Calculus Examples

$h (t) = 60 - 55 \sin (\frac{π}{10} t + \frac{π}{2})$

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate.

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Step 1.1.1

By the Sum Rule, the derivative of $60 - 55 \sin (\frac{π}{10} t + \frac{π}{2})$ with respect to $t$ is $\frac{d}{d t} [60] + \frac{d}{d t} [- 55 \sin (\frac{π}{10} t + \frac{π}{2})]$ .

$\frac{d}{d t} [60] + \frac{d}{d t} [- 55 \sin (\frac{π}{10} t + \frac{π}{2})]$

Step 1.1.2

Since $60$ is constant with respect to $t$ , the derivative of $60$ with respect to $t$ is $0$ .

$0 + \frac{d}{d t} [- 55 \sin (\frac{π}{10} t + \frac{π}{2})]$

Step 1.2

Evaluate $\frac{d}{d t} [- 55 \sin (\frac{π}{10} t + \frac{π}{2})]$ .

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Step 1.2.1

Combine $\frac{π}{10}$ and $t$ .

$0 + \frac{d}{d t} [- 55 \sin (\frac{π t}{10} + \frac{π}{2})]$

Step 1.2.2

Since $- 55$ is constant with respect to $t$ , the derivative of $- 55 \sin (\frac{π t}{10} + \frac{π}{2})$ with respect to $t$ is $- 55 \frac{d}{d t} [\sin (\frac{π t}{10} + \frac{π}{2})]$ .

$0 - 55 \frac{d}{d t} [\sin (\frac{π t}{10} + \frac{π}{2})]$

Step 1.2.3

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 $g (t) = \frac{π t}{10} + \frac{π}{2}$ .

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Step 1.2.3.1

To apply the Chain Rule, set $u$ as $\frac{π t}{10} + \frac{π}{2}$ .

$0 - 55 (\frac{d}{d u} [\sin (u)] \frac{d}{d t} [\frac{π t}{10} + \frac{π}{2}])$

Step 1.2.3.2

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

$0 - 55 (\cos (u) \frac{d}{d t} [\frac{π t}{10} + \frac{π}{2}])$

Step 1.2.3.3

Replace all occurrences of $u$ with $\frac{π t}{10} + \frac{π}{2}$ .

$0 - 55 (\cos (\frac{π t}{10} + \frac{π}{2}) \frac{d}{d t} [\frac{π t}{10} + \frac{π}{2}])$

Step 1.2.4

By the Sum Rule, the derivative of $\frac{π t}{10} + \frac{π}{2}$ with respect to $t$ is $\frac{d}{d t} [\frac{π t}{10}] + \frac{d}{d t} [\frac{π}{2}]$ .

$0 - 55 (\cos (\frac{π t}{10} + \frac{π}{2}) (\frac{d}{d t} [\frac{π t}{10}] + \frac{d}{d t} [\frac{π}{2}]))$

Step 1.2.5

Since $\frac{π}{10}$ is constant with respect to $t$ , the derivative of $\frac{π t}{10}$ with respect to $t$ is $\frac{π}{10} \frac{d}{d t} [t]$ .

$0 - 55 (\cos (\frac{π t}{10} + \frac{π}{2}) (\frac{π}{10} \frac{d}{d t} [t] + \frac{d}{d t} [\frac{π}{2}]))$

Step 1.2.6

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

$0 - 55 (\cos (\frac{π t}{10} + \frac{π}{2}) (\frac{π}{10} \cdot 1 + \frac{d}{d t} [\frac{π}{2}]))$

Step 1.2.7

Since $\frac{π}{2}$ is constant with respect to $t$ , the derivative of $\frac{π}{2}$ with respect to $t$ is $0$ .

$0 - 55 (\cos (\frac{π t}{10} + \frac{π}{2}) (\frac{π}{10} \cdot 1 + 0))$

Step 1.2.8

Multiply $\frac{π}{10}$ by $1$ .

