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AP Calculus AB : AP Calculus AB

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3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #21 : Implicit Differentiation

Use implicit differentiation to calculate the equation of the line tangent to the equation $2x^{2}+y^{2}=9$ at the point (2,1).

Possible Answers:

y=4x+9

y=-4x-9

y=-4x+9

y=4x-9

Correct answer:

y=-4x+9

Explanation:

Differentiate both sides of the equation: $\frac{\mathrm{d} }{\mathrm{d} x}(2x^{2}+y^{2}=9)$

Simplify: $4x+\frac{\mathrm{d} }{\mathrm{d} x}y^{2}=0$

Use implicit differentiation to differentiate the y term: $4x+2y\frac{dy}{dx}=0$

Subtract 4x from both sides of the equation: $2y\frac{dy}{dx}=-4x$

Divide both sides of the equation by 2y: $\frac{dy}{dx}=\frac{-4x}{2y}=\frac{-2x}{y}$

Plug in the appropriate values for x and y to find the slope of the tangent line: $m=\frac{-2(2)}{(1)}=-4$

Use slope-intercept form to solve for the equation of the tangent line: y=mx+b=y=-4x+b

Plug in the appropriate values of x and y into the equation, to find the equation of the tangent line: 1=(2)(-4)+b

Solve for b: 1=-8+b

b=9

Solution: y=-4x+9

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Example Question #21 : Implicit Differentiation

Find $\frac{\mathrm{d} z}{\mathrm{d} x}$ , where is a function of x.

Possible Answers:

$\frac{3-z}{x+2z}$

$\frac{3-2z}{x}$

$\frac{-z}{x+2z}$

Correct answer:

$\frac{3-z}{x+2z}$

Explanation:

To find $\frac{\mathrm{d} z}{\mathrm{d} x}$ we must use implicit differentiation, which is an application of the chain rule.
Taking $\frac{\mathrm{d} }{\mathrm{d} x}$ of both sides of the equation, we get

$z+x\frac{\mathrm{d} z}{\mathrm{d} x}+2z\frac{\mathrm{d} z}{\mathrm{d} x}=3$

and the derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$
Note that for every derivative of a function with z, the additional term $\frac{\mathrm{d} z}{\mathrm{d} x}$ appears; this is because of the chain rule, where z=g(x), so to speak, for the function it appears in.

Using algebra to solve, we get

$\frac{\mathrm{d} z}{\mathrm{d} x}=\frac{3-z}{x+2z}$

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Example Question #31 : Implicit Differentiation

Find the normal line of the curve $\frac{x^{2}}{2}+y^{2}=4$ at the point $(1,\frac{\sqrt{14}}{2})$

Possible Answers:

$y=x\sqrt{14}-\frac{\sqrt{14}}{2}$

$y=2x\sqrt{14}-\frac{\sqrt{14}}{2}$

$y=x\sqrt{14}-{\sqrt{14}}$

$y=x\sqrt{14}+\frac{\sqrt{14}}{2}$

Correct answer:

$y=x\sqrt{14}-\frac{\sqrt{14}}{2}$

Explanation:

$\frac{x^{2}}{2}+y^{2}=4$

Use implicit differentiation to calculate the slope of the tangent line:

$\frac{\mathrm{d} }{\mathrm{d} x}(\frac{x^{2}}{2}+y^{2}=4)$

$\frac{2x}{2}+2y\frac{dy}{dx}=0$

Simplify: $x+2y\frac{dy}{dx}=0$

Subtract x from both sides of the equation: $2y\frac{dy}{dx}=-x$

Divide both sides of the equation by 2y: $\frac{dy}{dx}=\frac{-x}{2y}$

Plug in the x and y values from the point $(1,\frac{\sqrt{14}}{2})$ into the equation, to calculate the slope of the tangent line: $\frac{dy}{dx}=\frac{-1}{\frac{2\sqrt{14}}{2}}=\frac{-1}{\sqrt{14}}=\frac{-\sqrt{14}}{14}$

Take the negative reciprocal of the slope of the tangent line to calculate the slope of the normal line: $m=-\frac{1}{\frac{-\sqrt{14}}{14}}=\sqrt{14}$

Plug the slope of the normal line into point slope form: $y-\frac{\sqrt{14}}{2}=(x-1)\sqrt{14}$

Add $\frac{\sqrt{14}}{2}$ to both sides to isolate y: $y=(x-1)\sqrt{14}+\frac{\sqrt{14}}{2}$

