Q.1 Invasive Species Problem: A BC Calculus Perspective
Note: All calculations requiring a calculator are performed in radian mode. Results are typically rounded to three decimal places.
Part A: Average Number of Acres Affected
The average value of a function \(f(t)\) over an interval \([a, b]\) is a fundamental application of the definite integral, defined by the formula \( \frac{1}{b-a} \int_a^b f(t) \, dt \). Here, we apply this to the function \(C(t) = 7.6 \arctan(0.2t)\) on the interval \([0, 4]\).
The setup for the calculation is:
While the integral of \(\arctan(u)\) can be found using integration by parts (a key BC technique), the non-elementary nature of this integral in a calculator-active question strongly indicates that we should use a numerical integrator.
Part B: Applying the Mean Value Theorem
This question is a direct application of the Mean Value Theorem (MVT) for derivatives. The MVT states that for a function continuous on \([a,b]\) and differentiable on \((a,b)\), there exists some \(c\) in \((a,b)\) such that \(f'(c) = \frac{f(b) - f(a)}{b-a}\). Here, we are finding that value \(c\) (denoted as \(t\)).
Step 1: Calculate the average rate of change of \(C\) on \([0, 4]\).
Step 2: Set the instantaneous rate of change, \(C'(t)\), equal to the average rate.
We are given \(C'(t) = \frac{38}{25 + t^2}\).
Solving this equation for \(t\) using a calculator's solver or by finding the intersection of the two functions:
Part C: End Behavior of the Rate of Change
"End behavior" refers to the limit of a function as its variable approaches infinity. We need to evaluate the limit of the rate of change, \(C'(t)\), as \(t \to \infty\). This is a concept foundational to the study of improper integrals and series convergence in BC Calculus.
As \(t \to \infty\), the denominator \(25 + t^2\) grows without bound, while the numerator remains constant. Therefore, the fraction approaches zero.
This result implies that, in the long run, the spread of the invasive species effectively stops, with the rate of new acres being affected approaching zero.
Part D: Absolute Maximum on a Closed Interval
We must find the absolute maximum of \(A(t) = C(t) - \int_4^t 0.1 \ln(x) \, dx\) on the closed interval \([4, 36]\). The standard procedure is the Candidates Test, which relies on the Extreme Value Theorem.
Step 1: Find the derivative, \(A'(t)\).
We apply the Fundamental Theorem of Calculus, Part 2, which states that \(\frac{d}{dt} \int_a^t f(x) \, dx = f(t)\).
Step 2: Find critical points.
Set \(A'(t) = 0\) and solve for \(t\) in the interval \((4, 36)\) using a calculator.
Step 3: Test the candidates (endpoints and critical points).
We evaluate \(A(t)\) at \(t=4\), \(t \approx 13.922\), and \(t=36\).
- \(A(4) = C(4) - \int_4^4 0.1 \ln(x) \, dx = C(4) = 7.6 \arctan(0.8) \approx 5.128\)
- \(A(13.922) = C(13.922) - \int_4^{13.922} 0.1 \ln(x) \, dx \approx 9.771 - 2.039 \approx 7.732\)
- \(A(36) = C(36) - \int_4^{36} 0.1 \ln(x) \, dx \approx 11.660 - 5.700 \approx 5.960\)
Step 4: Justify and Conclude.
By comparing the values from the Candidates Test, the maximum value is approximately 7.732.
The function \(A\) attains its maximum value at \(t \approx 13.922\) weeks. The justification is that we have tested all possible locations for an absolute extremum (the endpoints and the critical point) and found that the function's value is largest at this time.
Q.2 Polar Coordinates Problem: A BC Calculus Solution
Note: Your calculator should be in radian mode for all parts of this problem.
Part A: Rate of Change of \(r\) with respect to \(\theta\)
The "rate of change of \(r\) with respect to \(\theta\)" is simply the derivative, \(\frac{dr}{d\theta}\). We are given \(r(\theta) = 2\sin^2\theta = 2(\sin\theta)^2\). We apply the chain rule to differentiate.
Now, we evaluate this derivative at \(\theta = 1.3\).
Part B: Area Between Polar Curves
We want the area inside curve C (\(r_C = 2\sin^2\theta\)) but outside the semicircle (\(r_S = 1/2\)). The formula for the area between two polar curves is \( A = \frac{1}{2}\int_{\alpha}^{\beta} (r_{outer}^2 - r_{inner}^2) \, d\theta \).
Step 1: Find the points of intersection.
We set the radii equal to find the angles \(\theta\) where the curves intersect.
In the interval \(0 \le \theta \le \pi\), this occurs at \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\). These are our bounds of integration.
Step 2: Set up and evaluate the integral.
