 17th 2012, 09:10 PM
MarkFL
Re: 2 Related Rates Problems!!
I get the same value, which means the angle is increasing by 2/125 radians per second at that moment. This is about 0.92 deg./second.
• Oct 17th 2012, 09:13 PM
illicitkush
Re: 2 Related Rates Problems!!
And with the first problem. I know how to do the problem when the rate of the volume is increasing, but I don't know how to solve it when its decreasing. Can you help me on the first one please. I have a test tomorrow and need to know it please please.
• Oct 17th 2012, 09:22 PM
illicitkush
Re: 2 Related Rates Problems!!
When I use the equation, V=1/3*pi*r^2*h. I know to solve for r/h=6/9 which skeeter stated about. then i get r=2/3h.
New equation becomes: V=1/3*pi*(1/3h)^2*h. Now I dont know what to do.
• Oct 17th 2012, 09:31 PM
MarkFL
Re: 2 Related Rates Problems!!
Using the formula for the volume of a cone, we have:

$\displaystyle V=\frac{1}{3}\pi r^2h$

For any depth of water $\displaystyle h$ and radius of the circular surface, their ratio will remain constant, i.e.,

$\displaystyle \frac{r}{h}=\frac{6}{9}=\frac{2}{3}$

Hence $\displaystyle r=\frac{2h}{3}$ and so, substituting for $\displaystyle r$ into the volume, we find:

$\displaystyle V=\frac{1}{3}\pi \left(\frac{2h}{3} \right)^2h=\frac{4\pi}{27}h^3$

Now, differentiating with respect to time $\displaystyle t$, we find:

$\displaystyle \frac{dV}{dt}=\frac{4\pi}{9}h^2\cdot\frac{dh}{dt}$

We are told $\displaystyle \frac{dh}{dt}=-1\frac{\text{ft}}{\text{min}}$ and so to find the change in volume in cubic feet per minute when $\displaystyle h=6\text{ ft}$, we may write:

$\displaystyle \frac{dV}{dt}=\frac{4\pi}{9}(6\text{ ft})^2\cdot\left(-1\frac{\text{ft}}{\text{min}} \right)=-16\pi\frac{\text{ft}^3}{\text{min}}$