Calculus Examples

$\frac{d v}{d t} = 16 t - \sec^{2} (t)$ , $v (π) = - 3$

Step 1

Rewrite the equation.

$d v = (16 t - \sec^{2} (t)) d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d v = \int 16 t - \sec^{2} (t) d t$

Step 2.2

Apply the constant rule.

$v + C_{1} = \int 16 t - \sec^{2} (t) d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$v + C_{1} = \int 16 t d t + \int - \sec^{2} (t) d t$

Step 2.3.2

Since $16$ is constant with respect to $t$ , move $16$ out of the integral.

$v + C_{1} = 16 \int t d t + \int - \sec^{2} (t) d t$

Step 2.3.3

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

$v + C_{1} = 16 (\frac{1}{2} t^{2} + C_{2}) + \int - \sec^{2} (t) d t$

Step 2.3.4

Since $- 1$ is constant with respect to $t$ , move $- 1$ out of the integral.

$v + C_{1} = 16 (\frac{1}{2} t^{2} + C_{2}) - \int \sec^{2} (t) d t$

Step 2.3.5

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

$v + C_{1} = 16 (\frac{1}{2} t^{2} + C_{2}) - (\tan (t) + C_{3})$

Step 2.3.6

Simplify.

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Step 2.3.6.1

Simplify.

$v + C_{1} = 16 (\frac{1}{2}) t^{2} - \tan (t) + C_{4}$

Step 2.3.6.2

Simplify.

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Step 2.3.6.2.1

Combine $16$ and $\frac{1}{2}$ .

$v + C_{1} = \frac{16}{2} t^{2} - \tan (t) + C_{4}$

Step 2.3.6.2.2

Cancel the common factor of $16$ and $2$ .

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Step 2.3.6.2.2.1

Factor $2$ out of $16$ .

$v + C_{1} = \frac{2 \cdot 8}{2} t^{2} - \tan (t) + C_{4}$

Step 2.3.6.2.2.2

Cancel the common factors.

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Step 2.3.6.2.2.2.1

Factor $2$ out of $2$ .

$v + C_{1} = \frac{2 \cdot 8}{2 (1)} t^{2} - \tan (t) + C_{4}$

Step 2.3.6.2.2.2.2

Cancel the common factor.

$v + C_{1} = \frac{2 \cdot 8}{2 \cdot 1} t^{2} - \tan (t) + C_{4}$

Step 2.3.6.2.2.2.3

Rewrite the expression.

$v + C_{1} = \frac{8}{1} t^{2} - \tan (t) + C_{4}$

Step 2.3.6.2.2.2.4

Divide $8$ by $1$ .

$v + C_{1} = 8 t^{2} - \tan (t) + C_{4}$

Step 2.4

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

$v = 8 t^{2} - \tan (t) + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $π$ for $t$ and $- 3$ for $v$ in $v = 8 t^{2} - \tan (t) + K$ .

$- 3 = 8 π^{2} - \tan (π) + K$

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $8 π^{2} - \tan (π) + K = - 3$ .

$8 π^{2} - \tan (π) + K = - 3$

Step 4.2

Simplify the left side.

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Step 4.2.1

Simplify $8 π^{2} - \tan (π) + K$ .

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Step 4.2.1.1

Simplify each term.

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Step 4.2.1.1.1

Apply the 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.

$8 π^{2} - - \tan (0) + K = - 3$

Step 4.2.1.1.2

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

$8 π^{2} - - 0 + K = - 3$

Step 4.2.1.1.3

Multiply $- - 0$ .

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Step 4.2.1.1.3.1

Multiply $- 1$ by $0$ .

$8 π^{2} - 0 + K = - 3$

Step 4.2.1.1.3.2

Multiply $- 1$ by $0$ .

$8 π^{2} + 0 + K = - 3$

Step 4.2.1.2

Add $8 π^{2}$ and $0$ .

$8 π^{2} + K = - 3$

Step 4.3

Subtract $8 π^{2}$ from both sides of the equation.

$K = - 3 - 8 π^{2}$

Step 5

Substitute $- 3 - 8 π^{2}$ for $K$ in $v = 8 t^{2} - \tan (t) + K$ and simplify.

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Step 5.1

Substitute $- 3 - 8 π^{2}$ for $K$ .

$v = 8 t^{2} - \tan (t) - 3 - 8 π^{2}$