Simplify the equation: $y=x\sqrt{14}-\sqrt{14}+\frac{\sqrt{14}}{2}$

Simplify further: $y=x\sqrt{14}-\frac{\sqrt{14}}{2}$

Solution: $y=x\sqrt{14}-\frac{\sqrt{14}}{2}$

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Example Question #1 : Trapezoidal Sums

$\begin{align*}&\text{Using the method of trapezoidal sums approximate the integral:}\\&\int_{-4}^{1.4}(11sin(17cos(x)))dx\\&\text{Using }2\text{ intervals.}\end{align*}$

Possible Answers:

-63.97

-3.29

-10.84

-105.17

Correct answer:

-10.84

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{-4}^{1.4}(11sin(17cos(x)))dx\\&\text{So the interval is }[-4,1.4]\text{ the subintervals have length }\frac{1.4-(-4)}{2}=\frac{27}{10}\\&\text{and the x-values are:}\\&[-4,-\frac{13}{10},\frac{7}{5}]\\&\int_{-4}^{1.4}(11sin(17cos(x)))dx\approx\frac{27}{20}[(11sin(17cos((-4))))+2(11sin(17cos((-\frac{13}{10}))))+(11sin(17cos((\frac{7}{5}))))]\\&\int_{-4}^{1.4}(11sin(17cos(x)))dx\approx-10.84\end{align*}$

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Example Question #2 : Trapezoidal Sums

$\begin{align*}&\text{Calculate the trapezoidal sums integral approximation of :}\\&\int_{0}^{2}(6cos(20e^{(x)}))dx\\&\text{Using points over }2\text{ intervals.}\end{align*}$

Possible Answers:

-1.68

-0.60

-0.90

-5.20

Correct answer:

-5.20

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{0}^{2}(6cos(20e^{(x)}))dx\\&\text{So the interval is }[0,2]\text{ the subintervals have length }\frac{2-(0)}{2}=1\\&\text{and the x-values are:}\\&[0,1,2]\\&\int_{0}^{2}(6cos(20e^{(x)}))dx\approx\frac{1}{2}[(6cos(20e^{((0))}))+2(6cos(20e^{((1))}))+(6cos(20e^{((2))}))]\\&\int_{0}^{2}(6cos(20e^{(x)}))dx\approx-5.20\end{align*}$

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Example Question #3 : Trapezoidal Sums

$\begin{align*}&\text{Using }3\text{ intervals, and the method of trapezoidal sums approximate the integral:}\\&\int_{-1}^{8.9}(5sin(9e^{(x)}))dx\end{align*}$

Possible Answers:

35.68

6.37

4.20

114.16

Correct answer:

35.68

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{-1}^{8.9}(5sin(9e^{(x)}))dx\\&\text{So the interval is }[-1,8.9]\text{ the subintervals have length }\frac{8.9-(-1)}{3}=\frac{33}{10}\\&\text{and the x-values are:}\\&[-1,\frac{23}{10},\frac{28}{5},\frac{89}{10}]\\&\int_{-1}^{8.9}(5sin(9e^{(x)}))dx\approx\frac{33}{20}[(5sin(9e^{((-1))}))+2(5sin(9e^{((\frac{23}{10}))}))+2(5sin(9e^{((\frac{28}{5}))}))+(5sin(9e^{((\frac{89}{10}))}))]\\&\int_{-1}^{8.9}(5sin(9e^{(x)}))dx\approx35.68\end{align*}$

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Example Question #4 : Trapezoidal Sums

$\begin{align*}&\text{Using the method of trapezoidal sums approximate the integral:}\\&\int_{1}^{6.4}(-6cos(19x^{2}))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

-16.73

-33.45

-93.67

-138.84

Correct answer:

-16.73

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{1}^{6.4}(-6cos(19x^{2}))dx\\&\text{So the interval is }[1,6.4]\text{ the subintervals have length }\frac{6.4-(1)}{3}=\frac{9}{5}\\&\text{and the x-values are:}\\&[1,\frac{14}{5},\frac{23}{5},\frac{32}{5}]\\&\int_{1}^{6.4}(-6cos(19x^{2}))dx\approx\frac{9}{10}[(-6cos(19(1)^{2}))+2(-6cos(19(\frac{14}{5})^{2}))+2(-6cos(19(\frac{23}{5})^{2}))+(-6cos(19(\frac{32}{5})^{2}))]\\&\int_{1}^{6.4}(-6cos(19x^{2}))dx\approx-16.73\end{align*}$

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Example Question #5 : Trapezoidal Sums

