Precalculus Examples

$d = 8 \cos (\frac{3 π}{4} t)$

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 $\frac{3 π}{4}$ and $t$ .

$\frac{d}{d t} [8 \cos (\frac{3 π t}{4})]$

Step 1.1.2

Since $8$ is constant with respect to $t$ , the derivative of $8 \cos (\frac{3 π t}{4})$ with respect to $t$ is $8 \frac{d}{d t} [\cos (\frac{3 π t}{4})]$ .

$8 \frac{d}{d t} [\cos (\frac{3 π t}{4})]$

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 $f (t) = \cos (t)$ and $g (t) = \frac{3 π t}{4}$ .

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Step 1.2.1

To apply the Chain Rule, set $u$ as $\frac{3 π t}{4}$ .

$8 (\frac{d}{d u} [\cos (u)] \frac{d}{d t} [\frac{3 π t}{4}])$

Step 1.2.2

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

$8 (- \sin (u) \frac{d}{d t} [\frac{3 π t}{4}])$

Step 1.2.3

Replace all occurrences of $u$ with $\frac{3 π t}{4}$ .

$8 (- \sin (\frac{3 π t}{4}) \frac{d}{d t} [\frac{3 π t}{4}])$

Step 1.3

Differentiate.

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Step 1.3.1

Multiply $- 1$ by $8$ .

$- 8 (\sin (\frac{3 π t}{4}) \frac{d}{d t} [\frac{3 π t}{4}])$

Step 1.3.2

Since $\frac{3 π}{4}$ is constant with respect to $t$ , the derivative of $\frac{3 π t}{4}$ with respect to $t$ is $\frac{3 π}{4} \frac{d}{d t} [t]$ .

$- 8 \sin (\frac{3 π t}{4}) (\frac{3 π}{4} \frac{d}{d t} [t])$

Step 1.3.3

Simplify terms.

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Step 1.3.3.1

Combine $\frac{3 π}{4}$ and $- 8$ .

$\frac{3 π \cdot - 8}{4} \sin (\frac{3 π t}{4}) \frac{d}{d t} [t]$

Step 1.3.3.2

Multiply $- 8$ by $3$ .

$\frac{- 24 π}{4} \sin (\frac{3 π t}{4}) \frac{d}{d t} [t]$

Step 1.3.3.3

Combine $\frac{- 24 π}{4}$ and $\sin (\frac{3 π t}{4})$ .

$\frac{- 24 π \sin (\frac{3 π t}{4})}{4} \frac{d}{d t} [t]$

Step 1.3.3.4

Cancel the common factor of $- 24$ and $4$ .

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Step 1.3.3.4.1

Factor $4$ out of $- 24 π \sin (\frac{3 π t}{4})$ .

$\frac{4 (- 6 π \sin (\frac{3 π t}{4}))}{4} \frac{d}{d t} [t]$

Step 1.3.3.4.2

Cancel the common factors.

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Step 1.3.3.4.2.1

Factor $4$ out of $4$ .

$\frac{4 (- 6 π \sin (\frac{3 π t}{4}))}{4 (1)} \frac{d}{d t} [t]$

Step 1.3.3.4.2.2

Cancel the common factor.

$\frac{4 (- 6 π \sin (\frac{3 π t}{4}))}{4 \cdot 1} \frac{d}{d t} [t]$

Step 1.3.3.4.2.3

Rewrite the expression.

$\frac{- 6 π \sin (\frac{3 π t}{4})}{1} \frac{d}{d t} [t]$

Step 1.3.3.4.2.4

Divide $- 6 π \sin (\frac{3 π t}{4})$ by $1$ .

$- 6 π \sin (\frac{3 π t}{4}) \frac{d}{d t} [t]$

Step 1.3.4

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

$- 6 π \sin (\frac{3 π t}{4}) \cdot 1$

Step 1.3.5

Multiply $- 6$ by $1$ .

$- 6 π \sin (\frac{3 π t}{4})$

Step 2

Find the second derivative of the function.

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Step 2.1

Since $- 6 π$ is constant with respect to $t$ , the derivative of $- 6 π \sin (\frac{3 π t}{4})$ with respect to $t$ is $- 6 π \frac{d}{d t} [\sin (\frac{3 π t}{4})]$ .

$f'' (t) = - 6 π \frac{d}{d t} \sin (\frac{3 π t}{4})$

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 $f (t) = \sin (t)$ and $g (t) = \frac{3 π t}{4}$ .

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Step 2.2.1

To apply the Chain Rule, set $u$ as $\frac{3 π t}{4}$ .

