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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #10 : Euler's Method

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=y^2e^{x}$ at x=6 with the initial condition y(0)=0.01 and step size h= 0.5 .

Possible Answers:

(6,0.013)

(6,0.012)

(6,5.113)

(6,0.1089)

Correct answer:

(6,5.113)

Explanation:

Euler's method uses iterative equations to find a numerical solution to a differential equation. The following equations

$x_{n+1}=x_n+h$

$y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value. f(x_n,y_n) is the solution to the differential equation.

In this problem,

$f(x,y)=\frac{dy}{dx}=y^2e^{x}$

h= 0.5

Starting at the initial point (x_0,y_0)=(0,0.01)

$x_{1}=x_0+h=0+0.5=0.5$

$y_{1}=y_0+h*f(x_0,y_0)=0.01+0.5*(0.01)^2*e^{0}=0.01005$

(x_1,y_1)=(0.5,0.01005)

$x_{2}=x_1+h=0.5+0.5=1$

$y_{2}=y_1+h*f(x_1,y_1)=0.01005+0.5*(0.01005)*e^{1}=0.01013$

(x_2,y_2)=(1,0.01013)

We continue using Euler's method until x=6 . The results of Euler's method are in the table below.

Problem 12

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Example Question #11 : Euler's Method

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=ln\left (\frac{y}{x} \right )$ at x=3 with the initial condition y(1)=1 and step size h=0.5 .

Possible Answers:

(3,3)

(3,1)

(3,2.75)

(3,2)

Correct answer:

(3,2)

Explanation:

Euler's method uses iterative equations to find a numerical solution to a differential equation. The following equations

$x_{n+1}=x_n+h$

$y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value. f(x_n,y_n) is the solution to the differential equation.

In this problem,

$f(x)=\frac{dy}{dx}=ln\left (\frac{y}{x} \right )$

h=0.5

Starting at the initial point (x_0,y_0)=(1,1)

$x_{1}=x_0+h=1+0.5=1.5$

$y_{1}=y_0+h*f(x_0,y_0)=1+0.5*ln(1/1)=1$

(x_1,y_1)=(1.5,1)

$x_{2}=x_1+h=1.5+0.5=2$

$y_{2}=y_1+h*f(x_1,y_1)=1+0.5*ln(1/1.5)=1.333$

(x_2,y_2)=(2,1.333)

We continue using Euler's method until x=3 . The results of Euler's method are in the table below.

Problem 5

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Example Question #12 : Euler's Method

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=ln\left (\frac{y}{x} \right )$ at x=3 with the initial condition y(1)=1 and step size h=0.2 .

Possible Answers:

(3,2.5)

(3,2)

(3,1.25)

(3,3)

Correct answer:

(3,2.5)

Explanation:

Euler's method uses iterative equations to find a numerical solution to a differential equation. The following equations

$x_{n+1}=x_n+h$

$y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value. f(x_n,y_n) is the solution to the differential equation.

In this problem,

$f(x)=\frac{dy}{dx}=ln\left (\frac{y}{x} \right )$

h=0.2

Starting at the initial point (x_0,y_0)=(1,1)

$x_{1}=x_0+h=1+0.2=1.2$

$y_{1}=y_0+h*f(x_0,y_0)=1+0.2*ln(1/1)=1$

(x_1,y_1)=(1.2,1)

$x_{2}=x_1+h=1.2+0.2=1.4$

$y_{2}=y_1+h*f(x_1,y_1)=1+0.2*ln(1/1.2)=1.167$

(x_2,y_2)=(1.4,1.167)

We continue using Euler's method until x=3 . The results of Euler's method are in the table below.

Problem 6

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Example Question #13 : Euler's Method

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=e^{x}$ at x=5 with the initial condition y(0)=1 and step size h= 1 .

Possible Answers:

(5,32.19)

(5,86.79)

(5,65.49)

(5,94.77)

Correct answer:

(5,86.79)

Explanation:

Euler's method uses iterative equations to find a numerical solution to a differential equation. The following equations

$x_{n+1}=x_n+h$

$y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value. f(x_n,y_n) is the solution to the differential equation.

