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Calculus 1 : Calculus

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Example Questions

Example Question #721 : Differential Functions

Given that $\frac{dy}{dx}=\sin(x^2)$ $\displaystyle \frac{dy}{dx}=\sin(x^2)$ and y(1)=4 $\displaystyle y(1)=4$, use Euler's method to approximate y(2) $\displaystyle y(2)$ using three steps.

Possible Answers:

4.72 $\displaystyle 4.72$

4.61 $\displaystyle 4.61$

5.13 $\displaystyle 5.13$

4.44 $\displaystyle 4.44$

4.28 $\displaystyle 4.28$

Correct answer:

4.72 $\displaystyle 4.72$

Explanation:

When using Euler's method, the first step is to calculate step size:

$\Delta x =\frac{2-1}{3}=\frac{1}{3}$ $\displaystyle \Delta x =\frac{2-1}{3}=\frac{1}{3}$

Now, to approximate function values using Euler's method, utilize the following formula:

$y_{n+1}=y_n + \Delta x \cdot f'(x_n,y_n)$ $\displaystyle y_{n+1}=y_n + \Delta x \cdot f'(x_n,y_n)$

After that, it's merely a matter of taking the steps:

y_0=4 $\displaystyle y_0=4$

$y_1=4+\frac{1}{3}\sin(1^2)=4.28$ $\displaystyle y_1=4+\frac{1}{3}\sin(1^2)=4.28$

$y_2=4.28+\frac{1}{3}\sin((\frac{4}{3})^2)=4.61$ $\displaystyle y_2=4.28+\frac{1}{3}\sin((\frac{4}{3})^2)=4.61$

$y_3=4.61+\frac{1}{3}\sin((\frac{5}{3})^2)=4.72$ $\displaystyle y_3=4.61+\frac{1}{3}\sin((\frac{5}{3})^2)=4.72$

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Example Question #722 : Functions

Given that $\frac{dy}{dx}=4\sin(2\pi x)$ $\displaystyle \frac{dy}{dx}=4\sin(2\pi x)$ and y(1)=3 $\displaystyle y(1)=3$, use Euler's method to approximate y(2.2) $\displaystyle y(2.2)$ using three steps.

Possible Answers:

2.42 $\displaystyle 2.42$

1.68 $\displaystyle 1.68$

3.94 $\displaystyle 3.94$

5.46 $\displaystyle 5.46$

$\displaystyle 3$

Correct answer:

2.42 $\displaystyle 2.42$

Explanation:

When using Euler's method, the first step is to calculate step size:

$\Delta x =\frac{2.2-1}{3}=0.4$ $\displaystyle \Delta x =\frac{2.2-1}{3}=0.4$

Now, to approximate function values using Euler's method, utilize the following formula:

$y_{n+1}=y_n + \Delta x \cdot f'(x_n,y_n)$ $\displaystyle y_{n+1}=y_n + \Delta x \cdot f'(x_n,y_n)$

After that, it's merely a matter of taking the steps:

y_0=3 $\displaystyle y_0=3$

$y_1=3 +0.4(4sin(2\pi (1)))=3$ $\displaystyle y_1=3 +0.4(4sin(2\pi (1)))=3$

$y_2=3+0.4(4sin(2\pi (1.4)))=3.94$ $\displaystyle y_2=3+0.4(4sin(2\pi (1.4)))=3.94$

$y_3=3.94+0.4(4sin(2\pi (1.8)))=2.42$ $\displaystyle y_3=3.94+0.4(4sin(2\pi (1.8)))=2.42$

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Example Question #721 : Functions

Given that $\frac{dy}{dx}=3^{x^2}$ $\displaystyle \frac{dy}{dx}=3^{x^2}$ and y(1)=6 $\displaystyle y(1)=6$, use Euler's method to approximate y(1.3) $\displaystyle y(1.3)$ using three steps.

Possible Answers:

6.92 $\displaystyle 6.92$

7.56 $\displaystyle 7.56$

6.68 $\displaystyle 6.68$

7.17 $\displaystyle 7.17$

7.08 $\displaystyle 7.08$

Correct answer:

7.17 $\displaystyle 7.17$

Explanation:

When using Euler's method, the first step is to calculate step size:

$\Delta x =\frac{1.3-1}{3}=0.1$ $\displaystyle \Delta x =\frac{1.3-1}{3}=0.1$

Now, to approximate function values using Euler's method, utilize the following formula:

$y_{n+1}=y_n + \Delta x \cdot f'(x_n,y_n)$ $\displaystyle y_{n+1}=y_n + \Delta x \cdot f'(x_n,y_n)$

