Calculus Examples

$\frac{d s}{d t} = t^{- \frac{3}{2}}$ , $s (4) = - 1$

Step 1

Rewrite the equation.

$d s = t^{- \frac{3}{2}} d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d s = \int t^{- \frac{3}{2}} d t$

Step 2.2

Apply the constant rule.

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

Step 2.3

Integrate the right side.

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Step 2.3.1

By the Power Rule, the integral of $t^{- \frac{3}{2}}$ with respect to $t$ is $- 2 t^{- \frac{1}{2}}$ .

$s + C_{1} = - 2 t^{- \frac{1}{2}} + C_{2}$

Step 2.3.2

Simplify the answer.

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Step 2.3.2.1

Rewrite $- 2 t^{- \frac{1}{2}} + C_{2}$ as $- 2 \frac{1}{t^{\frac{1}{2}}} + C_{2}$ .

$s + C_{1} = - 2 \frac{1}{t^{\frac{1}{2}}} + C_{2}$

Step 2.3.2.2

Simplify.

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Step 2.3.2.2.1

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

$s + C_{1} = \frac{- 2}{t^{\frac{1}{2}}} + C_{2}$

Step 2.3.2.2.2

Move the negative in front of the fraction.

$s + C_{1} = - \frac{2}{t^{\frac{1}{2}}} + C_{2}$

Step 2.4

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

$s = - \frac{2}{t^{\frac{1}{2}}} + K$

Step 3

Use the initial condition to find the value of $K$ by substituting $4$ for $t$ and $- 1$ for $s$ in $s = - \frac{2}{t^{\frac{1}{2}}} + K$ .

$- 1 = - \frac{2}{4^{\frac{1}{2}}} + K$

Step 4

Solve for $K$ .

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Step 4.1

Rewrite the equation as $- \frac{2}{4^{\frac{1}{2}}} + K = - 1$ .

$- \frac{2}{4^{\frac{1}{2}}} + K = - 1$

Step 4.2

Simplify each term.

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Step 4.2.1

Simplify the denominator.

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Step 4.2.1.1

Rewrite $4$ as $2^{2}$ .

$- \frac{2}{{(2^{2})}^{\frac{1}{2}}} + K = - 1$

Step 4.2.1.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$- \frac{2}{2^{2 (\frac{1}{2})}} + K = - 1$

Step 4.2.1.3

Cancel the common factor of $2$ .

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Step 4.2.1.3.1

Cancel the common factor.

$- \frac{2}{2^{2 (\frac{1}{2})}} + K = - 1$

Step 4.2.1.3.2

Rewrite the expression.

$- \frac{2}{2^{1}} + K = - 1$

Step 4.2.1.4

Evaluate the exponent.

$- \frac{2}{2} + K = - 1$

Step 4.2.2

Cancel the common factor of $2$ .

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Step 4.2.2.1

Cancel the common factor.

$- \frac{2}{2} + K = - 1$

Step 4.2.2.2

Rewrite the expression.

$- 1 \cdot 1 + K = - 1$

Step 4.2.3

Multiply $- 1$ by $1$ .

$- 1 + K = - 1$

Step 4.3

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

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Step 4.3.1

Add $1$ to both sides of the equation.

$K = - 1 + 1$

Step 4.3.2

Add $- 1$ and $1$ .

$K = 0$

Step 5

Substitute $0$ for $K$ in $s = - \frac{2}{t^{\frac{1}{2}}} + K$ and simplify.

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Step 5.1

Substitute $0$ for $K$ .

$s = - \frac{2}{t^{\frac{1}{2}}} + 0$

Step 5.2

Add $- \frac{2}{t^{\frac{1}{2}}}$ and $0$ .

$s = - \frac{2}{t^{\frac{1}{2}}}$