$0 - 55 (\cos (\frac{π t}{10} + \frac{π}{2}) (\frac{π}{10} + 0))$

Step 1.2.9

Add $\frac{π}{10}$ and $0$ .

$0 - 55 (\cos (\frac{π t}{10} + \frac{π}{2}) \frac{π}{10})$

Step 1.2.10

Combine $\cos (\frac{π t}{10} + \frac{π}{2})$ and $\frac{π}{10}$ .

$0 - 55 \frac{\cos (\frac{π t}{10} + \frac{π}{2}) π}{10}$

Step 1.2.11

Combine $- 55$ and $\frac{\cos (\frac{π t}{10} + \frac{π}{2}) π}{10}$ .

$0 + \frac{- 55 (\cos (\frac{π t}{10} + \frac{π}{2}) π)}{10}$

Step 1.2.12

Cancel the common factor of $- 55$ and $10$ .

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Step 1.2.12.1

Factor $5$ out of $- 55 \cos (\frac{π t}{10} + \frac{π}{2}) π$ .

$0 + \frac{5 (- 11 \cos (\frac{π t}{10} + \frac{π}{2}) π)}{10}$

Step 1.2.12.2

Cancel the common factors.

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Step 1.2.12.2.1

Factor $5$ out of $10$ .

$0 + \frac{5 (- 11 \cos (\frac{π t}{10} + \frac{π}{2}) π)}{5 (2)}$

Step 1.2.12.2.2

Cancel the common factor.

$0 + \frac{5 (- 11 \cos (\frac{π t}{10} + \frac{π}{2}) π)}{5 \cdot 2}$

Step 1.2.12.2.3

Rewrite the expression.

$0 + \frac{- 11 \cos (\frac{π t}{10} + \frac{π}{2}) π}{2}$

Step 1.2.13

Move the negative in front of the fraction.

$0 - \frac{11 \cos (\frac{π t}{10} + \frac{π}{2}) π}{2}$

Step 1.3

Simplify.

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Step 1.3.1

Subtract $\frac{11 \cos (\frac{π t}{10} + \frac{π}{2}) π}{2}$ from $0$ .

$- \frac{11 \cos (\frac{π t}{10} + \frac{π}{2}) π}{2}$

Step 1.3.2

Reorder factors in $- \frac{11 \cos (\frac{π t}{10} + \frac{π}{2}) π}{2}$ .

$- \frac{11 π \cos (\frac{π t}{10} + \frac{π}{2})}{2}$

Step 2

Find the second derivative of the function.

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Step 2.1

Since $- \frac{11 π}{2}$ is constant with respect to $t$ , the derivative of $- \frac{11 π \cos (\frac{π t}{10} + \frac{π}{2})}{2}$ with respect to $t$ is $- \frac{11 π}{2} \frac{d}{d t} [\cos (\frac{π t}{10} + \frac{π}{2})]$ .

$h'' (t) = - \frac{11 π}{2} \cdot \frac{d}{d t} \cos (\frac{π t}{10} + \frac{π}{2})$

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 $g (t) = \frac{π t}{10} + \frac{π}{2}$ .

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Step 2.2.1

To apply the Chain Rule, set $u$ as $\frac{π t}{10} + \frac{π}{2}$ .

$h'' (t) = - \frac{11 π}{2} \cdot (\frac{d}{d u} (\cos (u)) \frac{d}{d t} (\frac{π t}{10} + \frac{π}{2}))$

Step 2.2.2

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

$h'' (t) = - \frac{11 π}{2} \cdot (- \sin (u) \frac{d}{d t} (\frac{π t}{10} + \frac{π}{2}))$

Step 2.2.3

Replace all occurrences of $u$ with $\frac{π t}{10} + \frac{π}{2}$ .

$h'' (t) = - \frac{11 π}{2} \cdot (- \sin (\frac{π t}{10} + \frac{π}{2}) \frac{d}{d t} (\frac{π t}{10} + \frac{π}{2}))$

Step 2.3

Differentiate.

