Verbal Stem FRQ Medley AP Calc AB April 28, 2020 Adapted from AP Central (with permission) At the beginning of 2010, a landfill contained 1400 tons of solid waste. The increasing function W models the total amount of solid waste stored at the landfill. Planners estimate that W will satisfy the differential equation dW/dt = 300 for the next 20 years. W is measured in tons, and t is measured in years from the start of 2010. (a) Use the line tangent to the graph of W at t = 0 to approximate the amount of solid waste that the landfill contains at the end of the first 3 months of 2010 (time dt = 0.25). (b) Find d^2W/dt^2 in terms of W. Use it to determine whether your answer in part (a) is an underestimate or an overestimate of the amount of solid waste that the landfill contains at time t = 0.25. (c) Find the particular solution W = W(t) to the differential equation condition W(0) = 1400 and dW/dt = 300. Two particles move along the x-axis. For 0 < t < 6, the position of particle P at time t is given by p(t) = 2cos(4t), while the position of particle R at time t is given by r(t) = t^3 – 6t^2 + 9t + 3. Find.
The Correct Answer and Explanation is:
(a) Tangent Line Approximation at t=0.25t = 0.25t=0.25
We are given:
- At t=0t = 0t=0, W(0)=1400W(0) = 1400W(0)=1400
- dWdt=300\frac{dW}{dt} = 300dtdW=300 is constant
The tangent line at t=0t = 0t=0 has the form:L(t)=W(0)+dWdt⋅t=1400+300tL(t) = W(0) + \frac{dW}{dt} \cdot t = 1400 + 300tL(t)=W(0)+dtdW⋅t=1400+300t
To approximate the value at t=0.25t = 0.25t=0.25 (3 months):L(0.25)=1400+300(0.25)=1400+75=1475 tonsL(0.25) = 1400 + 300(0.25) = 1400 + 75 = \boxed{1475 \text{ tons}}L(0.25)=1400+300(0.25)=1400+75=1475 tons
(b) Determine d2Wdt2\frac{d^2W}{dt^2}dt2d2W and Accuracy of Approximation
Since dWdt=300\frac{dW}{dt} = 300dtdW=300 is a constant, differentiate again:d2Wdt2=ddt(300)=0\frac{d^2W}{dt^2} = \frac{d}{dt}(300) = 0dt2d2W=dtd(300)=0
This implies that the rate of change is not increasing or decreasing. Because the second derivative is zero, the tangent line is exact, not an overestimate or an underestimate.
So, the approximation in part (a) is exact:Neither an overestimate nor an underestimate\boxed{\text{Neither an overestimate nor an underestimate}}Neither an overestimate nor an underestimate
(c) Find the Particular Solution W(t)W(t)W(t)
We are solving:dWdt=300withW(0)=1400\frac{dW}{dt} = 300 \quad \text{with} \quad W(0) = 1400dtdW=300withW(0)=1400
Integrate both sides:W(t)=∫300 dt=300t+CW(t) = \int 300 \, dt = 300t + CW(t)=∫300dt=300t+C
Use the initial condition W(0)=1400W(0) = 1400W(0)=1400:1400=300(0)+C⇒C=14001400 = 300(0) + C \Rightarrow C = 14001400=300(0)+C⇒C=1400
So the particular solution is:W(t)=300t+1400\boxed{W(t) = 300t + 1400}W(t)=300t+1400
Additional Question: Two Particles P and R
Given:
- p(t)=2cos(4t)p(t) = 2\cos(4t)p(t)=2cos(4t)
- r(t)=t3−6t2+9t+3r(t) = t^3 – 6t^2 + 9t + 3r(t)=t3−6t2+9t+3
(i) Velocity of each particle:
Differentiate both positions with respect to time:
- For particle P:
vP(t)=dpdt=−8sin(4t)v_P(t) = \frac{dp}{dt} = -8\sin(4t)vP(t)=dtdp=−8sin(4t)
- For particle R:
vR(t)=drdt=3t2−12t+9v_R(t) = \frac{dr}{dt} = 3t^2 – 12t + 9vR(t)=dtdr=3t2−12t+9
These velocity functions describe how each particle moves over time along the x-axis.
Explanation
This problem explores the use of tangent lines, differential equations, and particle motion to model real-world and theoretical scenarios. In part (a), the use of a tangent line to approximate a function value is a standard linearization technique. Since the rate of change dWdt\frac{dW}{dt}dtdW is constant, the tangent line not only approximates but exactly predicts the function’s value over the short time interval. When the second derivative of a function is zero, as in part (b), the function is linear. This confirms that the approximation has no error due to curvature, meaning it is neither an overestimate nor an underestimate.
Part (c) involves solving a basic differential equation. Since the rate of change is constant, integration results in a linear function. The initial value is used to determine the constant of integration, yielding the particular solution W(t)=300t+1400W(t) = 300t + 1400W(t)=300t+1400, which provides a complete model for the amount of waste over time.
The particle motion section introduces two functions modeling positions over time. By differentiating these functions, one obtains the velocities of the particles. Particle P moves with an oscillating velocity because of the sine function, indicating periodic motion. Particle R moves with a velocity determined by a quadratic expression, suggesting changes in direction and speed due to acceleration or deceleration over time.
Altogether, these problems test fundamental calculus concepts like linear approximation, solving differential equations, and interpreting derivative functions in motion contexts.
