Precalculus Examples

$d = 5 \sin (\frac{π}{4} t)$

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate using the Constant Multiple Rule.

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Step 1.1.1

Combine $\frac{π}{4}$ and $t$ .

$\frac{d}{d t} [5 \sin (\frac{π t}{4})]$

Step 1.1.2

Since $5$ is constant with respect to $t$ , the derivative of $5 \sin (\frac{π t}{4})$ with respect to $t$ is $5 \frac{d}{d t} [\sin (\frac{π t}{4})]$ .

$5 \frac{d}{d t} [\sin (\frac{π t}{4})]$

Step 1.2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = \sin (t)$ and $g (t) = \frac{π t}{4}$ .

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Step 1.2.1

To apply the Chain Rule, set $u$ as $\frac{π t}{4}$ .

$5 (\frac{d}{d u} [\sin (u)] \frac{d}{d t} [\frac{π t}{4}])$

Step 1.2.2

The derivative of $\sin (u)$ with respect to $u$ is $\cos (u)$ .

$5 (\cos (u) \frac{d}{d t} [\frac{π t}{4}])$

Step 1.2.3

Replace all occurrences of $u$ with $\frac{π t}{4}$ .

$5 (\cos (\frac{π t}{4}) \frac{d}{d t} [\frac{π t}{4}])$

Step 1.3

Differentiate.

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Step 1.3.1

Since $\frac{π}{4}$ is constant with respect to $t$ , the derivative of $\frac{π t}{4}$ with respect to $t$ is $\frac{π}{4} \frac{d}{d t} [t]$ .

$5 \cos (\frac{π t}{4}) (\frac{π}{4} \frac{d}{d t} [t])$

Step 1.3.2

Combine fractions.

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Step 1.3.2.1

Combine $\frac{π}{4}$ and $5$ .

$\frac{π \cdot 5}{4} \cos (\frac{π t}{4}) \frac{d}{d t} [t]$

Step 1.3.2.2

Combine $\frac{π \cdot 5}{4}$ and $\cos (\frac{π t}{4})$ .

$\frac{π \cdot 5 \cos (\frac{π t}{4})}{4} \frac{d}{d t} [t]$

Step 1.3.2.3

Move $5$ to the left of $π$ .

$\frac{5 \cdot π \cos (\frac{π t}{4})}{4} \frac{d}{d t} [t]$

Step 1.3.3

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$\frac{5 π \cos (\frac{π t}{4})}{4} \cdot 1$

Step 1.3.4

Multiply $\frac{5 π \cos (\frac{π t}{4})}{4}$ by $1$ .

$\frac{5 π \cos (\frac{π t}{4})}{4}$

Step 2

Find the second derivative of the function.

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Step 2.1

Since $\frac{5 π}{4}$ is constant with respect to $t$ , the derivative of $\frac{5 π \cos (\frac{π t}{4})}{4}$ with respect to $t$ is $\frac{5 π}{4} \frac{d}{d t} [\cos (\frac{π t}{4})]$ .

$f'' (t) = \frac{5 π}{4} \cdot \frac{d}{d t} (\cos (\frac{π t}{4}))$

Step 2.2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = \cos (t)$ and $g (t) = \frac{π t}{4}$ .

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Step 2.2.1

To apply the Chain Rule, set $u$ as $\frac{π t}{4}$ .

$f'' (t) = \frac{5 π}{4} \cdot (\frac{d}{d u} (\cos (u)) \frac{d}{d t} (\frac{π t}{4}))$

Step 2.2.2

The derivative of $\cos (u)$ with respect to $u$ is $- \sin (u)$ .

$f'' (t) = \frac{5 π}{4} \cdot (- \sin (u) \frac{d}{d t} \frac{π t}{4})$

Step 2.2.3

Replace all occurrences of $u$ with $\frac{π t}{4}$ .

$f'' (t) = \frac{5 π}{4} \cdot (- \sin (\frac{π t}{4}) \frac{d}{d t} \frac{π t}{4})$

Step 2.3

Differentiate using the Constant Multiple Rule.

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Step 2.3.1

Combine $\frac{5 π}{4}$ and $\sin (\frac{π t}{4})$ .