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Step 2.3.1

Multiply $- 1$ by $- 1$ .

$h'' (t) = 1 (\frac{11 π}{2}) \cdot (\sin (\frac{π t}{10} + \frac{π}{2}) \frac{d}{d t} (\frac{π t}{10} + \frac{π}{2}))$

Step 2.3.2

Combine fractions.

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Step 2.3.2.1

Multiply $\frac{11 π}{2}$ by $1$ .

$h'' (t) = \frac{11 π}{2} \cdot (\sin (\frac{π t}{10} + \frac{π}{2}) \frac{d}{d t} (\frac{π t}{10} + \frac{π}{2}))$

Step 2.3.2.2

Combine $\sin (\frac{π t}{10} + \frac{π}{2})$ and $\frac{11 π}{2}$ .

$h'' (t) = \frac{\sin (\frac{π t}{10} + \frac{π}{2}) (11 π)}{2} \cdot \frac{d}{d t} (\frac{π t}{10} + \frac{π}{2})$

Step 2.3.2.3

Move $11$ to the left of $\sin (\frac{π t}{10} + \frac{π}{2})$ .

$h'' (t) = \frac{11 \cdot (\sin (\frac{π t}{10} + \frac{π}{2}) π)}{2} \cdot \frac{d}{d t} (\frac{π t}{10} + \frac{π}{2})$

Step 2.3.3

By the Sum Rule, the derivative of $\frac{π t}{10} + \frac{π}{2}$ with respect to $t$ is $\frac{d}{d t} [\frac{π t}{10}] + \frac{d}{d t} [\frac{π}{2}]$ .

$h'' (t) = \frac{11 \sin (\frac{π t}{10} + \frac{π}{2}) π}{2} \cdot (\frac{d}{d t} (\frac{π t}{10}) + \frac{d}{d t} (\frac{π}{2}))$

Step 2.3.4

Since $\frac{π}{10}$ is constant with respect to $t$ , the derivative of $\frac{π t}{10}$ with respect to $t$ is $\frac{π}{10} \frac{d}{d t} [t]$ .

$h'' (t) = \frac{11 \sin (\frac{π t}{10} + \frac{π}{2}) π}{2} \cdot (\frac{π}{10} \cdot \frac{d}{d t} (t) + \frac{d}{d t} (\frac{π}{2}))$

Step 2.3.5

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

$h'' (t) = \frac{11 \sin (\frac{π t}{10} + \frac{π}{2}) π}{2} \cdot (\frac{π}{10} \cdot 1 + \frac{d}{d t} (\frac{π}{2}))$

Step 2.3.6

Multiply $\frac{π}{10}$ by $1$ .

$h'' (t) = \frac{11 \sin (\frac{π t}{10} + \frac{π}{2}) π}{2} \cdot (\frac{π}{10} + \frac{d}{d t} (\frac{π}{2}))$

Step 2.3.7

Since $\frac{π}{2}$ is constant with respect to $t$ , the derivative of $\frac{π}{2}$ with respect to $t$ is $0$ .

$h'' (t) = \frac{11 \sin (\frac{π t}{10} + \frac{π}{2}) π}{2} \cdot (\frac{π}{10} + 0)$

Step 2.3.8

Combine fractions.

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Step 2.3.8.1

Add $\frac{π}{10}$ and $0$ .

$h'' (t) = \frac{11 \sin (\frac{π t}{10} + \frac{π}{2}) π}{2} \cdot \frac{π}{10}$

Step 2.3.8.2

Multiply $\frac{11 \sin (\frac{π t}{10} + \frac{π}{2}) π}{2}$ by $\frac{π}{10}$ .

$h'' (t) = \frac{11 \sin (\frac{π t}{10} + \frac{π}{2}) π π}{2 \cdot 10}$

Step 2.4

Raise $π$ to the power of $1$ .

$h'' (t) = \frac{11 \sin (\frac{π t}{10} + \frac{π}{2}) (π π)}{2 \cdot 10}$

Step 2.5

Raise $π$ to the power of $1$ .

