Questions: Wed Jan 1 74 CALCULUS AB Section II Part B: No calculator is allowed for these problems. t 0 2 5 9 13 15 --------------------- h'(t) 0.4 0.5 0.7 1.0 1.1 1.3 3. A tank with a rectangular base measuring 10 inches by 20 inches is being filled with water at a variable rate. The depth of the water in the tank is given by a twice-differentiable function h of time, t, measured in minutes. The table above gives the rate of change, h'(t), of the depth of the water in the tank for selected values of t over the time interval 0 ≤ t ≤ 15. During this interval h''(t)>0. When t=5 minutes, the depth of the water is 6 inches. (Note: The volume of the water in the tank is given by V=lwh.) (a) Approximate the depth of the water in the tank at t=4 minutes using the tangent line approximation at t=5. Is your estimate greater than or less than the true value? Give a reason for your answer. (b) Find the rate of change of the volume of the water in the tank at t=2 minutes. Indicate units of measure. (c) Use a left Riemann sum with five subintervals indicated by the table to approximate ∫ from 0 to 15 of h'(t) dt. Using correct units, explain the meaning of ∫ from 0 to 15 of h'(t) dt in terms of the depth of the water in the tank. (d) Is the approximation in part (c) greater than or less than ∫ from 0 to 15 of h'(t) dt? Give a reason for your answer.

Wed Jan 1 74 CALCULUS AB Section II Part B: No calculator is allowed for these problems. t 0 2 5 9 13 15 --------------------- h'(t) 0.4 0.5 0.7 1.0 1.1 1.3 3. A tank with a rectangular base measuring 10 inches by 20 inches is being filled with water at a variable rate. The depth of the water in the tank is given by a twice-differentiable function h of time, t, measured in minutes. The table above gives the rate of change, h'(t), of the depth of the water in the tank for selected values of t over the time interval 0 ≤ t ≤ 15. During this interval h''(t)>0. When t=5 minutes, the depth of the water is 6 inches. (Note: The volume of the water in the tank is given by V=lwh.) (a) Approximate the depth of the water in the tank at t=4 minutes using the tangent line approximation at t=5. Is your estimate greater than or less than the true value? Give a reason for your answer. (b) Find the rate of change of the volume of the water in the tank at t=2 minutes. Indicate units of measure. (c) Use a left Riemann sum with five subintervals indicated by the table to approximate ∫ from 0 to 15 of h'(t) dt. Using correct units, explain the meaning of ∫ from 0 to 15 of h'(t) dt in terms of the depth of the water in the tank. (d) Is the approximation in part (c) greater than or less than ∫ from 0 to 15 of h'(t) dt? Give a reason for your answer.
Transcript text: Wed Jan 1 74 CALCULUS AB Section II Part B: No calculator is allowed for these problems. \begin{tabular}{|c|c|c|c|c|c|c|} \hline$t$ & 0 & 2 & 5 & 9 & 13 & 15 \\ \hline$h^{\prime}(t)$ & 0.4 & 0.5 & 0.7 & 1.0 & 1.1 & 1.3 \\ \hline \end{tabular} 3. A tank with a rectangular base measuring 10 inches by 20 inches is being filled with water at a variable rate. The depth of the water in the tank is given by a twice-differentiable function $h$ of time, $t$, measured in minutes. The table above gives the rate of change, $h^{\prime}(t)$, of the depth of the water in the tank for selected values of $t$ over the time interval $0 \leq t \leq 15$. During this interval $h^{\prime \prime}(t)>0$. When $t=5$ minutes, the depth of the water is 6 inches. (Note: The volume of the water in the tank is given by $V=l w h$.) (a) Approximate the depth of the water in the tank at $t=4$ minutes using the tangent line approximation at $t=5$. Is your estimate greater than or less than the true value? Give a reason for your answer. (b) Find the rate of change of the volume of the water in the tank at $t=2$ minutes. Indicate units of measure. (c) Use a left Riemann sum with five subintervals indicated by the table to approximate $\int_{0}^{15} h^{\prime}(t) d t$. Using correct units, explain the meaning of $\int_{0}^{15} h^{\prime}(t) d t$ in terms of the depth of the water in the tank. (d) Is the approximation in part (c) greater than or less than $\int_{0}^{15} h^{\prime}(t) d t$ ? Give a reason for your answer.
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Solution

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Solution Steps

Step 1: Approximate the Depth at \( t=4 \)

To approximate the depth of the water in the tank at \( t=4 \) minutes using the tangent line approximation at \( t=5 \), we use the formula: \[ h(4) \approx h(5) + h'(5) \cdot (4 - 5) \] Substituting the known values: \[ h(4) \approx 6 + 0.7 \cdot (-1) = 6 - 0.7 = 5.3 \] Since \( h''(t) > 0 \) indicates that \( h'(t) \) is increasing, the tangent line approximation will be less than the true value.

Step 2: Rate of Change of Volume at \( t=2 \)

The rate of change of the volume of water in the tank at \( t=2 \) minutes is calculated as: \[ \frac{dV}{dt} = A \cdot h'(2) \] where \( A = 10 \times 20 = 200 \) square inches and \( h'(2) = 0.5 \): \[ \frac{dV}{dt} = 200 \cdot 0.5 = 100.0 \text{ cubic inches per minute} \]

Step 3: Left Riemann Sum Approximation

To approximate \(\int_{0}^{15} h^{\prime}(t) \, dt\) using a left Riemann sum with the given data: \[ \int_{0}^{15} h^{\prime}(t) \, dt \approx \sum_{i=0}^{4} h'(t_i) \cdot (t_{i+1} - t_i) \] Calculating the left Riemann sum: \[ \int_{0}^{15} h^{\prime}(t) \, dt \approx (2 - 0) \cdot 0.4 + (5 - 2) \cdot 0.5 + (9 - 5) \cdot 0.7 + (13 - 9) \cdot 1.0 + (15 - 13) \cdot 1.1 \] This results in: \[ = 2 \cdot 0.4 + 3 \cdot 0.5 + 4 \cdot 0.7 + 4 \cdot 1.0 + 2 \cdot 1.1 = 0.8 + 1.5 + 2.8 + 4.0 + 2.2 = 11.3 \] Thus, the left Riemann sum approximation is \( 9.1 \) cubic inches.

Final Answer

  • Approximate depth at \( t=4 \): \( \boxed{h(4) = 5.3} \)
  • Rate of change of volume at \( t=2 \): \( \boxed{\frac{dV}{dt} = 100.0 \text{ cubic inches per minute}} \)
  • Left Riemann sum approximation: \( \boxed{\int_{0}^{15} h^{\prime}(t) \, dt \approx 9.1} \)
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