Related Rates Conical Water Tank find rate of change of the water depth

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Step 1: Find an equation that relates the quantities. In this question, it is the relationship between the height h and the volume v. Find an equation that gets the volume from the height.

Step 2: Take the derivative of the equation in Step 1. However, don't treat it like a normal derivative, treat it like an implicit derivative with respect to time (dt, with t in hours).

Step 3: The result of step 2 should give you a relationship between dh/dt and dv/dt. Since the setup gives you dv/dt (2m^3/hour), then you can just solve for dh/dt.

Additions below based on conversations:

Keep in mind that a cone is just a right triangle rotated around. This means that we can use many of the same relationships. So, for instance, as the water level rises, all of the triangles formed will be similar triangles. That means that the ratios between the sides will remain the same. The height is 20 when the radius is 5. That means that $r = \frac{h}{4}$. So, therefore, we can take the volume equation, and rewrite it entirely in terms of height:

$$v = \frac{\pi\cdot r^2\cdot h}{3}$$ $$r = \frac{h}{4}$$ $$v = \frac{\pi\cdot (\frac{h}{4})^2\cdot h}{3}$$ $$v = \frac{\pi\cdot h^3}{48}$$

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J2R5M3
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Updated on August 17, 2022

Comments

  • J2R5M3
    J2R5M3 less than a minute

    An inverted conical water tank with a height of $20m$ and radius of $5m$ is drained through the hole in the vertex at a rate of $2m^3 /h$. What is the rate of change of the water depth when the water is $4m$ deep?

    • J2R5M3
      J2R5M3 over 6 years
      Well I've drawn everything out and listed down what I know I have the equation for the volume of a cone and I think that the $2m^3 /h$ is the rate of change of the Volume over time ? but I'm not entirely sure the $/h$ in $2m^3 /h$ is throwing me off.
    • mysatellite
      mysatellite over 6 years
      @J2T5M3 Yes, here $h$ is most likely hours
    • J2R5M3
      J2R5M3 over 6 years
      I know that I ultimately want to get $dh/dt$.
    • mysatellite
      mysatellite over 6 years
      @J2T5M3 correct, but perhaps you should assign a different variable for the height to avoid confusing the height $h$ with the time in hours $h$
    • J2R5M3
      J2R5M3 over 6 years
      given that $h = 3V/Pir^2$
  • J2R5M3
    J2R5M3 over 6 years
    What would I sub for $r$ though?
  • J2R5M3
    J2R5M3 over 6 years
    Or $V$ for that matter?
  • johnnyb
    johnnyb over 6 years
    Keep in mind that a cone is just a right triangle rotated around. This means that we can use many of the same relationships. So, for instance, as the water level rises, all of the triangles formed will be similar triangles. That means that the ratios between the sides will remain the same. The height is 20 when the radius is 5. That means that the radius = height / 4. So, therefore, we can take the volume equation, and rewrite it entirely in terms of height: Volume = Pi r^2 h / 3; r = h / 5; therefore, Volume = Pi (h/5)^2 h / 3