$h'' (t) = \frac{11 \sin (\frac{π t}{10} + \frac{π}{2}) (π π)}{2 \cdot 10}$

Step 2.6

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

$h'' (t) = \frac{11 \sin (\frac{π t}{10} + \frac{π}{2}) π^{1 + 1}}{2 \cdot 10}$

Step 2.7

Add $1$ and $1$ .

$h'' (t) = \frac{11 \sin (\frac{π t}{10} + \frac{π}{2}) π^{2}}{2 \cdot 10}$

Step 2.8

Multiply $2$ by $10$ .

$h'' (t) = \frac{11 \sin (\frac{π t}{10} + \frac{π}{2}) π^{2}}{20}$

Step 2.9

Reorder factors in $11 \sin (\frac{π t}{10} + \frac{π}{2}) π^{2}$ .

$h'' (t) = \frac{11 π^{2} \sin (\frac{π t}{10} + \frac{π}{2})}{20}$

Step 3

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

$- \frac{11 π \cos (\frac{π t}{10} + \frac{π}{2})}{2} = 0$

Step 4

Set the numerator equal to zero.

$11 π \cos (\frac{π t}{10} + \frac{π}{2}) = 0$

Step 5

Solve the equation for $t$ .

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Step 5.1

Divide each term in $11 π \cos (\frac{π t}{10} + \frac{π}{2}) = 0$ by $11 π$ and simplify.

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Step 5.1.1

Divide each term in $11 π \cos (\frac{π t}{10} + \frac{π}{2}) = 0$ by $11 π$ .

$\frac{11 π \cos (\frac{π t}{10} + \frac{π}{2})}{11 π} = \frac{0}{11 π}$

Step 5.1.2

Simplify the left side.

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Step 5.1.2.1

Cancel the common factor of $11$ .

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Step 5.1.2.1.1

Cancel the common factor.

$\frac{11 π \cos (\frac{π t}{10} + \frac{π}{2})}{11 π} = \frac{0}{11 π}$

Step 5.1.2.1.2

Rewrite the expression.

$\frac{π \cos (\frac{π t}{10} + \frac{π}{2})}{π} = \frac{0}{11 π}$

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.

$\frac{π \cos (\frac{π t}{10} + \frac{π}{2})}{π} = \frac{0}{11 π}$

Step 5.1.2.2.2

Divide $\cos (\frac{π t}{10} + \frac{π}{2})$ by $1$ .

$\cos (\frac{π t}{10} + \frac{π}{2}) = \frac{0}{11 π}$

Step 5.1.3

Simplify the right side.

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Step 5.1.3.1

Cancel the common factor of $0$ and $11$ .

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Step 5.1.3.1.1

Factor $11$ out of $0$ .

$\cos (\frac{π t}{10} + \frac{π}{2}) = \frac{11 (0)}{11 π}$

Step 5.1.3.1.2

Cancel the common factors.

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Step 5.1.3.1.2.1

Factor $11$ out of $11 π$ .

$\cos (\frac{π t}{10} + \frac{π}{2}) = \frac{11 (0)}{11 (π)}$

Step 5.1.3.1.2.2

Cancel the common factor.

$\cos (\frac{π t}{10} + \frac{π}{2}) = \frac{11 \cdot 0}{11 π}$

Step 5.1.3.1.2.3

Rewrite the expression.

$\cos (\frac{π t}{10} + \frac{π}{2}) = \frac{0}{π}$

Step 5.1.3.2

Divide $0$ by $π$ .

$\cos (\frac{π t}{10} + \frac{π}{2}) = 0$

Step 5.2

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

$\frac{π t}{10} + \frac{π}{2} = \arccos (0)$

Step 5.3

Simplify the right side.

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Step 5.3.1

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

$\frac{π t}{10} + \frac{π}{2} = \frac{π}{2}$

Step 5.4

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

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Step 5.4.1

Subtract $\frac{π}{2}$ from both sides of the equation.

