2. The water volume inside a tank is changing at a rate of $$ \frac{dV}{dt} = -2.5e^{-0.5t+1} $$ where $V$ is the volume of in cubic feet (ft³) and $t$ is time measured in hours (h). The volume at $t = 2$ hours is 15 ft³. (a) How fast is the rate of change of the volume changing at the instant when $t = 2$ seconds? (b) Find $V(6)$ (c) Write you answer to part (b) in a sentence that clearly interprets the physical context of the situation.
Added by Robin J.
Close
Step 1
5e^{-0.5t+1} $$ We are given that the volume at $t=2$ hours is 15 ft³. We want to find the rate of change of the volume at $t=2$ seconds. First, we need to convert seconds to hours. Since there are 3600 seconds in an hour, 2 seconds is $\frac{2}{3600} = Show more…
Show all steps
Your feedback will help us improve your experience
Carson Merrill and 71 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
It takes 12 hours to drain a storage tank by opening the valve at the bottom. The depth $y$ of fluid in the tank $t$ hours after the valve is opened is given by the formula $$y=6\left(1-\frac{t}{12}\right)^{2} \mathrm{~m}$$ (a) Find the rate $d y / d t(\mathrm{~m} / \mathrm{h})$ at which the water level is changing at time $t$ (b) When is the fluid level in the tank falling fastest? slowest? What are the values of $d y / d t$ at these times? (c) Graph $y$ and $d y / d t$ together and discuss the behavior of $y$ in relation to the signs and values of $d y / d t$
Derivatives
Velocity and Other Rates of Change
A water tank has a rectangular base that is 2 meters wide and 5 meters long. The height in meters of water in the tank after t hours is h(t), and the volume in cubic meters of water in the tank after t hours is V(t). Water leaves the tank at a constant rate. V(2) = 40 and V'(t) = -3. (a) V(0) = (b) h(t) = (c) h'(t) = (d) How many hours until the height of water in the tank is 1 meter?
Madhur L.
It takes 12 hours to drain a storage tank by opening the valve at the bottom. The depth $y$ of fluid in the tank $t$ hours after the valve is opened is given by the formula $$y=6\left(1-\frac{t}{12}\right)^{2} \mathrm{m}$$ a. Find the rate $d y / d t(\mathrm{m} / \mathrm{h})$ at which the tank is draining at time $t$ b. When is the fluid level in the tank falling fastest? Slowest? What are the values of $d y / d t$ at these times? c. Graph $y$ and $d y / d t$ together and discuss the behavior of $y$ in relation to the signs and values of $d y / d t$
The Derivative as a Rate of Change
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD