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AP Calculus BC : Applications of Derivatives

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Example Questions

Example Question #21 : Euler's Method And L'hopital's Rule

$\begin{align*}&\text{The expression of a particular function is unknown;}\\&\text{however, we have an expression for its derivative.}\\&\text{Knowing that }f'(x)=13tan(5x)\text{ and }f(2)=9\\&\text{Approximate } f(11.9)\text{ using Euler's Method and }3\text{ steps.}\end{align*}$

Possible Answers:

36.08

901.94

6.5

36.81

180.39

Correct answer:

180.39

Explanation:

$\begin{align*}&\text{The general form of Euler's method, when a derivative function,}\\&\text{initial value, and step size are known, is:}\\&y_n=y_{n+1} +\Delta x f'(x_n,y_n)\\&\text{To calculate the step size find the distance between the final}\\&\text{and initial value and divide by the number of steps to be used:}\\&\Delta x = \frac{x_f-x_i}{Steps}\\&\text{For this problem, we are told: }f'(x)=13tan(5x);f(2)=9\\&\text{Knowing this, we may take the steps to estimate our function}\\&\text{value at our final x value:}\\&\Delta x = \frac{11.9-(2)}{3}=3.3\\&y_0=9;x_0=2\\&y_{1}=9+(3.3)(13tan(5(2)))=36.81\\&y_{2}=36.81+(3.3)(13tan(5(5.3)))=244.67\\&y_{3}=244.67+(3.3)(13tan(5(8.6)))=180.39\\&\\&\end{align*}$

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Example Question #22 : Euler's Method And L'hopital's Rule

$\begin{align*}&\text{The expression of a particular function is unknown;}\\&\text{however, we have an expression for its derivative.}\\&\text{Knowing that }f'(x)=-19ln(5x)\text{ and }f(1)=-3\\&\text{Approximate } f(3.8)\text{ using Euler's Method and }2\text{ steps.}\end{align*}$

Possible Answers:

-559.55

-22.38

-58.94

-45.81

-111.91

Correct answer:

-111.91

Explanation:

$\begin{align*}&\text{The general form of Euler's method, when a derivative function,}\\&\text{initial value, and step size are known, is:}\\&y_n=y_{n+1} +\Delta x f'(x_n,y_n)\\&\text{To calculate the step size find the distance between the final}\\&\text{and initial value and divide by the number of steps to be used:}\\&\Delta x = \frac{x_f-x_i}{Steps}\\&\text{For this problem, we are told: }f'(x)=-19ln(5x);f(1)=-3\\&\text{Knowing this, we may take the steps to estimate our function}\\&\text{value at our final x value:}\\&\Delta x = \frac{3.8-(1)}{2}=1.4\\&y_0=-3;x_0=1\\&y_{1}=-3+(1.4)(-19ln(5(1)))=-45.81\\&y_{2}=-45.81+(1.4)(-19ln(5(2.4)))=-111.91\\&\\&\end{align*}$

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Example Question #23 : Euler's Method And L'hopital's Rule

$\begin{align*}&\text{The expression of a particular function is unknown;}\\&\text{however, we have an expression for its derivative.}\\&\text{Knowing that }f'(x)=18\cdot 5^x\text{ and }f(0)=10\\&\text{Approximate } f(6.6)\text{ using Euler's Method and }3\text{ steps.}\end{align*}$

Possible Answers:

48531

97061.9

49.6

738721

16177

Correct answer:

48531

Explanation:

$\begin{align*}&\text{The general form of Euler's method, when a derivative function,}\\&\text{initial value, and step size are known, is:}\\&y_n=y_{n+1} +\Delta x f'(x_n,y_n)\\&\text{To calculate the step size find the distance between the final}\\&\text{and initial value and divide by the number of steps to be used:}\\&\Delta x = \frac{x_f-x_i}{Steps}\\&\text{For this problem, we are told: }f'(x)=18\cdot 5^x;f(0)=10\\&\text{Knowing this, we may take the steps to estimate our function}\\&\text{value at our final x value:}\\&\Delta x = \frac{6.6-(0)}{3}=2.2\\&y_0=10;x_0=0\\&y_{1}=10+(2.2)(18\cdot 5^{(0)})=49.6\\&y_{2}=49.6+(2.2)(18\cdot 5^{(2.2)})=1415.53\\&y_{3}=1415.53+(2.2)(18\cdot 5^{(4.4)})=48531\\&\\&\end{align*}$

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Example Question #24 : Euler's Method And L'hopital's Rule

$\begin{align*}&\text{The expression of a particular function is unknown;}\\&\text{however, we have an expression for its derivative.}\\&\text{Knowing that }f'(x)=-17e^{(5x)}\text{ and }f(-2)=-20\\&\text{Approximate } f(1.8)\text{ using Euler's Method and }2\text{ steps.}\end{align*}$

Possible Answers:

-39.59

-158.37

-7.92

Correct answer:

-39.59

Explanation:

$\begin{align*}&\text{The general form of Euler's method, when a derivative function,}\\&\text{initial value, and step size are known, is:}\\&y_n=y_{n+1} +\Delta x f'(x_n,y_n)\\&\text{To calculate the step size find the distance between the final}\\&\text{and initial value and divide by the number of steps to be used:}\\&\Delta x = \frac{x_f-x_i}{Steps}\\&\text{For this problem, we are told: }f'(x)=-17e^{(5x)};f(-2)=-20\\&\text{Knowing this, we may take the steps to estimate our function}\\&\text{value at our final x value:}\\&\Delta x = \frac{1.8-(-2)}{2}=1.9\\&y_0=-20;x_0=-2\\&y_{1}=-20+(1.9)(-17e^{(5{(-2)})})=-20\\&y_{2}=-20+(1.9)(-17e^{(5{(-0.1)})})=-39.59\\&\\&\end{align*}$

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Example Question #45 : Applications Of Derivatives

$\begin{align*}&\text{The expression of a particular function is unknown;}\\&\text{however, we have an expression for its derivative.}\\&\text{Knowing that }f'(x)=tan(x)\text{ and }f(3)=12\\&\text{Approximate } f(21)\text{ using Euler's Method and }4\text{ steps.}\end{align*}$

Possible Answers:

25.23

100.94

10.47

8.41

11.36

Correct answer:

25.23

Explanation:

$\begin{align*}&\text{The general form of Euler's method, when a derivative function,}\\&\text{initial value, and step size are known, is:}\\&y_n=y_{n+1} +\Delta x f'(x_n,y_n)\\&\text{To calculate the step size find the distance between the final}\\&\text{and initial value and divide by the number of steps to be used:}\\&\Delta x = \frac{x_f-x_i}{Steps}\\&\text{For this problem, we are told: }f'(x)=tan(x);f(3)=12\\&\text{Knowing this, we may take the steps to estimate our function}\\&\text{value at our final x value:}\\&\Delta x = \frac{21-(3)}{4}=4.5\\&y_0=12;x_0=3\\&y_{1}=12+(4.5)(tan{(3)})=11.36\\&y_{2}=11.36+(4.5)(tan{(7.5)})=23.54\\&y_{3}=23.54+(4.5)(tan{(12)})=20.67\\&y_{4}=20.67+(4.5)(tan{(16.5)})=25.23\\&\\&\end{align*}$

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Example Question #41 : Applications Of Derivatives

$\begin{align*}&\text{The expression of a particular function is unknown;}\\&\text{however, we have an expression for its derivative.}\\&\text{Knowing that }f'(x)=13sin(3x)\text{ and }f(2)=-15\\&\text{Approximate } f(11.4)\text{ using Euler's Method and }2\text{ steps.}\end{align*}$

Possible Answers:

-10.45

25.92

12.96

-32.07

103.68

Correct answer:

25.92

Explanation:

$\begin{align*}&\text{The general form of Euler's method, when a derivative function,}\\&\text{initial value, and step size are known, is:}\\&y_n=y_{n+1} +\Delta x f'(x_n,y_n)\\&\text{To calculate the step size find the distance between the final}\\&\text{and initial value and divide by the number of steps to be used:}\\&\Delta x = \frac{x_f-x_i}{Steps}\\&\text{For this problem, we are told: }f'(x)=13sin(3x);f(2)=-15\\&\text{Knowing this, we may take the steps to estimate our function}\\&\text{value at our final x value:}\\&\Delta x = \frac{11.4-(2)}{2}=4.7\\&y_0=-15;x_0=2\\&y_{1}=-15+(4.7)(13sin(3{(2)}))=-32.07\\&y_{2}=-32.07+(4.7)(13sin(3{(6.7)}))=25.92\\&\\&\end{align*}$

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Example Question #47 : Applications Of Derivatives

$\begin{align*}&\text{The expression of a particular function is unknown;}\\&\text{however, we have an expression for its derivative.}\\&\text{Knowing that }f'(x)=-14cos(4x)\text{ and }f(-1)=17\\&\text{Approximate } f(7.6)\text{ using Euler's Method and }2\text{ steps.}\end{align*}$

Possible Answers:

7.84

2.61

56.35

9.62

15.67

Correct answer:

