Calculus Examples

$\frac{d x}{d t} = 4 (x^{2} + 1)$ , $x (\frac{π}{4}) = 1$

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{x^{2} + 1}$ .

$\frac{1}{x^{2} + 1} \frac{d x}{d t} = \frac{1}{x^{2} + 1} (4 (x^{2} + 1))$

Step 1.2

Simplify.

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Step 1.2.1

Rewrite using the commutative property of multiplication.

$\frac{1}{x^{2} + 1} \frac{d x}{d t} = 4 \frac{1}{x^{2} + 1} (x^{2} + 1)$

Step 1.2.2

Combine $4$ and $\frac{1}{x^{2} + 1}$ .

$\frac{1}{x^{2} + 1} \frac{d x}{d t} = \frac{4}{x^{2} + 1} (x^{2} + 1)$

Step 1.2.3

Cancel the common factor of $x^{2} + 1$ .

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Step 1.2.3.1

Cancel the common factor.

$\frac{1}{x^{2} + 1} \frac{d x}{d t} = \frac{4}{x^{2} + 1} (x^{2} + 1)$

Step 1.2.3.2

Rewrite the expression.

$\frac{1}{x^{2} + 1} \frac{d x}{d t} = 4$

Step 1.3

Rewrite the equation.

$\frac{1}{x^{2} + 1} d x = 4 d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{x^{2} + 1} d x = \int 4 d t$

Step 2.2

Integrate the left side.

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Step 2.2.1

Simplify the expression.

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Step 2.2.1.1

Reorder $x^{2}$ and $1$ .

$\int \frac{1}{1 + x^{2}} d x = \int 4 d t$

Step 2.2.1.2

Rewrite $1$ as $1^{2}$ .

$\int \frac{1}{1^{2} + x^{2}} d x = \int 4 d t$

Step 2.2.2

The integral of $\frac{1}{1^{2} + x^{2}}$ with respect to $x$ is $\arctan (x) + C_{1}$ .

$\arctan (x) + C_{1} = \int 4 d t$

Step 2.3

Apply the constant rule.

$\arctan (x) + C_{1} = 4 t + C_{2}$

Step 2.4

Group the constant of integration on the right side as $K$ .

$\arctan (x) = 4 t + K$

Step 3

Take the inverse arctangent of both sides of the equation to extract $x$ from inside the arctangent.

$x = \tan (4 t + K)$

Step 4

Use the initial condition to find the value of $K$ by substituting $\frac{π}{4}$ for $t$ and $1$ for $x$ in $x = \tan (4 t + K)$ .

$1 = \tan (4 \frac{π}{4} + K)$

Step 5

Solve for $K$ .

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Step 5.1

Rewrite the equation as $\tan (4 (\frac{π}{4}) + K) = 1$ .

$\tan (4 (\frac{π}{4}) + K) = 1$

Step 5.2

Take the inverse tangent of both sides of the equation to extract $K$ from inside the tangent.

$4 (\frac{π}{4}) + K = \arctan (1)$

Step 5.3

Simplify the left side.

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Step 5.3.1

Cancel the common factor of $4$ .

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Step 5.3.1.1

Cancel the common factor.

$4 \frac{π}{4} + K = \arctan (1)$

Step 5.3.1.2

Rewrite the expression.

$π + K = \arctan (1)$

Step 5.4

Simplify the right side.

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Step 5.4.1

The exact value of $\arctan (1)$ is $\frac{π}{4}$ .

$π + K = \frac{π}{4}$

Step 5.5

Move all terms not containing $K$ to the right side of the equation.

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Step 5.5.1

Subtract $π$ from both sides of the equation.

$K = \frac{π}{4} - π$

Step 5.5.2

To write $- π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$K = \frac{π}{4} - π \cdot \frac{4}{4}$

Step 5.5.3

Combine $- π$ and $\frac{4}{4}$ .

$K = \frac{π}{4} + \frac{- π \cdot 4}{4}$

Step 5.5.4

Combine the numerators over the common denominator.

$K = \frac{π - π \cdot 4}{4}$

Step 5.5.5

Simplify the numerator.

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Step 5.5.5.1

Multiply $4$ by $- 1$ .