$\begin{align*}&\text{Using the method of trapezoidal sums approximate the integral:}\\&\int_{2}^{6.5}(17cos(2sin(2x)))dx\\&\text{Using }3\text{ intervals.}\end{align*}$

Possible Answers:

5.51

3.13

7.65

27.55

Correct answer:

27.55

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{2}^{6.5}(17cos(2sin(2x)))dx\\&\text{So the interval is }[2,6.5]\text{ the subintervals have length }\frac{6.5-(2)}{3}=\frac{3}{2}\\&\text{and the x-values are:}\\&[2,\frac{7}{2},5,\frac{13}{2}]\\&\int_{2}^{6.5}(17cos(2sin(2x)))dx\approx\frac{3}{4}[(17cos(2sin(2(2))))+2(17cos(2sin(2(\frac{7}{2}))))+2(17cos(2sin(2(5))))+(17cos(2sin(2(\frac{13}{2}))))]\\&\int_{2}^{6.5}(17cos(2sin(2x)))dx\approx27.55\end{align*}$

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Example Question #6 : Trapezoidal Sums

$\begin{align*}&\text{Using the method of trapezoidal sums approximate the integral:}\\&\int_{0}^{4.4}(-196cos(3x)^{2})dx\\&\text{Using }2\text{ intervals.}\end{align*}$

Possible Answers:

-2160.41

-88.69

-130.70

-744.97

Correct answer:

-744.97

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{0}^{4.4}(-196cos(3x)^{2})dx\\&\text{So the interval is }[0,4.4]\text{ the subintervals have length }\frac{4.4-(0)}{2}=\frac{11}{5}\\&\text{and the x-values are:}\\&[0,\frac{11}{5},\frac{22}{5}]\\&\int_{0}^{4.4}(-196cos(3x)^{2})dx\approx\frac{11}{10}[(-196cos(3(0))^{2})+2(-196cos(3(\frac{11}{5}))^{2})+(-196cos(3(\frac{22}{5}))^{2})]\\&\int_{0}^{4.4}(-196cos(3x)^{2})dx\approx-744.97\end{align*}$

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Example Question #7 : Trapezoidal Sums

$\begin{align*}&\text{Using }3\text{ intervals, and the method of trapezoidal sums approximate the integral:}\\&\int_{3}^{7.2}(11cos(6e^{(4x)}))dx\end{align*}$

Possible Answers:

-0.56

-18.44

-1.86

-0.29

Correct answer:

-1.86

Explanation:

$\begin{align*}&\text{A trapezoidal integral approximation over an interval [a, b]}\\&\text{with n subintervals follows the form:}\\&\int_a^b f(x)dx \approx\sum_{i=1}^{n}\frac{b-a}{n}\frac{f(x_{i+1})+f(x_i)}{2}=\frac{b-a}{2n}(f(x_1)+2f(x_2)+2f(x_3)+...+f(x_n))\\&\text{It is essentially a sum of n trapezoids, each with a base of}\\&\text{length :}\frac{b-a}{n}\text{ and variable paired heights :}f(x_i),f(x_{i+1})\text{, which depend on the function values at }x_i,x_{i+1}\\&\text{We're asked to approximate}\int_{3}^{7.2}(11cos(6e^{(4x)}))dx\\&\text{So the interval is }[3,7.2]\text{ the subintervals have length }\frac{7.2-(3)}{3}=\frac{7}{5}\\&\text{and the x-values are:}\\&[3,\frac{22}{5},\frac{29}{5},\frac{36}{5}]\\&\int_{3}^{7.2}(11cos(6e^{(4x)}))dx\approx\frac{7}{10}[(11cos(6e^{(4(3))}))+2(11cos(6e^{(4(\frac{22}{5}))}))+2(11cos(6e^{(4(\frac{29}{5}))}))+(11cos(6e^{(4(\frac{36}{5}))}))]\\&\int_{3}^{7.2}(11cos(6e^{(4x)}))dx\approx-1.86\end{align*}$

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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #21 : Implicit Differentiation

Example Question #21 : Implicit Differentiation

Example Question #31 : Implicit Differentiation

Example Question #1 : Trapezoidal Sums

Example Question #2 : Trapezoidal Sums

Example Question #3 : Trapezoidal Sums

Example Question #4 : Trapezoidal Sums

Example Question #5 : Trapezoidal Sums

Example Question #6 : Trapezoidal Sums

Example Question #7 : Trapezoidal Sums

All AP Calculus AB Resources

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