After that, it's merely a matter of taking the steps:

y_0=6 $\displaystyle y_0=6$

$y_1=6+0.1(3^{1^2})=6.3$ $\displaystyle y_1=6+0.1(3^{1^2})=6.3$

$y_2=6.3+0.1(3^{1.1^2})=6.68$ $\displaystyle y_2=6.3+0.1(3^{1.1^2})=6.68$

$y_3=6.68+0.1(3^{1.2^2})=7.17$ $\displaystyle y_3=6.68+0.1(3^{1.2^2})=7.17$

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Example Question #531 : How To Find Differential Functions

Given that $\frac{dy}{dx}=e^{e^{e^{x}}}$ $\displaystyle \frac{dy}{dx}=e^{e^{e^{x}}}$ and y(0)=0 $\displaystyle y(0)=0$, use Euler's method to approximate y(0.02) $\displaystyle y(0.02)$ using two steps.

Possible Answers:

$\displaystyle .132$

7.119 $\displaystyle 7.119$

0.308 $\displaystyle 0.308$

30.776 $\displaystyle 30.776$

13.24 $\displaystyle 13.24$

Correct answer:

0.308 $\displaystyle 0.308$

Explanation:

When using Euler's method, the first step is to calculate step size:

$\Delta x =\frac{0.02-0}{2}=0.01$ $\displaystyle \Delta x =\frac{0.02-0}{2}=0.01$

Now, to approximate function values using Euler's method, utilize the following formula:

$y_{n+1}=y_n + \Delta x \cdot f'(x_n,y_n)$ $\displaystyle y_{n+1}=y_n + \Delta x \cdot f'(x_n,y_n)$

After that, it's merely a matter of taking the steps:

$\frac{dy}{dx}=e^{e^{e^{x}}}$ $\displaystyle \frac{dy}{dx}=e^{e^{e^{x}}}$

$\displaystyle y_0=0;x_0=0$

$y_1=0+0.01e^{e^{e^{0}}}=0.152$ $\displaystyle y_1=0+0.01e^{e^{e^{0}}}=0.152$

$y_2=0.152+0.01e^{e^{e^{0.01}}}=0.308$ $\displaystyle y_2=0.152+0.01e^{e^{e^{0.01}}}=0.308$

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Example Question #725 : Functions

Given that $\frac{dy}{dx}=4^{x^5}$ $\displaystyle \frac{dy}{dx}=4^{x^5}$ and y(1)=0 $\displaystyle y(1)=0$, use Euler's method to approximate y(1.2) $\displaystyle y(1.2)$ using four steps.

Possible Answers:

2.088 $\displaystyle 2.088$

1.772 $\displaystyle 1.772$

3.191 $\displaystyle 3.191$

$\displaystyle 2$

0.812 $\displaystyle 0.812$

Correct answer:

1.772 $\displaystyle 1.772$

Explanation:

When using Euler's method, the first step is to calculate step size:

$\Delta x =\frac{1.2-1}{4}=0.05$ $\displaystyle \Delta x =\frac{1.2-1}{4}=0.05$

Now, to approximate function values using Euler's method, utilize the following formula:

$y_{n+1}=y_n + \Delta x \cdot f'(x_n,y_n)$ $\displaystyle y_{n+1}=y_n + \Delta x \cdot f'(x_n,y_n)$

After that, it's merely a matter of taking the steps:

$\frac{dy}{dx}=4^{x^5}$ $\displaystyle \frac{dy}{dx}=4^{x^5}$

$\displaystyle y_0=0;x_0=1$

$y_1=0+0.05(4^{1^5})=0.2$ $\displaystyle y_1=0+0.05(4^{1^5})=0.2$

$y_2=0.2+0.05(4^{1.05^5})=0.493$ $\displaystyle y_2=0.2+0.05(4^{1.05^5})=0.493$

$y_3=0.493+0.05(4^{1.1^5})=0.959$ $\displaystyle y_3=0.493+0.05(4^{1.1^5})=0.959$

$y_4=0.959+0.05(4^{1.15^5})=1.772$ $\displaystyle y_4=0.959+0.05(4^{1.15^5})=1.772$

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Example Question #531 : How To Find Differential Functions

Given that $\frac{dy}{dx}=\sin(\cos(x))$ $\displaystyle \frac{dy}{dx}=\sin(\cos(x))$ and y(3)=3 $\displaystyle y(3)=3$, use Euler's method to approximate y(3.6) $\displaystyle y(3.6)$ using three steps.

