Calculus Examples

$\frac{d r}{d t} = 6 t + \sec^{2} (t)$ , $r (- π) = 5$

Step 1

Rewrite the equation.

$d r = (6 t + \sec^{2} (t)) d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d r = \int 6 t + \sec^{2} (t) d t$

Step 2.2

Apply the constant rule.

$r + C_{1} = \int 6 t + \sec^{2} (t) d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$r + C_{1} = \int 6 t d t + \int \sec^{2} (t) d t$

Step 2.3.2

Since $6$ is constant with respect to $t$ , move $6$ out of the integral.

$r + C_{1} = 6 \int t d t + \int \sec^{2} (t) d t$

Step 2.3.3

By the Power Rule, the integral of $t$ with respect to $t$ is $\frac{1}{2} t^{2}$ .

$r + C_{1} = 6 (\frac{1}{2} t^{2} + C_{2}) + \int \sec^{2} (t) d t$

Step 2.3.4

Since the derivative of $\tan (t)$ is $\sec^{2} (t)$ , the integral of $\sec^{2} (t)$ is $\tan (t)$ .

$r + C_{1} = 6 (\frac{1}{2} t^{2} + C_{2}) + \tan (t) + C_{3}$

Step 2.3.5

Simplify.

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Step 2.3.5.1

Combine $\frac{1}{2}$ and $t^{2}$ .

$r + C_{1} = 6 (\frac{t^{2}}{2} + C_{2}) + \tan (t) + C_{3}$

Step 2.3.5.2

Simplify.

$r + C_{1} = 3 t^{2} + \tan (t) + C_{4}$

Step 2.4

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

$r = 3 t^{2} + \tan (t) + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $- π$ for $t$ and $5$ for $r$ in $r = 3 t^{2} + \tan (t) + K$ .

$5 = 3 {(- π)}^{2} + \tan (- π) + K$

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $3 {(- π)}^{2} + \tan (- π) + K = 5$ .

$3 {(- π)}^{2} + \tan (- π) + K = 5$

Step 4.2

Simplify the left side.

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Step 4.2.1

Simplify $3 {(- π)}^{2} + \tan (- π) + K$ .

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Step 4.2.1.1

Simplify each term.

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Step 4.2.1.1.1

Apply the product rule to $- π$ .

$3 ({(- 1)}^{2} π^{2}) + \tan (- π) + K = 5$

Step 4.2.1.1.2

Raise $- 1$ to the power of $2$ .

$3 (1 π^{2}) + \tan (- π) + K = 5$

Step 4.2.1.1.3

Multiply $π^{2}$ by $1$ .

$3 π^{2} + \tan (- π) + K = 5$

Step 4.2.1.1.4

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because tangent is negative in the second quadrant.

$3 π^{2} - \tan (0) + K = 5$

Step 4.2.1.1.5

The exact value of $\tan (0)$ is $0$ .

$3 π^{2} - 0 + K = 5$

Step 4.2.1.1.6

Multiply $- 1$ by $0$ .

$3 π^{2} + 0 + K = 5$

Step 4.2.1.2

Add $3 π^{2}$ and $0$ .

$3 π^{2} + K = 5$

Step 4.3

Subtract $3 π^{2}$ from both sides of the equation.

$K = 5 - 3 π^{2}$

Step 5

Substitute $5 - 3 π^{2}$ for $K$ in $r = 3 t^{2} + \tan (t) + K$ and simplify.

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Step 5.1

Substitute $5 - 3 π^{2}$ for $K$ .

$r = 3 t^{2} + \tan (t) + 5 - 3 π^{2}$