Question
upstudy study bank question image url

A conical tank has height 9 m and radius 3 m at the base. Water flows at a rate of \( 2 \mathrm{~m}^{3} / \mathrm{min} \). How fast is the water level rising when the level is 1 m and 2 m ? (Use symbolic notation and fractions where needed.) When the water level is 1 m , the water level is rising at a rate of When the water level is 2 m , the water level is rising at a rate of

Ask by Greene Wells. in the United States
Mar 15,2025

Upstudy AI Solution

Tutor-Verified Answer

Answer

When the water level is 1 m, the water level is rising at a rate of \( \frac{18}{\pi} \, \mathrm{m/min} \). When the water level is 2 m, the water level is rising at a rate of \( \frac{9}{2\pi} \, \mathrm{m/min} \).

Solution

Sign in to Unlock Answers for Free!

A Learning Platform Trusted by Millions of Real Students and Teachers.

star-icon Unlock

Answered by UpStudy AI and reviewed by a Professional Tutor

Extra Insights

To solve the problem, we need to relate the volume of the cone to the height of the water. The volume \( V \) of a cone is given by the formula: \[ V = \frac{1}{3} \pi r^2 h \] Since the tank's radius and height maintain a specific ratio due to the conical shape, we can express the radius \( r \) in terms of the height \( h \). Given the dimensions of the tank, we have the proportion: \[ \frac{r}{h} = \frac{3}{9} \quad \Rightarrow \quad r = \frac{h}{3} \] Substituting \( r \) into the volume formula gives: \[ V = \frac{1}{3} \pi \left(\frac{h}{3}\right)^2 h = \frac{1}{3} \pi \frac{h^2}{9} h = \frac{\pi}{27} h^3 \] Now, we can differentiate the volume with respect to time \( t \): \[ \frac{dV}{dt} = \frac{\pi}{27} \cdot 3h^2 \frac{dh}{dt} = \frac{\pi}{9} h^2 \frac{dh}{dt} \] Given that \( \frac{dV}{dt} = 2 \, \mathrm{m}^3/\mathrm{min} \), we need to find \( \frac{dh}{dt} \) when \( h = 1 \) m and \( h = 2 \) m. 1. For \( h = 1 \) m: \[ 2 = \frac{\pi}{9} (1)^2 \frac{dh}{dt} \] Solving for \( \frac{dh}{dt} \): \[ 2 = \frac{\pi}{9} \frac{dh}{dt} \quad \Rightarrow \quad \frac{dh}{dt} = \frac{2 \cdot 9}{\pi} = \frac{18}{\pi} \, \mathrm{m/min} \] 2. For \( h = 2 \) m: \[ 2 = \frac{\pi}{9} (2)^2 \frac{dh}{dt} \] Solving for \( \frac{dh}{dt} \): \[ 2 = \frac{\pi}{9} \cdot 4 \frac{dh}{dt} \quad \Rightarrow \quad \frac{dh}{dt} = \frac{2 \cdot 9}{4\pi} = \frac{18}{4\pi} = \frac{9}{2\pi} \, \mathrm{m/min} \] So, summarizing the findings: When the water level is 1 m, the water level is rising at a rate of \( \frac{18}{\pi} \, \mathrm{m/min} \). When the water level is 2 m, the water level is rising at a rate of \( \frac{9}{2\pi} \, \mathrm{m/min} \).

Related Questions

Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy