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Thread: Differential calculus :Related rates

  1. #1
    Member Vinod's Avatar
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    Question Differential calculus :Related rates

    An upright cylindrical tank full of water is tipped over at a constant (angular) speed.The height of the tank is at least twice its radius. How to prove that at the instant the tank has been tipped $45^{\circ}$, water is leaving the tank twice as fast as it did at the instant the tank was first tipped.
    HINT:-Think of how the water looks inside the tank as it is being tipped
    Last edited by Vinod; Jun 13th 2017 at 09:15 AM.
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  2. #2
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    Re: Differential calculus :Related rates

    The surface of the water is always parallel to the ground. That means that, if the tank makes angle \theta to the ground, the surface of the water makes angle \theta to the side of the tank. ("Alternate interior angles" in parallel lines.)

    You can use that to calculate the volume between the top of the water and the top of the tank. The volume of water left in the tank is the volume of the tank minus that volume. And, of course, the rate at which the water is coming out is the derivative of the that function.
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    Member Vinod's Avatar
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    Re: Differential calculus :Related rates

    Hello,
    Would you draw a picture of exact physical situation of this problem? Because i didn't understand some part of your answer. The volume of right circular cylinder is $\pi r^2$h.So the volume of the water and volume of the tank will be same.
    Last edited by Vinod; Jun 14th 2017 at 08:53 AM.
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    Re: Differential calculus :Related rates

    Quote Originally Posted by Vinod View Post
    Hello,
    Would you draw a picture of exact physical situation of this problem? Because i didn't understand some part of your answer. The volume of right circular cylinder is $\pi r^2$h.So the volume of the water and volume of the tank will be same.
    As the tank is being tilted, it is no longer a right circular cylinder of water. Now, the bottom is a right circular cylinder, but the top is slanted. It is like an angular slice to take the top of the cylinder off. The top surface would look like an oval, technically. That would be the case until the water drained to the point where the bottom of the tank was no longer completely covered with water.
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    Re: Differential calculus :Related rates

    volume left in the tank as a function of $\theta$ ...

    $V = \pi r^2 h - \pi r^3\tan{\theta}$

    $\dfrac{dV}{dt} = -\pi r^3 \cdot \sec^2{\theta} \cdot \dfrac{d\theta}{dt}$

    $\theta$ increases at a constant rate $\implies \dfrac{d\theta}{dt} = k$

    $\dfrac{dV}{dt} = -k \pi r^3 \cdot \sec^2{\theta}$

    $\dfrac{dV}{dt}\bigg|_{\theta = 0} = -k \pi r^3 \sec^2(0) = -k \pi r^3 \cdot (1)$

    $\dfrac{dV}{dt}\bigg|_{\theta = \frac{\pi}{4}} = -k \pi r^3 \sec^2\left(\dfrac{\pi}{4}\right) = -k \pi r^3 \cdot (2)$
    Attached Thumbnails Attached Thumbnails Differential calculus :Related rates-cylinder_rr.png  
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  6. #6
    Member Vinod's Avatar
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    Re: Differential calculus :Related rates

    Hello skeeter, how did you arrive at $\pi r^3 tan\theta$. I understood all of your other steps in your calculation.
    Last edited by Vinod; Jun 16th 2017 at 09:29 AM.
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    Re: Differential calculus :Related rates

    Quote Originally Posted by Vinod View Post
    Hello skeeter, how did you arrive at $\pi r^3 tan\theta$. I understood all of your other steps in your calculation.
    Consider the top of the tank to be a cylinder where the lowest point of where the water hits the tank is the bottom and the top of the tank is the top. The base is a cylinder of radius $r$ and height (as shown in the diagram) $2r\tan \theta$. So, the volume of the full cylinder is $2\pi r^3 \tan \theta$. But, half of the cylinder is empty (the diagonal line that skeeter drew cuts the cylinder exactly in half). So, the volume of the filled portion is $\pi r^3 \tan \theta$.
    Thanks from skeeter, Vinod and HallsofIvy
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  8. #8
    Member Vinod's Avatar
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    Re: Differential calculus :Related rates

    Hello SlipEternal, The tank has been tipped $45^\circ$,that's why you are saying half of the cylinder is empty or for any other reason? I think if the tank has been tipped $30^\circ$, the diagonal line won't cut the height of the cylinder exactly in half. Is my reasoning right?
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  9. #9
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    Re: Differential calculus :Related rates

    referencing the diagram ... tilted at any angle $\theta$ such that $2r\tan{\theta} \le H$, where $H$ is the height of the entire cylindrical tank, the upper cylinder formed between the red dashed line and the top of the tank is half full (or half empty)

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  10. #10
    Member Vinod's Avatar
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    Re: Differential calculus :Related rates

    Hello,
    If the height of the cylinder is at least thrice of its radius, then height of the cylinder would be $2rtan\theta$.In that case also, would the upper cylinder formed between the red dash line and the top of the tank would be half full[half empty]?
    Last edited by Vinod; Jun 17th 2017 at 08:19 AM.
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  11. #11
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    Re: Differential calculus :Related rates

    Read post #9 again ...

    Do you understand what it says?
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