$f'' (t) = - 6 π (\frac{d}{d u} (\sin (u)) \frac{d}{d t} (\frac{3 π t}{4}))$

Step 2.2.2

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

$f'' (t) = - 6 π (\cos (u) \frac{d}{d t} (\frac{3 π t}{4}))$

Step 2.2.3

Replace all occurrences of $u$ with $\frac{3 π t}{4}$ .

$f'' (t) = - 6 π (\cos (\frac{3 π t}{4}) \frac{d}{d t} (\frac{3 π t}{4}))$

Step 2.3

Differentiate using the Constant Multiple Rule.

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Step 2.3.1

Since $\frac{3 π}{4}$ is constant with respect to $t$ , the derivative of $\frac{3 π t}{4}$ with respect to $t$ is $\frac{3 π}{4} \frac{d}{d t} [t]$ .

$f'' (t) = - 6 π \cos (\frac{3 π t}{4}) (\frac{3 π}{4} \cdot \frac{d}{d t} (t))$

Step 2.3.2

Combine fractions.

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Step 2.3.2.1

Combine $\frac{3 π}{4}$ and $- 6$ .

$f'' (t) = \frac{3 π \cdot - 6}{4} \cdot (π \cos (\frac{3 π t}{4})) \frac{d}{d t} (t)$

Step 2.3.2.2

Multiply $- 6$ by $3$ .

$f'' (t) = \frac{- 18 π}{4} \cdot (π \cos (\frac{3 π t}{4})) \frac{d}{d t} (t)$

Step 2.3.2.3

Combine $\frac{- 18 π}{4}$ and $π$ .

$f'' (t) = \frac{- 18 π π}{4} \cdot \cos (\frac{3 π t}{4}) \frac{d}{d t} (t)$

Step 2.4

Raise $π$ to the power of $1$ .

$f'' (t) = \frac{- 18 (π π)}{4} \cdot \cos (\frac{3 π t}{4}) \frac{d}{d t} (t)$

Step 2.5

Raise $π$ to the power of $1$ .

$f'' (t) = \frac{- 18 (π π)}{4} \cdot \cos (\frac{3 π t}{4}) \frac{d}{d t} (t)$

Step 2.6

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

$f'' (t) = \frac{- 18 π^{1 + 1}}{4} \cdot \cos (\frac{3 π t}{4}) \frac{d}{d t} (t)$

Step 2.7

Simplify terms.

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Step 2.7.1

Add $1$ and $1$ .

$f'' (t) = \frac{- 18 π^{2}}{4} \cdot \cos (\frac{3 π t}{4}) \frac{d}{d t} (t)$

Step 2.7.2

Combine $\frac{- 18 π^{2}}{4}$ and $\cos (\frac{3 π t}{4})$ .

$f'' (t) = \frac{- 18 π^{2} \cos (\frac{3 π t}{4})}{4} \cdot \frac{d}{d t} (t)$

Step 2.7.3

Cancel the common factor of $- 18$ and $4$ .

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Step 2.7.3.1

Factor $2$ out of $- 18 π^{2} \cos (\frac{3 π t}{4})$ .

$f'' (t) = \frac{2 (- 9 π^{2} \cos (\frac{3 π t}{4}))}{4} \cdot \frac{d}{d t} (t)$

Step 2.7.3.2

Cancel the common factors.

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Step 2.7.3.2.1

Factor $2$ out of $4$ .

$f'' (t) = \frac{2 (- 9 π^{2} \cos (\frac{3 π t}{4}))}{2 (2)} \cdot \frac{d}{d t} (t)$

Step 2.7.3.2.2

Cancel the common factor.

$f'' (t) = \frac{2 (- 9 π^{2} \cos (\frac{3 π t}{4}))}{2 \cdot 2} \cdot \frac{d}{d t} (t)$

Step 2.7.3.2.3

Rewrite the expression.

$f'' (t) = \frac{- 9 π^{2} \cos (\frac{3 π t}{4})}{2} \cdot \frac{d}{d t} (t)$

Step 2.7.4

Move the negative in front of the fraction.

$f'' (t) = - \frac{9 π^{2} \cos (\frac{3 π t}{4})}{2} \cdot \frac{d}{d t} t$

Step 2.8

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

$f'' (t) = - \frac{9 π^{2} \cos (\frac{3 π t}{4})}{2} \cdot 1$

Step 2.9

Multiply $- 1$ by $1$ .

$f'' (t) = - \frac{9 π^{2} \cos (\frac{3 π t}{4})}{2}$

Step 3

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

$- 6 π \sin (\frac{3 π t}{4}) = 0$

Step 4

Divide each term in $- 6 π \sin (\frac{3 π t}{4}) = 0$ by $- 6 π$ and simplify.