$\frac{π t}{10} = \frac{π}{2} - \frac{π}{2}$

Step 5.4.2

Combine the numerators over the common denominator.

$\frac{π t}{10} = \frac{π - π}{2}$

Step 5.4.3

Subtract $π$ from $π$ .

$\frac{π t}{10} = \frac{0}{2}$

Step 5.4.4

Divide $0$ by $2$ .

$\frac{π t}{10} = 0$

Step 5.5

Set the numerator equal to zero.

$π t = 0$

Step 5.6

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

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Step 5.6.1

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

$\frac{π t}{π} = \frac{0}{π}$

Step 5.6.2

Simplify the left side.

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Step 5.6.2.1

Cancel the common factor of $π$ .

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Step 5.6.2.1.1

Cancel the common factor.

$\frac{π t}{π} = \frac{0}{π}$

Step 5.6.2.1.2

Divide $t$ by $1$ .

$t = \frac{0}{π}$

Step 5.6.3

Simplify the right side.

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Step 5.6.3.1

Divide $0$ by $π$ .

$t = 0$

Step 5.7

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.

$\frac{π t}{10} + \frac{π}{2} = 2 π - \frac{π}{2}$

Step 5.8

Solve for $t$ .

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Step 5.8.1

Simplify $2 π - \frac{π}{2}$ .

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Step 5.8.1.1

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

$\frac{π t}{10} + \frac{π}{2} = 2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 5.8.1.2

Combine fractions.

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Step 5.8.1.2.1

Combine $2 π$ and $\frac{2}{2}$ .

$\frac{π t}{10} + \frac{π}{2} = \frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 5.8.1.2.2

Combine the numerators over the common denominator.

$\frac{π t}{10} + \frac{π}{2} = \frac{2 π \cdot 2 - π}{2}$

Step 5.8.1.3

Simplify the numerator.

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Step 5.8.1.3.1

Multiply $2$ by $2$ .

$\frac{π t}{10} + \frac{π}{2} = \frac{4 π - π}{2}$

Step 5.8.1.3.2

Subtract $π$ from $4 π$ .

$\frac{π t}{10} + \frac{π}{2} = \frac{3 π}{2}$

Step 5.8.2

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

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Step 5.8.2.1

Subtract $\frac{π}{2}$ from both sides of the equation.

$\frac{π t}{10} = \frac{3 π}{2} - \frac{π}{2}$

Step 5.8.2.2

Combine the numerators over the common denominator.

$\frac{π t}{10} = \frac{3 π - π}{2}$

Step 5.8.2.3

Subtract $π$ from $3 π$ .

$\frac{π t}{10} = \frac{2 π}{2}$

Step 5.8.2.4

Cancel the common factor of $2$ .

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Step 5.8.2.4.1

Cancel the common factor.

$\frac{π t}{10} = \frac{2 π}{2}$

Step 5.8.2.4.2

Divide $π$ by $1$ .

$\frac{π t}{10} = π$

Step 5.8.3

Multiply both sides of the equation by $\frac{10}{π}$ .

$\frac{10}{π} \cdot \frac{π t}{10} = \frac{10}{π} π$

Step 5.8.4

Simplify both sides of the equation.

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Step 5.8.4.1

Simplify the left side.

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Step 5.8.4.1.1

Simplify $\frac{10}{π} \cdot \frac{π t}{10}$ .

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Step 5.8.4.1.1.1

Cancel the common factor of $10$ .

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Step 5.8.4.1.1.1.1

Cancel the common factor.

$\frac{10}{π} \cdot \frac{π t}{10} = \frac{10}{π} π$

Step 5.8.4.1.1.1.2

Rewrite the expression.

$\frac{1}{π} (π t) = \frac{10}{π} π$

Step 5.8.4.1.1.2

Cancel the common factor of $π$ .

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Step 5.8.4.1.1.2.1

Factor $π$ out of $π t$ .

$\frac{1}{π} (π (t)) = \frac{10}{π} π$

Step 5.8.4.1.1.2.2

Cancel the common factor.

