The 2011 AP Calculus AB Free Response Questions (FRQs) were a challenging yet rewarding experience for students. The questions covered a wide range of topics, from limits and derivatives to integrals and applications. In this article, we will provide a comprehensive review of the 2011 AP Calc AB FRQs, including solutions and strategies for solving each problem.
Question 1: Limit Evaluation
Question:
Find the limit:
$$\lim_{x\to0}\frac{\sin(3x)}{x}$$
Solution:
We can evaluate this limit using l’Hopital’s Rule. L’Hopital’s Rule states that if the limit of the numerator and denominator of a fraction is both 0 or both infinity, then the limit of the fraction is equal to the limit of the derivative of the numerator over the derivative of the denominator.
In this case, the limit of the numerator and denominator is both 0, so we can apply l’Hopital’s Rule. The derivative of the numerator is 3cos(3x), and the derivative of the denominator is 1. Therefore, the limit of the fraction is:
$$\lim_{x\to0}\frac{\sin(3x)}{x}=\lim_{x\to0}\frac{3\cos(3x)}{1}=\boxed{3}$$
Question 2: Derivative Calculation
Question:
Find the derivative of the function:
$$f(x)=\sqrt{x^2+1}$$
Solution:
We can find the derivative of this function using the chain rule. The chain rule states that if we have a function of the form f(g(x)), then the derivative of f(g(x)) is equal to f'(g(x)) * g'(x).
In this case, we have f(x) = sqrt(x^2+1), so f'(x) = 1/2 * (x^2+1)^(-1/2). And g(x) = x^2+1, so g'(x) = 2x. Therefore, the derivative of f(x) is:
$$f'(x)=\frac{1}{2}(x^2+1)^{-1/2}(2x)=\boxed{\frac{x}{\sqrt{x^2+1}}}$$
Question 3: Integral Evaluation
Question:
Find the integral:
$$\int_{0}^{\pi/2}\cos(x)dx$$
Solution:
We can evaluate this integral using the fundamental theorem of calculus. The fundamental theorem of calculus states that if we have a function f(x) and its antiderivative F(x), then the integral of f(x) from a to b is equal to F(b) – F(a).
In this case, the antiderivative of cos(x) is sin(x). Therefore, the integral of cos(x) from 0 to pi/2 is:
$$\int_{0}^{\pi/2}\cos(x)dx=F(\pi/2)-F(0)=\sin(\pi/2)-\sin(0)=\boxed{1}$$
Question 4: Application of Derivatives
Question:
A farmer has 100 feet of fencing to enclose a rectangular plot of land. What is the maximum area that the farmer can enclose?
Solution:
Let x be the length of the plot and y be the width of the plot. The area of the plot is A = xy. The farmer has 100 feet of fencing, so the perimeter of the plot is 2x + 2y = 100.
We can use the perimeter equation to solve for y in terms of x:
$$2y = 100-2x$$
$$y = 50-x$$
We can substitute this expression for y into the area equation:
$$A = xy = x(50-x) = 50x-x^2$$
To find the maximum area, we need to find the critical points of the area function. The critical points are the points where the first derivative of the function is equal to zero.
The first derivative of the area function is:
$$A'(x) = 50-2x$$
Setting the first derivative equal to zero, we get:
$$50-2x = 0$$
$$x = 25$$
The critical point is x = 25. To determine if this critical point is a maximum, we need to look at the second derivative of the area function.
The second derivative of the area function is:
$$A”(x) = -2$$
The second derivative is negative, which means that the critical point is a maximum. Therefore, the maximum area that the farmer can enclose is:
$$A(25) = 50(25)-25^2 = \boxed{625}$$ square feet.
Conclusion
The 2011 AP Calculus AB FRQs were a challenging yet rewarding experience for students. The questions covered a wide range of topics, and students who were well-prepared were able to demonstrate their understanding of the subject. We hope that this review has been helpful for students who are preparing for the AP Calculus AB exam.