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Step 4.1

Divide each term in $- 6 π \sin (\frac{3 π t}{4}) = 0$ by $- 6 π$ .

$\frac{- 6 π \sin (\frac{3 π t}{4})}{- 6 π} = \frac{0}{- 6 π}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $- 6$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{- 6 π \sin (\frac{3 π t}{4})}{- 6 π} = \frac{0}{- 6 π}$

Step 4.2.1.2

Rewrite the expression.

$\frac{π \sin (\frac{3 π t}{4})}{π} = \frac{0}{- 6 π}$

Step 4.2.2

Cancel the common factor of $π$ .

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Step 4.2.2.1

Cancel the common factor.

$\frac{π \sin (\frac{3 π t}{4})}{π} = \frac{0}{- 6 π}$

Step 4.2.2.2

Divide $\sin (\frac{3 π t}{4})$ by $1$ .

$\sin (\frac{3 π t}{4}) = \frac{0}{- 6 π}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Cancel the common factor of $0$ and $- 6$ .

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Step 4.3.1.1

Factor $6$ out of $0$ .

$\sin (\frac{3 π t}{4}) = \frac{6 (0)}{- 6 π}$

Step 4.3.1.2

Cancel the common factors.

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Step 4.3.1.2.1

Factor $6$ out of $- 6 π$ .

$\sin (\frac{3 π t}{4}) = \frac{6 (0)}{6 (- π)}$

Step 4.3.1.2.2

Cancel the common factor.

$\sin (\frac{3 π t}{4}) = \frac{6 \cdot 0}{6 (- π)}$

Step 4.3.1.2.3

Rewrite the expression.

$\sin (\frac{3 π t}{4}) = \frac{0}{- π}$

Step 4.3.2

Divide $0$ by $- π$ .

$\sin (\frac{3 π t}{4}) = 0$

Step 5

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

$\frac{3 π t}{4} = \arcsin (0)$

Step 6

Simplify the right side.

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Step 6.1

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

$\frac{3 π t}{4} = 0$

Step 7

Set the numerator equal to zero.

$3 π t = 0$

Step 8

Divide each term in $3 π t = 0$ by $3 π$ and simplify.

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Step 8.1

Divide each term in $3 π t = 0$ by $3 π$ .

$\frac{3 π t}{3 π} = \frac{0}{3 π}$

Step 8.2

Simplify the left side.

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Step 8.2.1

Cancel the common factor of $3$ .

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Step 8.2.1.1

Cancel the common factor.

$\frac{3 π t}{3 π} = \frac{0}{3 π}$

Step 8.2.1.2

Rewrite the expression.

$\frac{π t}{π} = \frac{0}{3 π}$

Step 8.2.2

Cancel the common factor of $π$ .

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Step 8.2.2.1

Cancel the common factor.

$\frac{π t}{π} = \frac{0}{3 π}$

Step 8.2.2.2

Divide $t$ by $1$ .

$t = \frac{0}{3 π}$

Step 8.3

Simplify the right side.

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Step 8.3.1

Cancel the common factor of $0$ and $3$ .

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Step 8.3.1.1

Factor $3$ out of $0$ .

$t = \frac{3 (0)}{3 π}$

Step 8.3.1.2

Cancel the common factors.

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Step 8.3.1.2.1

Factor $3$ out of $3 π$ .

$t = \frac{3 (0)}{3 (π)}$

Step 8.3.1.2.2

Cancel the common factor.

$t = \frac{3 \cdot 0}{3 π}$

Step 8.3.1.2.3

Rewrite the expression.

$t = \frac{0}{π}$

Step 8.3.2

Divide $0$ by $π$ .

$t = 0$

Step 9

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

$\frac{3 π t}{4} = π - 0$

Step 10

Solve for $t$ .

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Step 10.1

Multiply both sides of the equation by $\frac{4}{3 π}$ .

$\frac{4}{3 π} \cdot \frac{3 π t}{4} = \frac{4}{3 π} (π - 0)$

Step 10.2

Simplify both sides of the equation.

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Step 10.2.1

Simplify the left side.

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Step 10.2.1.1

Simplify $\frac{4}{3 π} \cdot \frac{3 π t}{4}$ .

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Step 10.2.1.1.1

Cancel the common factor of $4$ .

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Step 10.2.1.1.1.1

Cancel the common factor.

$\frac{4}{3 π} \cdot \frac{3 π t}{4} = \frac{4}{3 π} (π - 0)$

Step 10.2.1.1.1.2

Rewrite the expression.

$\frac{1}{3 π} (3 π t) = \frac{4}{3 π} (π - 0)$

Step 10.2.1.1.2

Cancel the common factor of $3 π$ .