$f'' (t) = - \frac{5 π \sin (\frac{π t}{4})}{4} \cdot \frac{d}{d t} \frac{π t}{4}$

Step 2.3.2

Since $\frac{π}{4}$ is constant with respect to $t$ , the derivative of $\frac{π t}{4}$ with respect to $t$ is $\frac{π}{4} \frac{d}{d t} [t]$ .

$f'' (t) = - \frac{5 π \sin (\frac{π t}{4})}{4} \cdot (\frac{π}{4} \cdot \frac{d}{d t} (t))$

Step 2.3.3

Multiply $\frac{π}{4}$ by $\frac{5 π \sin (\frac{π t}{4})}{4}$ .

$f'' (t) = - \frac{π (5 π \sin (\frac{π t}{4}))}{4 \cdot 4} \cdot \frac{d}{d t} t$

Step 2.4

Raise $π$ to the power of $1$ .

$f'' (t) = - \frac{5 (π π) \sin (\frac{π t}{4})}{4 \cdot 4} \cdot \frac{d}{d t} t$

Step 2.5

Raise $π$ to the power of $1$ .

$f'' (t) = - \frac{5 (π π) \sin (\frac{π t}{4})}{4 \cdot 4} \cdot \frac{d}{d t} t$

Step 2.6

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (t) = - \frac{5 π^{1 + 1} \sin (\frac{π t}{4})}{4 \cdot 4} \cdot \frac{d}{d t} t$

Step 2.7

Simplify the expression.

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Step 2.7.1

Add $1$ and $1$ .

$f'' (t) = - \frac{5 π^{2} \sin (\frac{π t}{4})}{4 \cdot 4} \cdot \frac{d}{d t} t$

Step 2.7.2

Multiply $4$ by $4$ .

$f'' (t) = - \frac{5 π^{2} \sin (\frac{π t}{4})}{16} \cdot \frac{d}{d t} t$

Step 2.8

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 1$ .

$f'' (t) = - \frac{5 π^{2} \sin (\frac{π t}{4})}{16} \cdot 1$

Step 2.9

Multiply $- 1$ by $1$ .

$f'' (t) = - \frac{5 π^{2} \sin (\frac{π t}{4})}{16}$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$\frac{5 π \cos (\frac{π t}{4})}{4} = 0$

Step 4

Set the numerator equal to zero.

$5 π \cos (\frac{π t}{4}) = 0$

Step 5

Solve the equation for $t$ .

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Step 5.1

Divide each term in $5 π \cos (\frac{π t}{4}) = 0$ by $5 π$ and simplify.

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Step 5.1.1

Divide each term in $5 π \cos (\frac{π t}{4}) = 0$ by $5 π$ .

$\frac{5 π \cos (\frac{π t}{4})}{5 π} = \frac{0}{5 π}$

Step 5.1.2

Simplify the left side.

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Step 5.1.2.1

Cancel the common factor of $5$ .

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Step 5.1.2.1.1

Cancel the common factor.

$\frac{5 π \cos (\frac{π t}{4})}{5 π} = \frac{0}{5 π}$

Step 5.1.2.1.2

Rewrite the expression.

$\frac{π \cos (\frac{π t}{4})}{π} = \frac{0}{5 π}$

Step 5.1.2.2

Cancel the common factor of $π$ .

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Step 5.1.2.2.1

Cancel the common factor.

$\frac{π \cos (\frac{π t}{4})}{π} = \frac{0}{5 π}$

Step 5.1.2.2.2

Divide $\cos (\frac{π t}{4})$ by $1$ .

$\cos (\frac{π t}{4}) = \frac{0}{5 π}$

Step 5.1.3

Simplify the right side.

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Step 5.1.3.1

Cancel the common factor of $0$ and $5$ .

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Step 5.1.3.1.1

Factor $5$ out of $0$ .

$\cos (\frac{π t}{4}) = \frac{5 (0)}{5 π}$

Step 5.1.3.1.2

Cancel the common factors.

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Step 5.1.3.1.2.1

Factor $5$ out of $5 π$ .

$\cos (\frac{π t}{4}) = \frac{5 (0)}{5 (π)}$

Step 5.1.3.1.2.2

Cancel the common factor.

$\cos (\frac{π t}{4}) = \frac{5 \cdot 0}{5 π}$

Step 5.1.3.1.2.3

Rewrite the expression.