The setup for the area is:
Evaluating this integral with a calculator:
Part C: Point Farthest from the y-axis
The distance from the y-axis is the absolute value of the x-coordinate. For a polar curve, \(x = r\cos\theta\). We want to maximize \(|x|\).
Step 1: Express x as a function of \(\theta\).
For curve C, \(x(\theta) = (2\sin^2\theta)\cos\theta\). The problem restricts us to \(0 \le \theta \le \pi/2\), where \(x \ge 0\), so we just need to maximize \(x(\theta)\).
Step 2: Find the critical points.
We must find where \( \frac{dx}{d\theta} = 0 \). The problem graciously provides the derivative:
In the open interval \((0, \pi/2)\), \(\sin\theta \neq 0\). So we solve:
Step 3: Justify the maximum.
We use the Candidates Test. The candidates are the endpoints \(\theta=0\), \(\theta=\pi/2\) and the critical point \(\theta = \arctan(\sqrt{2})\).
- \(x(0) = 2\sin^2(0)\cos(0) = 0\)
- \(x(\pi/2) = 2\sin^2(\pi/2)\cos(\pi/2) = 2(1)(0) = 0\)
- At \(\theta = \arctan(\sqrt{2})\), we have \(\tan\theta = \sqrt{2}\). From a right triangle, \(\sin\theta = \frac{\sqrt{2}}{\sqrt{3}}\) and \(\cos\theta = \frac{1}{\sqrt{3}}\). So, \(x(\arctan(\sqrt{2})) = 2\left(\frac{\sqrt{2}}{\sqrt{3}}\right)^2\left(\frac{1}{\sqrt{3}}\right) = 2\left(\frac{2}{3}\right)\left(\frac{1}{\sqrt{3}}\right) = \frac{4}{3\sqrt{3}} > 0\).
Since the value at the critical point is positive and the values at the endpoints are 0, the maximum occurs at the critical point. The value of \(\theta\) is \(\arctan(\sqrt{2})\).
Part D: Related Rates with Polar Coordinates
The particle's distance from the origin is simply \(r\). We want to find the rate at which this distance changes with respect to time, which is \(\frac{dr}{dt}\). This is a related rates problem.
Step 1: Relate r and \(\theta\), then differentiate with respect to time \(t\).
The relationship is \(r = 2\sin^2\theta\). We differentiate both sides with respect to \(t\), using the chain rule.
Notice that the term \(4\sin\theta\cos\theta\) is exactly \(\frac{dr}{d\theta}\) from Part A. The chain rule connects these rates beautifully: \(\frac{dr}{dt} = \frac{dr}{d\theta} \cdot \frac{d\theta}{dt}\).
Step 2: Substitute the given values.
We are given \(\frac{d\theta}{dt} = 15\) and we need to find \(\frac{dr}{dt}\) at the instant when \(\theta = 1.3\).
From Part A, we already calculated \(4\sin(1.3)\cos(1.3) \approx 1.037\).
Q.3 Reading Rate Problem: A BC Calculus Solution
Part A: Approximate \(R'(1)\)
The instantaneous rate of change, \(R'(1)\), can be approximated by the average rate of change over the smallest given interval containing \(t=1\). This interval is \([0, 2]\). The average rate of change is the slope of the secant line connecting the endpoints of this interval.
The setup for the calculation is:
Substituting the values from the table:
Units: The units of \(R(t)\) are words/minute, and the units of \(t\) are minutes. The units of the derivative, \(R'(t)\), are the units of \(R\) divided by the units of \(t\). This represents the acceleration of the student's reading.
Part B: Existence of \(c\) such that \(R(c) = 155\)
This question is a direct application of the Intermediate Value Theorem (IVT). The IVT guarantees that a continuous function will take on all values between its endpoint values.
Justification using the Intermediate Value Theorem:
- Continuity: The problem states that \(R(t)\) is a differentiable function. A core theorem in calculus is that differentiability implies continuity. Therefore, \(R(t)\) is continuous on the closed interval \([0, 10]\).
- Intermediate Value: We examine the table for an interval where 155 lies between the function values at the endpoints. On the interval \([8, 10]\), we have \(R(8) = 150\) and \(R(10) = 162\). The value 155 is between 150 and 162.
- Conclusion: Since \(R\) is continuous on \([8, 10]\) and \(R(8) < 155 < R(10)\), the IVT guarantees that yes, there must be a value \(c\) in the open interval \((8, 10)\) such that \(R(c) = 155\).
Part C: Trapezoidal Sum Approximation
The definite integral \(\int_{0}^{10} R(t) dt\) represents the total number of words read over the 10-minute period. We will approximate this using a trapezoidal sum over the three non-uniform subintervals given: \([0, 2]\), \([2, 8]\), and \([8, 10]\).