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Step 10.2.1.1.2.1

Factor $3 π$ out of $3 π t$ .

$\frac{1}{3 π} (3 π (t)) = \frac{4}{3 π} (π - 0)$

Step 10.2.1.1.2.2

Cancel the common factor.

$\frac{1}{3 π} (3 π t) = \frac{4}{3 π} (π - 0)$

Step 10.2.1.1.2.3

Rewrite the expression.

$t = \frac{4}{3 π} (π - 0)$

Step 10.2.2

Simplify the right side.

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Step 10.2.2.1

Simplify $\frac{4}{3 π} (π - 0)$ .

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Step 10.2.2.1.1

Subtract $0$ from $π$ .

$t = \frac{4}{3 π} π$

Step 10.2.2.1.2

Cancel the common factor of $π$ .

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Step 10.2.2.1.2.1

Factor $π$ out of $3 π$ .

$t = \frac{4}{π \cdot 3} π$

Step 10.2.2.1.2.2

Cancel the common factor.

$t = \frac{4}{π \cdot 3} π$

Step 10.2.2.1.2.3

Rewrite the expression.

$t = \frac{4}{3}$

Step 11

The solution to the equation $\frac{3 π t}{4} = 0$ .

$t = 0, \frac{4}{3}$

Step 12

Evaluate the second derivative at $t = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \frac{9 π^{2} \cos (\frac{3 π (0)}{4})}{2}$

Step 13

Evaluate the second derivative.

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Step 13.1

Cancel the common factor of $0$ and $4$ .

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Step 13.1.1

Factor $4$ out of $3 π (0)$ .

$- \frac{9 π^{2} \cos (\frac{4 (3 π \cdot (0))}{4})}{2}$

Step 13.1.2

Cancel the common factors.

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Step 13.1.2.1

Factor $4$ out of $4$ .

$- \frac{9 π^{2} \cos (\frac{4 (3 π \cdot (0))}{4 (1)})}{2}$

Step 13.1.2.2

Cancel the common factor.

$- \frac{9 π^{2} \cos (\frac{4 (3 π \cdot (0))}{4 \cdot 1})}{2}$

Step 13.1.2.3

Rewrite the expression.

$- \frac{9 π^{2} \cos (\frac{3 π \cdot (0)}{1})}{2}$

Step 13.1.2.4

Divide $3 π \cdot (0)$ by $1$ .

$- \frac{9 π^{2} \cos (3 π \cdot (0))}{2}$

Step 13.2

Simplify the numerator.

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Step 13.2.1

Combine exponents.

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Step 13.2.1.1

Multiply $0$ by $3$ .

$- \frac{9 π^{2} \cos (0 π)}{2}$

Step 13.2.1.2

Multiply $0$ by $π$ .

$- \frac{9 π^{2} \cos (0)}{2}$

Step 13.2.2

The exact value of $\cos (0)$ is $1$ .

$- \frac{9 π^{2} \cdot 1}{2}$

Step 13.2.3

Multiply $9$ by $1$ .

$- \frac{9 π^{2}}{2}$

Step 14

$t = 0$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$t = 0$ is a local maximum

Step 15

Find the y-value when $t = 0$ .

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Step 15.1

Replace the variable $t$ with $0$ in the expression.

$f (0) = 8 \cos (\frac{3 π}{4} \cdot (0))$

Step 15.2

Simplify the result.

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Step 15.2.1

Multiply $\frac{3 π}{4}$ by $0$ .

$f (0) = 8 \cos (0)$

Step 15.2.2

The exact value of $\cos (0)$ is $1$ .

$f (0) = 8 \cdot 1$

Step 15.2.3

Multiply $8$ by $1$ .

$f (0) = 8$

Step 15.2.4

The final answer is $8$ .

$y = 8$

Step 16

Evaluate the second derivative at $t = \frac{4}{3}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \frac{9 π^{2} \cos (\frac{3 π (\frac{4}{3})}{4})}{2}$

Step 17

Evaluate the second derivative.

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Step 17.1

Simplify the numerator.

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Step 17.1.1

Combine $\frac{4}{3}$ and $3$ .

$- \frac{9 π^{2} \cos (\frac{\frac{4 \cdot 3}{3} π}{4})}{2}$

Step 17.1.2

Combine $\frac{4 \cdot 3}{3}$ and $π$ .

$- \frac{9 π^{2} \cos (\frac{\frac{4 \cdot 3 π}{3}}{4})}{2}$

Step 17.2

Multiply $4$ by $3$ .