$\cos (\frac{π t}{4}) = \frac{0}{π}$

Step 5.1.3.2

Divide $0$ by $π$ .

$\cos (\frac{π t}{4}) = 0$

Step 5.2

Take the inverse cosine of both sides of the equation to extract $t$ from inside the cosine.

$\frac{π t}{4} = \arccos (0)$

Step 5.3

Simplify the right side.

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Step 5.3.1

The exact value of $\arccos (0)$ is $\frac{π}{2}$ .

$\frac{π t}{4} = \frac{π}{2}$

Step 5.4

Multiply both sides of the equation by $\frac{4}{π}$ .

$\frac{4}{π} \cdot \frac{π t}{4} = \frac{4}{π} \cdot \frac{π}{2}$

Step 5.5

Simplify both sides of the equation.

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Step 5.5.1

Simplify the left side.

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Step 5.5.1.1

Simplify $\frac{4}{π} \cdot \frac{π t}{4}$ .

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Step 5.5.1.1.1

Cancel the common factor of $4$ .

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Step 5.5.1.1.1.1

Cancel the common factor.

$\frac{4}{π} \cdot \frac{π t}{4} = \frac{4}{π} \cdot \frac{π}{2}$

Step 5.5.1.1.1.2

Rewrite the expression.

$\frac{1}{π} (π t) = \frac{4}{π} \cdot \frac{π}{2}$

Step 5.5.1.1.2

Cancel the common factor of $π$ .

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Step 5.5.1.1.2.1

Factor $π$ out of $π t$ .

$\frac{1}{π} (π (t)) = \frac{4}{π} \cdot \frac{π}{2}$

Step 5.5.1.1.2.2

Cancel the common factor.

$\frac{1}{π} (π t) = \frac{4}{π} \cdot \frac{π}{2}$

Step 5.5.1.1.2.3

Rewrite the expression.

$t = \frac{4}{π} \cdot \frac{π}{2}$

Step 5.5.2

Simplify the right side.

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Step 5.5.2.1

Simplify $\frac{4}{π} \cdot \frac{π}{2}$ .

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Step 5.5.2.1.1

Cancel the common factor of $2$ .

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Step 5.5.2.1.1.1

Factor $2$ out of $4$ .

$t = \frac{2 (2)}{π} \cdot \frac{π}{2}$

Step 5.5.2.1.1.2

Cancel the common factor.

$t = \frac{2 \cdot 2}{π} \cdot \frac{π}{2}$

Step 5.5.2.1.1.3

Rewrite the expression.

$t = \frac{2}{π} π$

Step 5.5.2.1.2

Cancel the common factor of $π$ .

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Step 5.5.2.1.2.1

Cancel the common factor.

$t = \frac{2}{π} π$

Step 5.5.2.1.2.2

Rewrite the expression.

$t = 2$

Step 5.6

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$\frac{π t}{4} = 2 π - \frac{π}{2}$

Step 5.7

Solve for $t$ .

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Step 5.7.1

Multiply both sides of the equation by $\frac{4}{π}$ .

$\frac{4}{π} \cdot \frac{π t}{4} = \frac{4}{π} (2 π - \frac{π}{2})$

Step 5.7.2

Simplify both sides of the equation.

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Step 5.7.2.1

Simplify the left side.

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Step 5.7.2.1.1

Simplify $\frac{4}{π} \cdot \frac{π t}{4}$ .

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Step 5.7.2.1.1.1

Cancel the common factor of $4$ .

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Step 5.7.2.1.1.1.1

Cancel the common factor.

$\frac{4}{π} \cdot \frac{π t}{4} = \frac{4}{π} (2 π - \frac{π}{2})$

Step 5.7.2.1.1.1.2

Rewrite the expression.

$\frac{1}{π} (π t) = \frac{4}{π} (2 π - \frac{π}{2})$

Step 5.7.2.1.1.2

Cancel the common factor of $π$ .

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Step 5.7.2.1.1.2.1

Factor $π$ out of $π t$ .

$\frac{1}{π} (π (t)) = \frac{4}{π} (2 π - \frac{π}{2})$

Step 5.7.2.1.1.2.2

Cancel the common factor.

$\frac{1}{π} (π t) = \frac{4}{π} (2 π - \frac{π}{2})$

Step 5.7.2.1.1.2.3

Rewrite the expression.