Possible Answers:

2.665 $\displaystyle 2.665$

2.500 $\displaystyle 2.500$

3.112 $\displaystyle 3.112$

2.833 $\displaystyle 2.833$

3.254 $\displaystyle 3.254$

Correct answer:

2.500 $\displaystyle 2.500$

Explanation:

When using Euler's method, the first step is to calculate step size:

$\Delta x =\frac{3.6-3}{3}=0.2$ $\displaystyle \Delta x =\frac{3.6-3}{3}=0.2$

Now, to approximate function values using Euler's method, utilize the following formula:

$y_{n+1}=y_n + \Delta x \cdot f'(x_n,y_n)$ $\displaystyle y_{n+1}=y_n + \Delta x \cdot f'(x_n,y_n)$

After that, it's merely a matter of taking the steps:

y_0=3 $\displaystyle y_0=3$

$y_1=3+0.2\sin(\cos(3))=2.833$ $\displaystyle y_1=3+0.2\sin(\cos(3))=2.833$

$y_2=2.833+0.2\sin(\cos(3.2))=2.665$ $\displaystyle y_2=2.833+0.2\sin(\cos(3.2))=2.665$

$y_3=2.665+0.2\sin(\cos(3.4))=2.500$ $\displaystyle y_3=2.665+0.2\sin(\cos(3.4))=2.500$

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Example Question #541 : Other Differential Functions

Given that $\frac{dy}{dx}=\ln(x^3)$ $\displaystyle \frac{dy}{dx}=\ln(x^3)$ and y(1)=13 $\displaystyle y(1)=13$, use Euler's method to approximate y(1.9) $\displaystyle y(1.9)$ using three steps.

Possible Answers:

15.44 $\displaystyle 15.44$

13.66 $\displaystyle 13.66$

14.12 $\displaystyle 14.12$

$\displaystyle 13$

14.89 $\displaystyle 14.89$

Correct answer:

13.66 $\displaystyle 13.66$

Explanation:

When using Euler's method, the first step is to calculate step size:

$\Delta x =\frac{1.9-1}{3}=0.3$ $\displaystyle \Delta x =\frac{1.9-1}{3}=0.3$

Now, to approximate function values using Euler's method, utilize the following formula:

$y_{n+1}=y_n + \Delta x \cdot f'(x_n,y_n)$ $\displaystyle y_{n+1}=y_n + \Delta x \cdot f'(x_n,y_n)$

After that, it's merely a matter of taking the steps:

$\displaystyle y_0=13,x_0=1$

$y_1=13+0.3(\ln(1^3))=13$ $\displaystyle y_1=13+0.3(\ln(1^3))=13$

$y_2=13+0.3(\ln(1.3^3))=13.24$ $\displaystyle y_2=13+0.3(\ln(1.3^3))=13.24$

$y_3=13.24+0.3(\ln(1.6^3))=13.66$ $\displaystyle y_3=13.24+0.3(\ln(1.6^3))=13.66$

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Example Question #541 : Other Differential Functions

Given that $\frac{dy}{dx}=2^{\ln(x^4)}$ $\displaystyle \frac{dy}{dx}=2^{\ln(x^4)}$ and y(1)=3 $\displaystyle y(1)=3$, use Euler's method to approximate y(1.8) $\displaystyle y(1.8)$ using four steps.

Possible Answers:

4.46 $\displaystyle 4.46$

4.04 $\displaystyle 4.04$

5.22 $\displaystyle 5.22$

4.93 $\displaystyle 4.93$

4.78 $\displaystyle 4.78$

Correct answer:

4.78 $\displaystyle 4.78$

Explanation:

When using Euler's method, the first step is to calculate step size:

$\Delta x =\frac{1.8-1}{4}=0.2$ $\displaystyle \Delta x =\frac{1.8-1}{4}=0.2$

Now, to approximate function values using Euler's method, utilize the following formula:

$y_{n+1}=y_n + \Delta x \cdot f'(x_n,y_n)$ $\displaystyle y_{n+1}=y_n + \Delta x \cdot f'(x_n,y_n)$

After that, it's merely a matter of taking the steps:

$\displaystyle y_0=3;x_0=1$

$y_1=3+0.2(2^{\ln(1^4)})=3.2$ $\displaystyle y_1=3+0.2(2^{\ln(1^4)})=3.2$

$y_2=3.2+0.2(2^{\ln(1.2^4)})=3.53$ $\displaystyle y_2=3.2+0.2(2^{\ln(1.2^4)})=3.53$

$y_3=3.53+0.2(2^{\ln(1.4^4)})=4.04$ $\displaystyle y_3=3.53+0.2(2^{\ln(1.4^4)})=4.04$

$y_4=4.04+0.2(2^{\ln(1.6^4)})=4.78$ $\displaystyle y_4=4.04+0.2(2^{\ln(1.6^4)})=4.78$

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Example Question #722 : Functions

Given that $\frac{dy}{dx}=\ln(2^{x})$ $\displaystyle \frac{dy}{dx}=\ln(2^{x})$ and y(0)=6 $\displaystyle y(0)=6$, use Euler's method to approximate y(1.5) $\displaystyle y(1.5)$ using three steps.

