NO CALCULATOR IS ALLOWED FOR THIS QUESTION

AP Calculus AB Test Booklet Unit Progress Check: FRQ Part B Name NO CALCULATOR IS ALLOWED FOR THIS QUESTION: Show all of your work even though the question may not explicitly remind you to do so. Clearly label any functions, graphs, tables, or other objects that you use. Justifications require that you give mathematical reasons and that you verify the needed conditions under which relevant theorems, properties, definitions, or tests are applied. Your work will be scored on the correctness and completeness of your methods as well as your answers. Answers without supporting work will usually not receive credit. Unless otherwise specified, answers (numeric or algebraic) need not be simplified. If your answer is given as a decimal approximation, it should be correct to three places after the decimal point. Unless otherwise specified, the domain of function f is assumed to be the set of all real numbers for which f(x) is a real number. W(t) = {cos(t) for 0 < t < 6 W(t) = 10 – (t – 6)^2 for 6 < t < 10 The depth of a river at a certain point is modeled by the function W defined above, where W(t) is measured in feet and time is measured in hours. Find W'(8). Using correct units, explain the meaning of W'(8) in the context of the problem. Please respond on separate paper, following directions from your teacher. The graph of W is concave down for 3 < t < 3.5. Use the line tangent to the graph of W at t = 3 to show that W(3.5) < 9. Please respond on separate paper, following directions from your teacher. W(t) = √(t+1). Find lim (t→2) W(t).

The Correct Answer and Explanation is:

1. Find W′(8)W'(8)W′(8). Then, using correct units, explain the meaning of W′(8)W'(8)W′(8) in context.

For 6<t<106 < t < 106<t<10, we are told that W(t)=10−(t−6)2W(t) = 10 – (t – 6)^2W(t)=10−(t−6)2

To find W′(t)W'(t)W′(t), differentiate using the chain rule: W′(t)=ddt[10−(t−6)2]=0−2(t−6)=−2(t−6)W'(t) = \frac{d}{dt} \left[10 – (t – 6)^2 \right] = 0 – 2(t – 6) = -2(t – 6)W′(t)=dtd​[10−(t−6)2]=0−2(t−6)=−2(t−6)

Now substitute t=8t = 8t=8 into this derivative: W′(8)=−2(8−6)=−2(2)=−4W'(8) = -2(8 – 6) = -2(2) = -4W′(8)=−2(8−6)=−2(2)=−4

Answer: W′(8)=−4W'(8) = -4W′(8)=−4

Interpretation in context:
The derivative W′(8)W'(8)W′(8) represents the rate of change of the river’s depth at time t=8t = 8t=8 hours. Since W(t)W(t)W(t) is measured in feet and ttt is in hours, the units of W′(8)W'(8)W′(8) are feet per hour. A value of −4-4−4 means that the river’s depth is decreasing at a rate of 4 feet per hour at 8 hours after the start of measurement.

2. Use the line tangent to the graph of WWW at t=3t = 3t=3 to show that W(3.5)<9W(3.5) < 9W(3.5)<9, given that the graph is concave down for 3<t<3.53 < t < 3.53<t<3.5.

Since the graph is concave down in the interval 3<t<3.53 < t < 3.53<t<3.5, any tangent line in that interval lies above the graph of the function. This means the function value at t=3.5t = 3.5t=3.5 is less than the value predicted by the tangent line at t=3t = 3t=3.

We are also told that: W(t)=cos⁡(t) for 0<t<6W(t) = \cos(t) \text{ for } 0 < t < 6W(t)=cos(t) for 0<t<6

So: W(3)=cos⁡(3),W′(t)=−sin⁡(t)⇒W′(3)=−sin⁡(3)W(3) = \cos(3), \quad W'(t) = -\sin(t) \Rightarrow W'(3) = -\sin(3)W(3)=cos(3),W′(t)=−sin(t)⇒W′(3)=−sin(3)

Using the equation of the tangent line at t=3t = 3t=3: L(t)=W(3)+W′(3)(t−3)=cos⁡(3)−sin⁡(3)(t−3)L(t) = W(3) + W'(3)(t – 3) = \cos(3) – \sin(3)(t – 3)L(t)=W(3)+W′(3)(t−3)=cos(3)−sin(3)(t−3)

Now substitute t=3.5t = 3.5t=3.5: L(3.5)=cos⁡(3)−sin⁡(3)(0.5)L(3.5) = \cos(3) – \sin(3)(0.5)L(3.5)=cos(3)−sin(3)(0.5)

Use calculator approximations: cos⁡(3)≈−0.989,sin⁡(3)≈0.141\cos(3) \approx -0.989, \quad \sin(3) \approx 0.141cos(3)≈−0.989,sin(3)≈0.141 L(3.5)≈−0.989−(0.141)(0.5)=−0.989−0.0705=−1.0595L(3.5) \approx -0.989 – (0.141)(0.5) = -0.989 – 0.0705 = -1.0595L(3.5)≈−0.989−(0.141)(0.5)=−0.989−0.0705=−1.0595

So: W(3.5)<L(3.5)≈−1.0595W(3.5) < L(3.5) \approx -1.0595W(3.5)<L(3.5)≈−1.0595

This shows that W(3.5)<9W(3.5) < 9W(3.5)<9, because the tangent line gives an estimate lower than 9, and the function lies below the tangent due to concavity.

3. Find lim⁡t→2t+1\lim_{t \to 2} \sqrt{t + 1}limt→2​t+1​

We are asked to evaluate: lim⁡t→2t+1\lim_{t \to 2} \sqrt{t + 1}t→2lim​t+1​

This is a direct substitution limit. Since the expression t+1\sqrt{t + 1}t+1​ is continuous at t=2t = 2t=2, we can plug in the value directly: 2+1=3\sqrt{2 + 1} = \sqrt{3}2+1​=3​

Answer: lim⁡t→2t+1=3\lim_{t \to 2} \sqrt{t + 1} = \sqrt{3}limt→2​t+1​=3​

The value of this limit is approximately 1.732, but unless a decimal is required, it is acceptable to leave it as 3\sqrt{3}3​.

Summary

• W′(8)=−4W'(8) = -4W′(8)=−4, meaning the river depth is decreasing at 4 feet per hour at t=8t = 8t=8.
• W(3.5)<9W(3.5) < 9W(3.5)<9 is justified using the concave-down property and the tangent line at t=3t = 3t=3.
• lim⁡t→2t+1=3\lim_{t \to 2} \sqrt{t + 1} = \sqrt{3}limt→2​t+1​=3​, a direct substitution due to continuity.