The formula for the area of a trapezoid is \(\frac{1}{2}(b_1 + b_2)h\), where \(h\) is the width of the interval (\(\Delta t\)).
Substitute the values from the table into the setup:
Part D: Total Words Read by the Teacher
To find the total number of words the teacher read, we must integrate the teacher's reading rate, \(W(t)\), from \(t=0\) to \(t=10\). This requires evaluating a definite integral using the Fundamental Theorem of Calculus.
First, find the antiderivative of \(W(t)\) using the power rule for integration:
Now, evaluate the antiderivative at the limits of integration:
Q.4 Graphical Analysis Problem: A BC Calculus Solution
We are given the function \(g(x) = \int_6^x f(t) \, dt\), which establishes a critical link between \(g\) and \(f\): \(f\) is the derivative of \(g\).
Part A: Find \(g'(8)\)
The cornerstone of this problem is the Second Fundamental Theorem of Calculus (FTC). It states that if a function \(g\) is defined as \(g(x) = \int_a^x f(t) \, dt\), then its derivative is \(g'(x) = f(x)\).
The point \((8, f(8))\) lies on the line segment connecting \((6, 0)\) and \((12, 3)\). The slope of this segment is \(m = \frac{3-0}{12-6} = \frac{1}{2}\). Using the point-slope form, the equation of the line is \(y - 0 = \frac{1}{2}(x - 6)\). At \(x=8\), \(y = \frac{1}{2}(8-6) = 1\).
Part B: Find Points of Inflection for g
A point of inflection on the graph of \(g\) occurs where its concavity changes, which means its second derivative, \(g''(x)\), changes sign.
- At \(\boldsymbol{x=-3}\), \(f\) changes from increasing to decreasing, so \(f'\) changes from positive to negative.
- At \(\boldsymbol{x=0}\), \(f\) changes from decreasing to increasing, so \(f'\) changes from negative to positive.
- At \(\boldsymbol{x=3}\), \(f\) changes from increasing to decreasing, so \(f'\) changes from positive to negative.
- At \(\boldsymbol{x=6}\), \(f\) changes from decreasing to increasing, so \(f'\) changes from negative to positive.
Part C: Find \(g(12)\) and \(g(0)\)
The value of \(g(x)\) is the net signed area under the curve of \(f\) from the starting point \(t=6\) to the endpoint \(t=x\).
Calculation for g(12):
This integral represents the area of the triangle from \(x=6\) to \(x=12\), which has a base of 6 and a height of 3.
Calculation for g(0):
The integral \(\int_0^6 f(t) \, dt\) is the area of the semicircle above the x-axis with radius \(r=3\).
Part D: Absolute Minimum of g
The Extreme Value Theorem guarantees that a continuous function on a closed interval attains an absolute minimum and maximum. These extrema can only occur at the endpoints of the interval or at critical points within the interval. We use the Candidates Test to find the absolute minimum.
- Find Critical Points: Critical points of \(g\) occur where \(g'(x) = f(x) = 0\). From the graph, this happens at \(x = -6, 0,\) and \(6\). The critical points in the open interval \((-6, 12)\) are \(x=0\) and \(x=6\).
- Identify Candidates: The candidates for the location of the absolute minimum are the endpoints \(-6\) and \(12\), and the critical points \(0\) and \(6\).
- Evaluate g at Each Candidate:
- \(g(-6) = \int_6^{-6} f(t) \, dt = -\int_{-6}^6 f(t) \, dt = -[-\frac{1}{2}\pi(3)^2 + \frac{1}{2}\pi(3)^2] = 0\)
- \(g(0) = -\frac{9\pi}{2} \approx -14.137\) (from Part C)
- \(g(6) = \int_6^6 f(t) \, dt = 0\)
- \(g(12) = 9\) (from Part C)
- Conclusion: Comparing the values \(\{0, -9\pi/2, 0, 9\}\), the lowest value is \(-9\pi/2\). Therefore, the absolute minimum of \(g\) on the interval \([-6, 12]\) occurs at \(x=0\).
Q.5 Differential Equation and Approximation Problem
Part A: Find \(f''(1)\)
To find the second derivative, we must differentiate the given first derivative, \(\frac{dy}{dx} = (3-x)y^2\), with respect to \(x\). This requires the product rule and implicit differentiation.
Step 1: Apply the product rule.
Let \(u = (3-x)\) and \(v = y^2\). Then \(u' = -1\) and \(v' = 2y \cdot \frac{dy}{dx}\) (by the chain rule).
Step 2: Evaluate at the point \((1, -1)\).
First, we need the value of the first derivative, \(\frac{dy}{dx}\), at \((1,-1)\).
Now substitute \(x=1\), \(y=-1\), and \(\frac{dy}{dx}=2\) into the expression for the second derivative.