Possible Answers:

6.347 $\displaystyle 6.347$

6.782 $\displaystyle 6.782$

6.520 $\displaystyle 6.520$

6.173 $\displaystyle 6.173$

6.999 $\displaystyle 6.999$

Correct answer:

6.520 $\displaystyle 6.520$

Explanation:

When using Euler's method, the first step is to calculate step size:

$\Delta x =\frac{1.5-0}{3}=0.5$ $\displaystyle \Delta x =\frac{1.5-0}{3}=0.5$

Now, to approximate function values using Euler's method, utilize the following formula:

$y_{n+1}=y_n + \Delta x \cdot f'(x_n,y_n)$ $\displaystyle y_{n+1}=y_n + \Delta x \cdot f'(x_n,y_n)$

After that, it's merely a matter of taking the steps:

$\frac{dy}{dx}=\ln(2^{x}),y_0=6,x_0=0$ $\displaystyle \frac{dy}{dx}=\ln(2^{x}),y_0=6,x_0=0$

$y_1=6+0.5(\ln(2^{0}))=6$ $\displaystyle y_1=6+0.5(\ln(2^{0}))=6$

$y_2=6+0.5(\ln(2^{0.5}))=6.173$ $\displaystyle y_2=6+0.5(\ln(2^{0.5}))=6.173$

$y_3=6.173+0.5(\ln(2^{1}))=6.520$ $\displaystyle y_3=6.173+0.5(\ln(2^{1}))=6.520$

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Example Question #541 : How To Find Differential Functions

Given that $\frac{dy}{dx}=\sin(x^2)$ $\displaystyle \frac{dy}{dx}=\sin(x^2)$ and y(0)=0 $\displaystyle y(0)=0$, use Euler's method to approximate y(2) $\displaystyle y(2)$ using five steps.

Possible Answers:

0.920 $\displaystyle 0.920$

0.763 $\displaystyle 0.763$

0.700 $\displaystyle 0.700$

1.424 $\displaystyle 1.424$

0.219 $\displaystyle 0.219$

Correct answer:

0.920 $\displaystyle 0.920$

Explanation:

When using Euler's method, the first step is to calculate step size:

$\Delta x =\frac{2-0}{5}=0.4$ $\displaystyle \Delta x =\frac{2-0}{5}=0.4$

Now, to approximate function values using Euler's method, utilize the following formula:

$y_{n+1}=y_n + \Delta x \cdot f'(x_n,y_n)$ $\displaystyle y_{n+1}=y_n + \Delta x \cdot f'(x_n,y_n)$

After that, it's merely a matter of taking the steps:

$\frac{dy}{dx}=\sin(x^2);y_0=0;x_0=0$ $\displaystyle \frac{dy}{dx}=\sin(x^2);y_0=0;x_0=0$

$y_1=0+0.4\sin(0^2)=0$ $\displaystyle y_1=0+0.4\sin(0^2)=0$

$y_2=0+0.4\sin(0.4^2)=0.064$ $\displaystyle y_2=0+0.4\sin(0.4^2)=0.064$

$y_3=0.064+0.4\sin(0.8^2)=0.303$ $\displaystyle y_3=0.064+0.4\sin(0.8^2)=0.303$

$y_4=0.303+0.4\sin(1.2^2)=0.700$ $\displaystyle y_4=0.303+0.4\sin(1.2^2)=0.700$

$y_5=0.700+0.4\sin(1.6^2)=0.920$ $\displaystyle y_5=0.700+0.4\sin(1.6^2)=0.920$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #721 : Differential Functions

Example Question #722 : Functions

Example Question #721 : Functions

Example Question #531 : How To Find Differential Functions

Example Question #725 : Functions

Example Question #531 : How To Find Differential Functions

Example Question #541 : Other Differential Functions

Example Question #541 : Other Differential Functions

Example Question #722 : Functions

Example Question #541 : How To Find Differential Functions

All Calculus 1 Resources

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