Calculus Examples

$\frac{d v}{d t} = \frac{3}{t \sqrt{t^{2} - 1}}$ , $v (2) = 0$

Step 1

Rewrite the equation.

$d v = \frac{3}{t \sqrt{t^{2} - 1}} d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d v = \int \frac{3}{t \sqrt{t^{2} - 1}} d t$

Step 2.2

Apply the constant rule.

$v + C_{1} = \int \frac{3}{t \sqrt{t^{2} - 1}} d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

Since $3$ is constant with respect to $t$ , move $3$ out of the integral.

$v + C_{1} = 3 \int \frac{1}{t \sqrt{t^{2} - 1}} d t$

Step 2.3.2

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

$v + C_{1} = 3 \int \frac{1}{t \sqrt{t^{2} - 1^{2}}} d t$

Step 2.3.3

The integral of $\frac{1}{t \sqrt{t^{2} - 1^{2}}}$ with respect to $t$ is $arcsec (| t |)$

$v + C_{1} = 3 (arcsec (| t |) + C_{2})$

Step 2.3.4

Simplify.

$v + C_{1} = 3 arcsec (| t |) + C_{2}$

Step 2.4

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

$v = 3 arcsec (| t |) + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $2$ for $t$ and $0$ for $v$ in $v = 3 arcsec (| t |) + K$ .

$0 = 3 arcsec (| 2 |) + K$

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $3 arcsec (| 2 |) + K = 0$ .

$3 arcsec (| 2 |) + K = 0$

Step 4.2

Simplify the left side.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$3 arcsec (2) + K = 0$

Step 4.2.1.2

The exact value of $arcsec (2)$ is $\frac{π}{3}$ .

$3 \frac{π}{3} + K = 0$

Step 4.2.1.3

Cancel the common factor of $3$ .

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Step 4.2.1.3.1

Cancel the common factor.

$3 \frac{π}{3} + K = 0$

Step 4.2.1.3.2

Rewrite the expression.

$π + K = 0$

Step 4.3

Subtract $π$ from both sides of the equation.

$K = - π$

Step 5

Substitute $- π$ for $K$ in $v = 3 arcsec (| t |) + K$ and simplify.

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Step 5.1

Substitute $- π$ for $K$ .

$v = 3 arcsec (| t |) - π$