Home / AP Calc BC : Exam question -MCQs and FRQ / AP Calculus BC 2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions  Exam Style Questions - FRQ

AP Calculus BC 2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions  Exam Style Questions – FRQ

AP Calculus BC 2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions  Exam Style Questions - FRQ

Question

(a) Topic-3.1 The Chain Rule

(b) Topic-2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions

(c) Topic-1.11 Defining Continuity at a Point

(d) Topic-6.9 Integrating Using Substitution

The function f is defined by \(f(x)=\sqrt{25-x^{2}}\) for -5 ≤ x ≤ 5.

(a) Find f'(x).
(b) Write an equation for the line tangent to the graph of f at x = -3.

(c) Let g be the function defined by \(g(x)=\left\{\begin{matrix}
f(x) &for -5\leq x\leq -3 & & \\x+7
&for -3<x\leq 5. & &
\end{matrix}\right.\)
Is g continuous at x = -3 ? Use the definition of continuity to explain your answer
(d) Find the value of \(\int_{0}^{5}x\sqrt{25-x^{2}}dx.\)

▶️Answer/Explanation

\(
\textbf{4(a) } f'(x) = \frac{1}{2} (25 – x^2)^{-1/2}(-2x) = \frac{-x}{\sqrt{25 – x^2}}, \quad -5 < x < 5
\)

\(
\textbf{4(b) } f'(-3) = \frac{3}{\sqrt{25 – 9}} = \frac{3}{4}
\)
\(
f(-3) = \sqrt{25 – 9} = 4
\)
\(
\text{An equation for the tangent line is } y = 4 + \frac{3}{4}(x + 3).
\)

\(
\textbf{4(c) } \lim_{x \to -3} g(x) = \lim_{x \to -3} f(x) = \lim_{x \to -3} \sqrt{25 – x^2} = 4
\)
\(
\lim_{x \to -3^+} g(x) = \lim_{x \to -3^+} (x + 7) = 4
\)
\(
\text{Therefore, } \lim_{x \to -3} g(x) = 4.
\)
\(
g(-3) = f(-3) = 4
\)
\(
\text{So, } \lim_{x \to -3} g(x) = g(-3).
\)
\(
\text{Therefore, } g \text{ is continuous at } x = -3.
\)

\(
\textbf{4(d)} \text{Let } u = 25 – x^2 \implies du = -2x \, dx
\)
\(
\int_{0}^{5} x \sqrt{25 – x^2} \, dx = -\frac{1}{2} \int_{25}^{0} \sqrt{u} \, du
\)
\(
= -\left[-\frac{1}{2} \cdot \frac{2}{3} u^{3/2}\right]_{u = 0}^{u = 25}
\)
\(
= -\frac{1}{3} \left[0 – 125\right] = \frac{125}{3}.
\)

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