$- \frac{9 π^{2} \cos (\frac{\frac{12 π}{3}}{4})}{2}$

Step 17.3

Reduce the expression by cancelling the common factors.

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Step 17.3.1

Reduce the expression $\frac{12 π}{3}$ by cancelling the common factors.

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Step 17.3.1.1

Factor $3$ out of $12 π$ .

$- \frac{9 π^{2} \cos (\frac{\frac{3 (4 π)}{3}}{4})}{2}$

Step 17.3.1.2

Factor $3$ out of $3$ .

$- \frac{9 π^{2} \cos (\frac{\frac{3 (4 π)}{3 (1)}}{4})}{2}$

Step 17.3.1.3

Cancel the common factor.

$- \frac{9 π^{2} \cos (\frac{\frac{3 (4 π)}{3 \cdot 1}}{4})}{2}$

Step 17.3.1.4

Rewrite the expression.

$- \frac{9 π^{2} \cos (\frac{\frac{4 π}{1}}{4})}{2}$

Step 17.3.2

Divide $4 π$ by $1$ .

$- \frac{9 π^{2} \cos (\frac{4 π}{4})}{2}$

Step 17.4

Simplify the numerator.

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Step 17.4.1

Cancel the common factor of $4$ .

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Step 17.4.1.1

Cancel the common factor.

$- \frac{9 π^{2} \cos (\frac{4 π}{4})}{2}$

Step 17.4.1.2

Divide $π$ by $1$ .

$- \frac{9 π^{2} \cos (π)}{2}$

Step 17.4.2

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

$- \frac{9 π^{2} (- \cos (0))}{2}$

Step 17.4.3

The exact value of $\cos (0)$ is $1$ .

$- \frac{9 π^{2} (- 1 \cdot 1)}{2}$

Step 17.4.4

Multiply $- 1$ by $1$ .

$- \frac{9 π^{2} \cdot - 1}{2}$

Step 17.4.5

Multiply $- 1$ by $9$ .

$- \frac{- 9 π^{2}}{2}$

Step 17.5

Move the negative in front of the fraction.

$- - \frac{9 π^{2}}{2}$

Step 17.6

Multiply $- - \frac{9 π^{2}}{2}$ .

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Step 17.6.1

Multiply $- 1$ by $- 1$ .

$1 \frac{9 π^{2}}{2}$

Step 17.6.2

Multiply $\frac{9 π^{2}}{2}$ by $1$ .

$\frac{9 π^{2}}{2}$

Step 18

$t = \frac{4}{3}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$t = \frac{4}{3}$ is a local minimum

Step 19

Find the y-value when $t = \frac{4}{3}$ .

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Step 19.1

Replace the variable $t$ with $\frac{4}{3}$ in the expression.

$f (\frac{4}{3}) = 8 \cos (\frac{3 π}{4} \cdot (\frac{4}{3}))$

Step 19.2

Simplify the result.

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Step 19.2.1

Cancel the common factor of $3$ .

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Step 19.2.1.1

Factor $3$ out of $3 π$ .

$f (\frac{4}{3}) = 8 \cos (\frac{3 (π)}{4} \cdot \frac{4}{3})$

Step 19.2.1.2

Cancel the common factor.

$f (\frac{4}{3}) = 8 \cos (\frac{3 π}{4} \cdot \frac{4}{3})$

Step 19.2.1.3

Rewrite the expression.

$f (\frac{4}{3}) = 8 \cos (\frac{π}{4} \cdot 4)$

Step 19.2.2

Cancel the common factor of $4$ .

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Step 19.2.2.1

Cancel the common factor.

$f (\frac{4}{3}) = 8 \cos (\frac{π}{4} \cdot 4)$

Step 19.2.2.2

Rewrite the expression.

$f (\frac{4}{3}) = 8 \cos (π)$

Step 19.2.3

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

$f (\frac{4}{3}) = 8 (- \cos (0))$

Step 19.2.4

The exact value of $\cos (0)$ is $1$ .

$f (\frac{4}{3}) = 8 (- 1 \cdot 1)$

Step 19.2.5

Multiply $8 (- 1 \cdot 1)$ .

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Step 19.2.5.1

Multiply $- 1$ by $1$ .

$f (\frac{4}{3}) = 8 \cdot - 1$

Step 19.2.5.2

Multiply $8$ by $- 1$ .

$f (\frac{4}{3}) = - 8$

Step 19.2.6

The final answer is $- 8$ .

$y = - 8$

Step 20

These are the local extrema for $f (t) = 8 \cos (\frac{3 π}{4} t)$ .

$(0, 8)$ is a local maxima

$(\frac{4}{3}, - 8)$ is a local minima

Step 21