$t = \frac{4}{π} (2 π - \frac{π}{2})$

Step 5.7.2.2

Simplify the right side.

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Step 5.7.2.2.1

Simplify $\frac{4}{π} (2 π - \frac{π}{2})$ .

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Step 5.7.2.2.1.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$t = \frac{4}{π} (2 π \cdot \frac{2}{2} - \frac{π}{2})$

Step 5.7.2.2.1.2

Combine $2 π$ and $\frac{2}{2}$ .

$t = \frac{4}{π} (\frac{2 π \cdot 2}{2} - \frac{π}{2})$

Step 5.7.2.2.1.3

Combine the numerators over the common denominator.

$t = \frac{4}{π} \cdot \frac{2 π \cdot 2 - π}{2}$

Step 5.7.2.2.1.4

Cancel the common factor of $2$ .

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Step 5.7.2.2.1.4.1

Factor $2$ out of $4$ .

$t = \frac{2 (2)}{π} \cdot \frac{2 π \cdot 2 - π}{2}$

Step 5.7.2.2.1.4.2

Cancel the common factor.

$t = \frac{2 \cdot 2}{π} \cdot \frac{2 π \cdot 2 - π}{2}$

Step 5.7.2.2.1.4.3

Rewrite the expression.

$t = \frac{2}{π} (2 π \cdot 2 - π)$

Step 5.7.2.2.1.5

Multiply $2$ by $2$ .

$t = \frac{2}{π} (4 π - π)$

Step 5.7.2.2.1.6

Subtract $π$ from $4 π$ .

$t = \frac{2}{π} (3 π)$

Step 5.7.2.2.1.7

Cancel the common factor of $π$ .

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Step 5.7.2.2.1.7.1

Factor $π$ out of $3 π$ .

$t = \frac{2}{π} (π \cdot 3)$

Step 5.7.2.2.1.7.2

Cancel the common factor.

$t = \frac{2}{π} (π \cdot 3)$

Step 5.7.2.2.1.7.3

Rewrite the expression.

$t = 2 \cdot 3$

Step 5.7.2.2.1.8

Multiply $2$ by $3$ .

$t = 6$

Step 5.8

The solution to the equation $\frac{π t}{4} = \frac{π}{2}$ .

$t = 2, 6$

Step 6

Evaluate the second derivative at $t = 2$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \frac{5 π^{2} \sin (\frac{π (2)}{4})}{16}$

Step 7

Evaluate the second derivative.

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Step 7.1

Cancel the common factor of $2$ and $4$ .

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Step 7.1.1

Factor $2$ out of $π (2)$ .

$- \frac{5 π^{2} \sin (\frac{2 π}{4})}{16}$

Step 7.1.2

Cancel the common factors.

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Step 7.1.2.1

Factor $2$ out of $4$ .

$- \frac{5 π^{2} \sin (\frac{2 π}{2 \cdot 2})}{16}$

Step 7.1.2.2

Cancel the common factor.

$- \frac{5 π^{2} \sin (\frac{2 π}{2 \cdot 2})}{16}$

Step 7.1.2.3

Rewrite the expression.

$- \frac{5 π^{2} \sin (\frac{π}{2})}{16}$

Step 7.2

Simplify the numerator.

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Step 7.2.1

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$- \frac{5 π^{2} \cdot 1}{16}$

Step 7.2.2

Multiply $5$ by $1$ .

$- \frac{5 π^{2}}{16}$

Step 8

$t = 2$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$t = 2$ is a local maximum

Step 9

Find the y-value when $t = 2$ .

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Step 9.1

Replace the variable $t$ with $2$ in the expression.

$f (2) = 5 \sin (\frac{π}{4} \cdot (2))$

Step 9.2

Simplify the result.

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Step 9.2.1

Cancel the common factor of $2$ .

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Step 9.2.1.1

Factor $2$ out of $4$ .

$f (2) = 5 \sin (\frac{π}{2 (2)} \cdot 2)$

Step 9.2.1.2

Cancel the common factor.

$f (2) = 5 \sin (\frac{π}{2 \cdot 2} \cdot 2)$

Step 9.2.1.3

Rewrite the expression.

$f (2) = 5 \sin (\frac{π}{2})$

Step 9.2.2

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$f (2) = 5 \cdot 1$

Step 9.2.3

Multiply $5$ by $1$ .