Part B: Second-Degree Taylor Polynomial
The formula for a second-degree Taylor polynomial for a function \(f\) about \(x=a\) is:
For our problem, \(a=1\). We have the necessary components from our initial condition and Part A:
- \(f(1) = -1\) (given)
- \(f'(1) = 2\) (calculated in Part A)
- \(f''(1) = -9\) (calculated in Part A)
Substituting these values into the formula gives our polynomial:
Part C: Lagrange Error Bound
The Lagrange error bound for an \(n\)-th degree Taylor polynomial approximation is given by \(|R_n(x)| \le \frac{M}{(n+1)!}|x-a|^{n+1}\), where \(M\) is an upper bound for \(|f^{(n+1)}(z)|\) on the interval between \(a\) and \(x\).
- For our second-degree polynomial (\(n=2\)), the error is bounded by the third derivative term.
- We are given \(|f'''(x)| \le 60\) for \(1 \le x \le 1.1\). This means we can use \(M=60\).
- Our center is \(a=1\) and we are approximating at \(x=1.1\).
We set up the inequality:
Thus, we have shown that the approximation of \(f(1.1)\) using \(P_2(1.1)\) differs from the true value by at most 0.01.
Part D: Euler's Method
Euler's method is a numerical technique to approximate solutions to differential equations by taking small steps along tangent lines.
Step 1: Determine the setup.
- Starting point: \((x_0, y_0) = (1, -1)\).
- Target: \(f(1.4)\).
- Number of steps: 2.
- Step size: \(\Delta x = \frac{\text{end} - \text{start}}{\text{steps}} = \frac{1.4 - 1}{2} = 0.2\).
Step 2: Perform the first step (from x=1 to x=1.2).
Our first approximated point is \((1.2, -0.6)\).
Step 3: Perform the second step (from x=1.2 to x=1.4).
The approximation for \(f(1.4)\) using Euler's method is -0.4704.
Q.6 Taylor Series and Convergence Problem
Part A: Interval of Convergence
We use the Ratio Test to find the radius of convergence for the series \(f(x) = \sum_{n=1}^{\infty} \frac{(x-4)^{n+1}}{(n+1)3^n}\).
Step 1: Set up the Ratio Test limit.
Let \(a_n = \frac{(x-4)^{n+1}}{(n+1)3^n}\). We evaluate \(L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\).
Step 2: Simplify and evaluate the limit.
Step 3: Find the radius and interval of convergence.
The series converges absolutely when \(L < 1\).
The radius of convergence is \(R=3\). This gives an initial interval of \(1 < x < 7\).
Step 4: Test the endpoints.
- At \(x=1\): The series becomes \( \sum_{n=1}^{\infty} \frac{(-3)^{n+1}}{(n+1)3^n} = \sum_{n=1}^{\infty} \frac{3(-1)^{n+1}}{n+1} \). This is 3 times the alternating harmonic series, which converges by the Alternating Series Test.
- At \(x=7\): The series becomes \( \sum_{n=1}^{\infty} \frac{3^{n+1}}{(n+1)3^n} = \sum_{n=1}^{\infty} \frac{3}{n+1} \). This is 3 times the harmonic series, which diverges.
Part B: Taylor Series for the Derivative, \(f'(x)\)
We can find the series for \(f'(x)\) by differentiating the series for \(f(x)\) term-by-term.
Differentiate the general term:
So, the Taylor series for \(f'(x)\) is \(f'(x) = \sum_{n=1}^{\infty} \frac{(x-4)^n}{3^n}\).
The first three nonzero terms correspond to n=1, 2, and 3:
The general term is:
Part C: Show \(f'(x)\) is the Sum of a Geometric Series
The series for \(f'(x)\) found in Part B can be rewritten as:
This is a geometric series with first term \(a = \frac{x-4}{3}\) and common ratio \(r = \frac{x-4}{3}\). The sum of a convergent geometric series is given by the formula \(S = \frac{a}{1-r}\).
Multiplying the numerator and denominator by 3 simplifies the expression:
This shows that for all \(x\) in the interval of convergence, the series for \(f'(x)\) sums to the given function.
Part D: Convergence of \(f'(x)\) at \(x=8\)
We are asked if the Taylor series for \(f'(x)\) converges to \(f'(x)\) at \(x=8\).
The series for \(f'(x)\), as identified in Part C, is a geometric series with common ratio \(r = \frac{x-4}{3}\). A geometric series converges if and only if its common ratio \(|r|\) is less than 1.
At \(x=8\), the common ratio is:
Since \(|r| = \frac{4}{3} \ge 1\), the geometric series diverges at \(x=8\). Therefore, the series cannot converge to the function value \(f'(8)\).