$K = \frac{π - 4 π}{4}$

Step 5.5.5.2

Subtract $4 π$ from $π$ .

$K = \frac{- 3 π}{4}$

Step 5.5.6

Move the negative in front of the fraction.

$K = - \frac{3 π}{4}$

Step 5.6

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$π + K = π + \frac{π}{4}$

Step 5.7

Solve for $K$ .

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Step 5.7.1

Simplify $π + \frac{π}{4}$ .

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Step 5.7.1.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$π + K = π \cdot \frac{4}{4} + \frac{π}{4}$

Step 5.7.1.2

Combine fractions.

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Step 5.7.1.2.1

Combine $π$ and $\frac{4}{4}$ .

$π + K = \frac{π \cdot 4}{4} + \frac{π}{4}$

Step 5.7.1.2.2

Combine the numerators over the common denominator.

$π + K = \frac{π \cdot 4 + π}{4}$

Step 5.7.1.3

Simplify the numerator.

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Step 5.7.1.3.1

Move $4$ to the left of $π$ .

$π + K = \frac{4 \cdot π + π}{4}$

Step 5.7.1.3.2

Add $4 π$ and $π$ .

$π + K = \frac{5 π}{4}$

Step 5.7.2

Move all terms not containing $K$ to the right side of the equation.

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Step 5.7.2.1

Subtract $π$ from both sides of the equation.

$K = \frac{5 π}{4} - π$

Step 5.7.2.2

To write $- π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$K = \frac{5 π}{4} - π \cdot \frac{4}{4}$

Step 5.7.2.3

Combine $- π$ and $\frac{4}{4}$ .

$K = \frac{5 π}{4} + \frac{- π \cdot 4}{4}$

Step 5.7.2.4

Combine the numerators over the common denominator.

$K = \frac{5 π - π \cdot 4}{4}$

Step 5.7.2.5

Simplify the numerator.

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Step 5.7.2.5.1

Multiply $4$ by $- 1$ .

$K = \frac{5 π - 4 π}{4}$

Step 5.7.2.5.2

Subtract $4 π$ from $5 π$ .

$K = \frac{π}{4}$

Step 5.8

Find the period of $\tan (4 (\frac{π}{4}) + K)$ .

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Step 5.8.1

The period of the function can be calculated using $\frac{π}{| b |}$ .

$\frac{π}{| b |}$

Step 5.8.2

Replace $b$ with $1$ in the formula for period.

$\frac{π}{| 1 |}$

Step 5.8.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Step 5.8.4

Divide $π$ by $1$ .

$π$

Step 5.9

Add $π$ to every negative angle to get positive angles.

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Step 5.9.1

Add $π$ to $- \frac{3 π}{4}$ to find the positive angle.

$- \frac{3 π}{4} + π$

Step 5.9.2

To write $π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$π \cdot \frac{4}{4} - \frac{3 π}{4}$

Step 5.9.3

Combine fractions.

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Step 5.9.3.1

Combine $π$ and $\frac{4}{4}$ .

$\frac{π \cdot 4}{4} - \frac{3 π}{4}$

Step 5.9.3.2

Combine the numerators over the common denominator.

$\frac{π \cdot 4 - 3 π}{4}$

Step 5.9.4

Simplify the numerator.

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Step 5.9.4.1

Move $4$ to the left of $π$ .

$\frac{4 \cdot π - 3 π}{4}$

Step 5.9.4.2

Subtract $3 π$ from $4 π$ .

$\frac{π}{4}$

Step 5.9.5

List the new angles.

$K = \frac{π}{4}$

Step 5.10

The period of the $\tan (4 (\frac{π}{4}) + K)$ function is $π$ so values will repeat every $π$ radians in both directions.

$K = \frac{π}{4} + π n, \frac{π}{4} + π n$ , for any integer $n$

Step 5.11

Consolidate the answers.

$K = \frac{π}{4} + π n$

Step 6

Substitute $\frac{π}{4} + π n$ for $K$ in $x = \tan (4 t + K)$ and simplify.

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Step 6.1

Substitute $\frac{π}{4} + π n$ for $K$ .

$x = \tan (4 t + \frac{π}{4} + π n)$