[Technology Exercise]


Draining a tank 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(1 - t/12)² m.


b. When is the fluid level in the tank falling fastest? Slowest? What are the values of dy/dt at these times?

Verified step by step guidance

1. To determine when the fluid level is falling fastest or slowest, we need to find the derivative of the depth function y with respect to time t, which is dy/dt. This will give us the rate of change of the fluid depth over time.
2. The given formula for the depth y is y = 6(1 - t/12)². First, apply the chain rule to differentiate this function with respect to t. Let u = 1 - t/12, then y = 6u². The derivative dy/dt is found by differentiating y with respect to u and then multiplying by du/dt.
3. Differentiate y = 6u² with respect to u to get dy/du = 12u. Then, differentiate u = 1 - t/12 with respect to t to get du/dt = -1/12.
4. Now, substitute back to find dy/dt: dy/dt = 12u * (-1/12) = -u. Since u = 1 - t/12, dy/dt = -(1 - t/12).
5. To find when the fluid level is falling fastest or slowest, analyze the expression for dy/dt. The fluid level falls fastest when dy/dt is at its minimum value, and slowest when dy/dt is at its maximum value. Evaluate dy/dt at the endpoints of the interval [0, 12] to determine these times.

Key Concepts

Derivative

The derivative of a function represents the rate of change of the function with respect to a variable. In this context, dy/dt indicates how the depth of the fluid in the tank changes over time. Calculating the derivative helps determine when the fluid level is falling fastest or slowest by identifying critical points where the rate of change is maximized or minimized.

Critical Points

Critical points occur where the derivative of a function is zero or undefined, indicating potential maxima, minima, or points of inflection. To find when the fluid level is falling fastest or slowest, we need to find the critical points of dy/dt. These points help identify the times when the rate of change of the fluid level is at its extreme values.

Second Derivative Test

The second derivative test helps determine the nature of critical points found using the first derivative. By evaluating the second derivative at these points, we can ascertain whether they correspond to local maxima, minima, or points of inflection. In this problem, applying the second derivative test to dy/dt will reveal when the fluid level is falling fastest (maximum rate) or slowest (minimum rate).