In this problem,

$f(x,y)=\frac{dy}{dx}=e^{x}$

h= 1

Starting at the initial point (x_0,y_0)=(0,1)

$x_{1}=x_0+h=0+1=1$

$y_{1}=y_0+h*f(x_0,y_0)=1+1*e^{0}=2$

(x_1,y_1)=(1,2)

$x_{2}=x_1+h=1+1=2$

$y_{2}=y_1+h*f(x_1,y_1)=2+1*e^1=4.718$

(x_2,y_2)=(2,4.718)

We continue using Euler's method until x=5 . The results of Euler's method are in the table below.

Problem 9

Note: Using a step size of gives an answer that is not close to the correct answer, y(5)=148.413 . This is because the step size is too small.

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Example Question #14 : Euler's Method

Use Euler's method to find the solution to the differential equation $\frac{dy}{dx}=e^{x}$ at x=5 with the initial condition y(0)=1 and step size h= 0.5 .

Possible Answers:

(5,129.26)

(5,114.62)

(5,132.77)

(5,69.61)

Correct answer:

(5,114.62)

Explanation:

Euler's method uses iterative equations to find a numerical solution to a differential equation. The following equations

$x_{n+1}=x_n+h$

$y_{n+1}=y_n+h*f(x_n,y_n)$

are solved starting at the initial condition and ending at the desired value. f(x_n,y_n) is the solution to the differential equation.

In this problem,

$f(x,y)=\frac{dy}{dx}=e^{x}$

h= 1

Starting at the initial point (x_0,y_0)=(0,1)

$x_{1}=x_0+h=0+0.5=0.5$

$y_{1}=y_0+h*f(x_0,y_0)=1+0.5*e^{0}=1.5$

(x_1,y_1)=(0.5,1.5)

$x_{2}=x_1+h=0.5+0.5=1$

$y_{2}=y_1+h*f(x_1,y_1)=1.5+0.5*e^{0.5}=2.324$

(x_2,y_2)=(1,2.324)

We continue using Euler's method until x=5 . The results of Euler's method are in the table below.

Problem 10

Note: Using a larger step size got us closer the correct answer, y(5)=148.413 .

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Example Question #1651 : Calculus Ii

Euler's explicit method is often used to find a numerical solution to a differential equation.Given the step size and initial condition, it utilizes the tangent line equation to find the consecutive value of the solution function.

Find an approximation to a given differential equation using Euler's explicit method:

$\frac{dy}{dx}=y-x$

Use step size $\Delta x=1$ , and range of from 0 to 3.

Initial condition is .

In the answer there should be three consecutive values of solution function for given range and step size of x.

Possible Answers:

$\newline y(1)=2\newline y(2)=7\newline y(3)=12$

$\newline y(1)=4\newline y(2)=9\newline y(3)=17$

$\newline y(1)=2\newline y(2)=5\newline y(3)=7$

$\newline y(1)=1\newline y(2)=3\newline y(3)=5$

$\newline y(1)=2\newline y(2)=5\newline y(3)=12$

Correct answer:

$\newline y(1)=2\newline y(2)=5\newline y(3)=12$

Explanation:

Recall formula of the tangent line to a two-dimensional function from Calculus 1:

$y(x)=y(x_0)+y^{'}(x_0)(x-x_0);$

For the Euler's method, this equation can be rewritten as:

$y(i+1)=y(i)+y'(i)\Delta x;$

Where is the next value of the solution function, is the derivative of the previous step.

We are given a differential equation $\frac{dy}{dx}=y-x$ . Therefore, if we know the value of and , we can find the . Consecutively, we can plug it into the Euler's method equation and find the .

Here is how this process is executed for the y(1):

1) Identify the initial condition and step size. This is the only information that Euler's method uses to calculate the approximation of solution function, other than the equation itself. In this case, we can say, that for the first step,

$\newline y(i)=y(0)=1;\newline x(i)=x(0)=0;$

Note, that for the whole process, step size remains the same, $\Delta x=1$

2) Calculate the derivative of function at point using given differential equation. In this case,

$\frac{dy}{dx}(i)=y(i)+x(i)=y(0)+0=1$

3) Note that in step 2 we found . Now we have all components to plug in to the Euler's method formula. Again, for our first step:

$y(i+1)=y(1)=y(i)+y'(i)\Delta x=1+1*1=2$

The same algorithm is used for y(2) and y(3). Note, that if we know step size and range of x, we will always know . For each consecutive step, is the value of the solution function, found in the previous step.