$\frac{1}{π} (π t) = \frac{10}{π} π$

Step 5.8.4.1.1.2.3

Rewrite the expression.

$t = \frac{10}{π} π$

Step 5.8.4.2

Simplify the right side.

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Step 5.8.4.2.1

Cancel the common factor of $π$ .

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Step 5.8.4.2.1.1

Cancel the common factor.

$t = \frac{10}{π} π$

Step 5.8.4.2.1.2

Rewrite the expression.

$t = 10$

Step 5.9

The solution to the equation $\frac{π t}{10} + \frac{π}{2} = \frac{π}{2}$ .

$t = 0, 10$

Step 6

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.

$\frac{11 π^{2} \sin (\frac{π (0)}{10} + \frac{π}{2})}{20}$

Step 7

Evaluate the second derivative.

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Step 7.1

Cancel the common factor of $0$ and $10$ .

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Step 7.1.1

Factor $10$ out of $π (0)$ .

$\frac{11 π^{2} \sin (\frac{10 (π \cdot (0))}{10} + \frac{π}{2})}{20}$

Step 7.1.2

Cancel the common factors.

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Step 7.1.2.1

Factor $10$ out of $10$ .

$\frac{11 π^{2} \sin (\frac{10 (π \cdot (0))}{10 (1)} + \frac{π}{2})}{20}$

Step 7.1.2.2

Cancel the common factor.

$\frac{11 π^{2} \sin (\frac{10 (π \cdot (0))}{10 \cdot 1} + \frac{π}{2})}{20}$

Step 7.1.2.3

Rewrite the expression.

$\frac{11 π^{2} \sin (\frac{π \cdot (0)}{1} + \frac{π}{2})}{20}$

Step 7.1.2.4

Divide $π \cdot (0)$ by $1$ .

$\frac{11 π^{2} \sin (π \cdot (0) + \frac{π}{2})}{20}$

Step 7.2

Simplify the numerator.

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Step 7.2.1

Multiply $π$ by $0$ .

$\frac{11 π^{2} \sin (0 + \frac{π}{2})}{20}$

Step 7.2.2

Add $0$ and $\frac{π}{2}$ .

$\frac{11 π^{2} \sin (\frac{π}{2})}{20}$

Step 7.2.3

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

$\frac{11 π^{2} \cdot 1}{20}$

Step 7.2.4

Multiply $11$ by $1$ .

$\frac{11 π^{2}}{20}$

Step 8

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

$t = 0$ is a local minimum

Step 9

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

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Step 9.1

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

$h (0) = 60 - 55 \sin (\frac{π}{10} \cdot (0) + \frac{π}{2})$

Step 9.2

Simplify the result.

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Step 9.2.1

Simplify each term.

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Step 9.2.1.1

Multiply $\frac{π}{10}$ by $0$ .

$h (0) = 60 - 55 \sin (0 + \frac{π}{2})$

Step 9.2.1.2

Add $0$ and $\frac{π}{2}$ .

$h (0) = 60 - 55 \sin (\frac{π}{2})$

Step 9.2.1.3

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

$h (0) = 60 - 55 \cdot 1$

Step 9.2.1.4

Multiply $- 55$ by $1$ .

$h (0) = 60 - 55$

Step 9.2.2

Subtract $55$ from $60$ .

$h (0) = 5$

Step 9.2.3

The final answer is $5$ .

$y = 5$

Step 10

Evaluate the second derivative at $t = 10$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{11 π^{2} \sin (\frac{π (10)}{10} + \frac{π}{2})}{20}$

Step 11

Evaluate the second derivative.

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Step 11.1

Cancel the common factor of $10$ .

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Step 11.1.1

Cancel the common factor.

$\frac{11 π^{2} \sin (\frac{π \cdot 10}{10} + \frac{π}{2})}{20}$

Step 11.1.2

Divide $π$ by $1$ .

