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Math Help - 2 Related Rates Problems!!

  1. #1
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    Unhappy 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.

    Thank you in advance!
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  2. #2
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    Re: 2 Related Rates Problems!!

    Quote Originally Posted by illicitkush View Post
    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 \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.
    ...
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    Re: 2 Related Rates Problems!!

    Quote Originally Posted by skeeter View Post
    ...
    I can't get the equation for the 2nd problem. How do I create the equation?
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  4. #4
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    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?
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    Re: 2 Related Rates Problems!!

    Well sin? I just don't know how to build the function itself.
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    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:

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

    Now, differentiate with respect to t. You are given \frac{dx}{dt}. Then solve for \frac{d\theta}{dt}, and now use the Pythagorean theorem to find the hypotenuse in terms of x. Once you have done this, then use x=4t, and plug in the given value of t.
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    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?
    Last edited by illicitkush; October 17th 2012 at 09:29 PM.
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  8. #8
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    Re: 2 Related Rates Problems!!

    On the left side you should get:

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

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

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

    and so you have:

    \sec^2(\theta)\cdot\frac{d\theta}{dt}=\frac{1}{200  }\cdot\frac{dx}{dt}
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    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)
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  10. #10
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    Re: 2 Related Rates Problems!!

    I would use \frac{1}{\sec^2(\theta)}=\cos^2(\theta)

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

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

    So now we have:

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

    Now, as you observed, use \frac{dx}{dt}=4 and let x=4t to get...
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    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?
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  12. #12
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    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.
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    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.
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    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.
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  15. #15
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    Re: 2 Related Rates Problems!!

    Using the formula for the volume of a cone, we have:

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

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

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

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

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

    Now, differentiating with respect to time t, we find:

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

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

    \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}}
    Thanks from illicitkush
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