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Example Question #1 : Euler's Method And L'hopital's Rule

Evaluate:

$\lim_{x\rightarrow \infty } \frac{x \sin^{2} x}{2^{x} - 1}$

Possible Answers:

The limit does not exist.

$\infty$

$\frac{1}{2}$

Correct answer:

Explanation:

Let's examine the limit

$\lim_{x\rightarrow \infty } \frac{x }{2^{x} - 1}$

first.

$\lim_{x\rightarrow \infty }x= \lim_{x\rightarrow \infty } \left ( 2^{x} - 1 \right )= \infty$

$\frac{\mathrm{d} }{\mathrm{d} x}x = 1$

and

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 2^{x} - 1 \right ) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( e^{ x \ln 2} - 1 \right )$

$= \frac{\mathrm{d} }{\mathrm{d} x}e^{ x \ln 2} - \frac{\mathrm{d} }{\mathrm{d} x} 1$

$=\ln 2 \cdot e^{ x \ln 2} - 0 = \ln 2 \cdot 2^{x}$ ,

so by L'Hospital's Rule,

$\lim_{x\rightarrow \infty } \frac{x }{2^{x} - 1} = \lim_{x\rightarrow \infty } \frac{1 }{\ln 2 \cdot 2^{x}}$

Since $\lim_{x\rightarrow \infty } \ln 2 \cdot 2^{x} = \infty$ ,

$\lim_{x\rightarrow \infty } \frac{x }{2^{x} - 1} = \lim_{x\rightarrow \infty } \frac{1 }{\ln 2 \cdot 2^{x}} = 0$

Now, for each x > 0 , $0 \leq \sin^{2} x \leq 1$ ; therefore,

$0 \leq \frac{x \sin^{2} x}{2^{x} - 1} \leq \frac{x }{2^{x} - 1}$

By the Squeeze Theorem,

$0 \leq \lim_{x\rightarrow \infty } \frac{x \sin^{2} x}{2^{x} - 1} \leq \lim_{x\rightarrow \infty } \frac{x }{2^{x} - 1} = 0$

and

$\lim_{x\rightarrow \infty } \frac{x \sin^{2} x}{2^{x} - 1} = 0$

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Example Question #1 : L'hospital's Rule

Evaluate:

$\lim_{x\rightarrow \infty } \frac{5^{x} - 1}{3^{x} - 1}$

Possible Answers:

The limit does not exist.

$\infty$

$\frac{5}{3}$

$\frac{\ln 5}{\ln 3}$

$\ln \frac{5}{3}$

Correct answer:

$\infty$

Explanation:

$\lim_{x\rightarrow \infty }\left ( 5^{x} - 1 \right ) = \lim_{x\rightarrow \infty }\left ( 3^{x} - 1 \right ) = \infty$

Therefore, by L'Hospital's Rule, we can find $\lim_{x\rightarrow \infty } \frac{5^{x} - 1}{3^{x} - 1}$ by taking the derivatives of the expressions in both the numerator and the denominator:

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 5^{x} - 1 \right ) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( e^{x \ln 5} - 1 \right )$

$= \frac{\mathrm{d} }{\mathrm{d} x} e^{x \ln 5 } - \frac{\mathrm{d} }{\mathrm{d} x}1$

$=\ln 5\cdot e^{x \ln 5 } - 0 = \ln 5\cdot 5^{x }$

Similarly,

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 3^{x} - 1 \right ) = \ln 3 \cdot 3 ^{x}$

$\lim_{x\rightarrow \infty } \frac{5^{x} - 1}{3^{x} - 1} = \lim_{x\rightarrow \infty } \frac{\ln 5 \cdot 5^{x} }{\ln 3 \cdot 3^{x} }$

$=\frac{\ln 5 }{\ln 3 } \cdot \lim_{x\rightarrow \infty } \frac{ 5^{x} }{ 3^{x} }= \frac{\ln 5 }{\ln 3 } \cdot \lim_{x\rightarrow \infty }\left ( \frac{ 5 }{ 3 } \right )^{x}$

But $\lim_{x\rightarrow \infty }a^{x} = \infty$ for any a > 1 , so

$\lim_{x\rightarrow \infty } \frac{5^{x} - 1}{3^{x} - 1}= \frac{\ln 5 }{\ln 3 } \cdot \lim_{x\rightarrow \infty }\left ( \frac{ 5 }{ 3 } \right )^{x} = \infty$

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Example Question #2 : Euler's Method And L'hopital's Rule

Evaluate:

$\lim_{x\rightarrow 0 } \frac{7^{x} - 1}{2^{x} - 1}$

Possible Answers:

The limit does not exist.

