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.
failed

Solution

failed
failed

Solution Steps

Step 1: Approximate the Depth at t=4 t=4

To approximate the depth of the water in the tank at t=4 t=4 minutes using the tangent line approximation at t=5 t=5 , we use the formula: h(4)≈h(5)+h′(5)⋅(4−5) h(4) \approx h(5) + h'(5) \cdot (4 - 5) Substituting the known values: h(4)≈6+0.7⋅(−1)=6−0.7=5.3 h(4) \approx 6 + 0.7 \cdot (-1) = 6 - 0.7 = 5.3 Since h′′(t)>0 h''(t) > 0 indicates that h′(t) 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 t=2

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

Step 3: Left Riemann Sum Approximation

To approximate ∫015h′(t) dt\int_{0}^{15} h^{\prime}(t) \, dt using a left Riemann sum with the given data: ∫015h′(t) dt≈∑i=04h′(ti)⋅(ti+1−ti) \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: ∫015h′(t) dt≈(2−0)⋅0.4+(5−2)⋅0.5+(9−5)⋅0.7+(13−9)⋅1.0+(15−13)⋅1.1 \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⋅0.4+3⋅0.5+4⋅0.7+4⋅1.0+2⋅1.1=0.8+1.5+2.8+4.0+2.2=11.3 = 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 9.1 cubic inches.

Final Answer

  • Approximate depth at t=4 t=4 : h(4)=5.3 \boxed{h(4) = 5.3}
  • Rate of change of volume at t=2 t=2 : dVdt=100.0 cubic inches per minute \boxed{\frac{dV}{dt} = 100.0 \text{ cubic inches per minute}}
  • Left Riemann sum approximation: ∫015h′(t) dt≈9.1 \boxed{\int_{0}^{15} h^{\prime}(t) \, dt \approx 9.1}
Was this solution helpful?
failed
Unhelpful
failed
Helpful

Questions asked by other users

failedAnswered
1821 Test 9 Test 1821 Test 9 Time Remaining: 0/48:54
failedAnswered
Use the metric line to determine which of the following the nurse will consider when converting 0.25 decigrams (dg) to milligrams (mg). Select all that apply. "deci" is the starting place on the metric line. The base unit is the desired place. Gram is the base unit of measurement. "deka" is the starting place on the metric line. "milli" is the desired place.
failedAnswered
Part B Which other sentence from the story serves a similar purpose? (4. "Unfortunately, Felicia had waited until the day before the assignment was due to start it." (paragraph 4) (B) "Nervous but hopeful, Felicia leaned toward the girl." (paragraph 6) (c) "Now, as the memory sketched itself across her mind, Felicia smiled." (paragraph 10) (D) "She nodded to her cell phone sitting silently in her purse on the passenger seat, and smiled, grateful for a number of reasons." (paragraph 10)
failedAnswered
Question 16 1 pts Coolers Inc. has decided to launch a new energy drink that will have the fewest calories among its competitors. To understand the market potential for the new drink and the demographics and attitudes of consumers who are likely to buy the product, Coolers should most likely use archival research descriptive research exploratory research experimental research causal research
failedAnswered
Solve this equation. -x/24 = -4