7.84

Explanation:

$\begin{align*}&\text{The general form of Euler's method, when a derivative function,}\\&\text{initial value, and step size are known, is:}\\&y_n=y_{n+1} +\Delta x f'(x_n,y_n)\\&\text{To calculate the step size find the distance between the final}\\&\text{and initial value and divide by the number of steps to be used:}\\&\Delta x = \frac{x_f-x_i}{Steps}\\&\text{For this problem, we are told: }f'(x)=-14cos(4x);f(-1)=17\\&\text{Knowing this, we may take the steps to estimate our function}\\&\text{value at our final x value:}\\&\Delta x = \frac{7.6-(-1)}{2}=4.3\\&y_0=17;x_0=-1\\&y_{1}=17+(4.3)(-14cos(4{(-1)}))=56.35\\&y_{2}=56.35+(4.3)(-14cos(4{(3.3)}))=7.84\\&\\&\end{align*}$

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Example Question #221 : Derivatives

$\begin{align*}&\text{The expression of a particular function is unknown;}\\&\text{however, we have an expression for its derivative.}\\&\text{Knowing that }f'(x)=-6x^{2}\text{ and }f(-1)=-20\\&\text{Approximate } f(11.9)\text{ using Euler's Method and }3\text{ steps.}\end{align*}$

Possible Answers:

-1816.97

-363.39

-869.66

-45.8

-9084.85

Correct answer:

-1816.97

Explanation:

$\begin{align*}&\text{The general form of Euler's method, when a derivative function,}\\&\text{initial value, and step size are known, is:}\\&y_n=y_{n+1} +\Delta x f'(x_n,y_n)\\&\text{To calculate the step size find the distance between the final}\\&\text{and initial value and divide by the number of steps to be used:}\\&\Delta x = \frac{x_f-x_i}{Steps}\\&\text{For this problem, we are told: }f'(x)=-6x^{2};f(-1)=-20\\&\text{Knowing this, we may take the steps to estimate our function}\\&\text{value at our final x value:}\\&\Delta x = \frac{11.9-(-1)}{3}=4.3\\&y_0=-20;x_0=-1\\&y_{1}=-20+(4.3)(-6{(-1)}^{2})=-45.8\\&y_{2}=-45.8+(4.3)(-6{(3.3)}^{2})=-326.76\\&y_{3}=-326.76+(4.3)(-6{(7.6)}^{2})=-1816.97\\&\\&\end{align*}$

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Example Question #42 : Applications Of Derivatives

$\begin{align*}&\text{The expression of a particular function is unknown;}\\&\text{however, we have an expression for its derivative.}\\&\text{Knowing that }f'(x)=-2cos(6x)\text{ and }f(-1)=-11\\&\text{Approximate } f(6.2)\text{ using Euler's Method and }4\text{ steps.}\end{align*}$

Possible Answers:

-36.81

-6.13

-12.27

-14.46

-12.76

Correct answer:

-12.27

Explanation:

$\begin{align*}&\text{The general form of Euler's method, when a derivative function,}\\&\text{initial value, and step size are known, is:}\\&y_n=y_{n+1} +\Delta x f'(x_n,y_n)\\&\text{To calculate the step size find the distance between the final}\\&\text{and initial value and divide by the number of steps to be used:}\\&\Delta x = \frac{x_f-x_i}{Steps}\\&\text{For this problem, we are told: }f'(x)=-2cos(6x);f(-1)=-11\\&\text{Knowing this, we may take the steps to estimate our function}\\&\text{value at our final x value:}\\&\Delta x = \frac{6.2-(-1)}{4}=1.8\\&y_0=-11;x_0=-1\\&y_{1}=-11+(1.8)(-2cos(6{(-1)}))=-14.46\\&y_{2}=-14.46+(1.8)(-2cos(6{(0.8)}))=-14.77\\&y_{3}=-14.77+(1.8)(-2cos(6{(2.6)}))=-11.19\\&y_{4}=-11.19+(1.8)(-2cos(6{(4.4)}))=-12.27\\&\\&\end{align*}$

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AP Calculus BC : Applications of Derivatives

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #21 : Euler's Method And L'hopital's Rule

Example Question #22 : Euler's Method And L'hopital's Rule

Example Question #23 : Euler's Method And L'hopital's Rule

Example Question #24 : Euler's Method And L'hopital's Rule

Example Question #45 : Applications Of Derivatives

Example Question #41 : Applications Of Derivatives

Example Question #47 : Applications Of Derivatives

Example Question #221 : Derivatives

Example Question #42 : Applications Of Derivatives

All AP Calculus BC Resources

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