$f (2) = 5$

Step 9.2.4

The final answer is $5$ .

$y = 5$

Step 10

Evaluate the second derivative at $t = 6$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \frac{5 π^{2} \sin (\frac{π (6)}{4})}{16}$

Step 11

Evaluate the second derivative.

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Step 11.1

Cancel the common factor of $6$ and $4$ .

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Step 11.1.1

Factor $2$ out of $π (6)$ .

$- \frac{5 π^{2} \sin (\frac{2 (π \cdot (3))}{4})}{16}$

Step 11.1.2

Cancel the common factors.

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Step 11.1.2.1

Factor $2$ out of $4$ .

$- \frac{5 π^{2} \sin (\frac{2 (π \cdot (3))}{2 (2)})}{16}$

Step 11.1.2.2

Cancel the common factor.

$- \frac{5 π^{2} \sin (\frac{2 (π \cdot (3))}{2 \cdot 2})}{16}$

Step 11.1.2.3

Rewrite the expression.

$- \frac{5 π^{2} \sin (\frac{π \cdot (3)}{2})}{16}$

Step 11.2

Simplify the numerator.

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Step 11.2.1

Move $3$ to the left of $π$ .

$- \frac{5 π^{2} \sin (\frac{3 \cdot π}{2})}{16}$

Step 11.2.2

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

$- \frac{5 π^{2} (- \sin (\frac{π}{2}))}{16}$

Step 11.2.3

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$- \frac{5 π^{2} (- 1 \cdot 1)}{16}$

Step 11.2.4

Multiply $- 1$ by $1$ .

$- \frac{5 π^{2} \cdot - 1}{16}$

Step 11.2.5

Multiply $- 1$ by $5$ .

$- \frac{- 5 π^{2}}{16}$

Step 11.3

Move the negative in front of the fraction.

$- - \frac{5 π^{2}}{16}$

Step 11.4

Multiply $- - \frac{5 π^{2}}{16}$ .

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Step 11.4.1

Multiply $- 1$ by $- 1$ .

$1 \frac{5 π^{2}}{16}$

Step 11.4.2

Multiply $\frac{5 π^{2}}{16}$ by $1$ .

$\frac{5 π^{2}}{16}$

Step 12

$t = 6$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$t = 6$ is a local minimum

Step 13

Find the y-value when $t = 6$ .

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Step 13.1

Replace the variable $t$ with $6$ in the expression.

$f (6) = 5 \sin (\frac{π}{4} \cdot (6))$

Step 13.2

Simplify the result.

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Step 13.2.1

Cancel the common factor of $2$ .

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Step 13.2.1.1

Factor $2$ out of $4$ .

$f (6) = 5 \sin (\frac{π}{2 (2)} \cdot 6)$

Step 13.2.1.2

Factor $2$ out of $6$ .

$f (6) = 5 \sin (\frac{π}{2 \cdot 2} \cdot (2 \cdot 3))$

Step 13.2.1.3

Cancel the common factor.

$f (6) = 5 \sin (\frac{π}{2 \cdot 2} \cdot (2 \cdot 3))$

Step 13.2.1.4

Rewrite the expression.

$f (6) = 5 \sin (\frac{π}{2} \cdot 3)$

Step 13.2.2

Combine $\frac{π}{2}$ and $3$ .

$f (6) = 5 \sin (\frac{π \cdot 3}{2})$

Step 13.2.3

Move $3$ to the left of $π$ .

$f (6) = 5 \sin (\frac{3 \cdot π}{2})$

Step 13.2.4

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

$f (6) = 5 (- \sin (\frac{π}{2}))$

Step 13.2.5

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$f (6) = 5 (- 1 \cdot 1)$

Step 13.2.6

Multiply $5 (- 1 \cdot 1)$ .

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Step 13.2.6.1

Multiply $- 1$ by $1$ .

$f (6) = 5 \cdot - 1$

Step 13.2.6.2

Multiply $5$ by $- 1$ .

$f (6) = - 5$

Step 13.2.7

The final answer is $- 5$ .

$y = - 5$

Step 14

These are the local extrema for $f (t) = 5 \sin (\frac{π}{4} t)$ .

$(2, 5)$ is a local maxima

$(6, - 5)$ is a local minima

Step 15