$\infty$

$\log_{2}7$

$\ln \frac{7}{2}$

$\frac{1}{2}\ln 7$

Correct answer:

$\log_{2}7$

Explanation:

$\lim_{x\rightarrow 0 }\left ( 7^{x} - 1 \right ) = 7^{0} - 1 = 1- 1 = 0$

and

$\lim_{x\rightarrow 0 }\left ( 2^{x} - 1 \right ) = 2^{0} - 1 = 1- 1 = 0$

Therefore, by L'Hospital's Rule, we can find $\lim_{x\rightarrow 0 } \frac{7^{x} - 1}{2^{x} - 1}$ by taking the derivatives of the expressions in both the numerator and the denominator:

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 7^{x} - 1 \right ) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( e^{x \ln 7} - 1 \right )$

$= \frac{\mathrm{d} }{\mathrm{d} x} e^{x \ln 7 } - \frac{\mathrm{d} }{\mathrm{d} x}1$

$=\ln 7\cdot e^{x \ln 7 } - 0 = \ln 7\cdot 7^{x }$

Similarly,

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 2^{x} - 1 \right ) = \ln 2 \cdot 2 ^{x}$

$\lim_{x\rightarrow 0 } \frac{7^{x} - 1}{2^{x} - 1} = \lim_{x\rightarrow 0 } \frac{ \ln 7\cdot 7^{x }}{ \ln 2\cdot 2^{x }}$

$= \frac{ \ln 7 \cdot 7^{0 }}{ \ln 2\cdot 2^{0 }} = \frac{ \ln 7\cdot 1}{ \ln 2\cdot 1} = \frac{ \ln 7}{ \ln 2 }= \log_{2}7$

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Example Question #3 : Euler's Method And L'hopital's Rule

Evaluate:

$\lim_{x\rightarrow 0^{+} } \frac{x \sin x}{2^{x} - 1}$

Possible Answers:

$\ln 2$

$\frac{1}{\ln2}$

Correct answer:

Explanation:

$\lim_{x\rightarrow 0^{+} } x \sin x= 0 \cdot \sin 0 = 0$

and

$\lim_{x\rightarrow 0^{+} } \left ( 2 ^{x} - 1 \right )= 2^{0} -1 = 0$

Therefore, by L'Hospital's Rule, we can find $\lim_{x\rightarrow 0^{+} } \frac{x \sin x}{2^{x} - 1}$ by taking the derivatives of the expressions in both the numerator and the denominator:

$\frac{\mathrm{d} }{\mathrm{d} x}x \sin x = \frac{\mathrm{d} }{\mathrm{d} x} x \cdot \sin x + \frac{\mathrm{d} }{\mathrm{d} x}\sin x \cdot x$

$= 1 \cdot \sin x + \cos x \cdot x = \sin x + x \cos x$

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( 2^{x} - 1 \right ) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( e^{x \ln 2} - 1 \right )$

$= \frac{\mathrm{d} }{\mathrm{d} x} e^{x \ln 2} - \frac{\mathrm{d} }{\mathrm{d} x} 1$

$= \ln 2\cdot e^{x \ln 2} - 0 = \ln 2\cdot 2^{x }$

$\lim_{x\rightarrow 0^{+} } \frac{x \sin x}{2^{x} - 1} = \lim_{x\rightarrow 0^{+} } \frac{\sin x + x \cos x}{ \ln 2\cdot 2^{x }}$

$= \frac{\sin 0 + 0 \cos 0}{ \ln 2\cdot 2^{0 }} = \frac{ 0 + 0 }{ \ln 2\cdot 1} = 0$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #10 : Euler's Method

Example Question #11 : Euler's Method

Example Question #12 : Euler's Method

Example Question #13 : Euler's Method

Example Question #14 : Euler's Method

Example Question #1651 : Calculus Ii

Example Question #1 : Euler's Method And L'hopital's Rule

Example Question #1 : L'hospital's Rule

Example Question #2 : Euler's Method And L'hopital's Rule

Example Question #3 : Euler's Method And L'hopital's Rule

All Calculus 2 Resources

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