$\frac{11 π^{2} \sin (π + \frac{π}{2})}{20}$

Step 11.2

Simplify the numerator.

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Step 11.2.1

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

$\frac{11 π^{2} \sin (π \cdot \frac{2}{2} + \frac{π}{2})}{20}$

Step 11.2.2

Combine $π$ and $\frac{2}{2}$ .

$\frac{11 π^{2} \sin (\frac{π \cdot 2}{2} + \frac{π}{2})}{20}$

Step 11.2.3

Combine the numerators over the common denominator.

$\frac{11 π^{2} \sin (\frac{π \cdot 2 + π}{2})}{20}$

Step 11.2.4

Simplify the numerator.

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Step 11.2.4.1

Move $2$ to the left of $π$ .

$\frac{11 π^{2} \sin (\frac{2 \cdot π + π}{2})}{20}$

Step 11.2.4.2

Add $2 π$ and $π$ .

$\frac{11 π^{2} \sin (\frac{3 π}{2})}{20}$

Step 11.2.5

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.

$\frac{11 π^{2} (- \sin (\frac{π}{2}))}{20}$

Step 11.2.6

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

$\frac{11 π^{2} (- 1 \cdot 1)}{20}$

Step 11.2.7

Multiply $- 1$ by $1$ .

$\frac{11 π^{2} \cdot - 1}{20}$

Step 11.2.8

Multiply $- 1$ by $11$ .

$\frac{- 11 π^{2}}{20}$

Step 11.3

Move the negative in front of the fraction.

$- \frac{11 π^{2}}{20}$

Step 12

$t = 10$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$t = 10$ is a local maximum

Step 13

Find the y-value when $t = 10$ .

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Step 13.1

Replace the variable $t$ with $10$ in the expression.

$h (10) = 60 - 55 \sin (\frac{π}{10} \cdot (10) + \frac{π}{2})$

Step 13.2

Simplify the result.

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Step 13.2.1

Simplify each term.

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Step 13.2.1.1

Cancel the common factor of $10$ .

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Step 13.2.1.1.1

Cancel the common factor.

$h (10) = 60 - 55 \sin (\frac{π}{10} \cdot 10 + \frac{π}{2})$

Step 13.2.1.1.2

Rewrite the expression.

$h (10) = 60 - 55 \sin (π + \frac{π}{2})$

Step 13.2.1.2

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

$h (10) = 60 - 55 \sin (π \cdot \frac{2}{2} + \frac{π}{2})$

Step 13.2.1.3

Combine $π$ and $\frac{2}{2}$ .

$h (10) = 60 - 55 \sin (\frac{π \cdot 2}{2} + \frac{π}{2})$

Step 13.2.1.4

Combine the numerators over the common denominator.

$h (10) = 60 - 55 \sin (\frac{π \cdot 2 + π}{2})$

Step 13.2.1.5

Simplify the numerator.

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Step 13.2.1.5.1

Move $2$ to the left of $π$ .

$h (10) = 60 - 55 \sin (\frac{2 \cdot π + π}{2})$

Step 13.2.1.5.2

Add $2 π$ and $π$ .

$h (10) = 60 - 55 \sin (\frac{3 π}{2})$

Step 13.2.1.6

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.

$h (10) = 60 - 55 (- \sin (\frac{π}{2}))$

Step 13.2.1.7

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

$h (10) = 60 - 55 (- 1 \cdot 1)$

Step 13.2.1.8

Multiply $- 55 (- 1 \cdot 1)$ .

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Step 13.2.1.8.1

Multiply $- 1$ by $1$ .

$h (10) = 60 - 55 \cdot - 1$

Step 13.2.1.8.2

Multiply $- 55$ by $- 1$ .

$h (10) = 60 + 55$

Step 13.2.2

Add $60$ and $55$ .

$h (10) = 115$

Step 13.2.3

The final answer is $115$ .

$y = 115$

Step 14

These are the local extrema for $h (t) = 60 - 55 \sin (\frac{π}{10} t + \frac{π}{2})$ .

$(0, 5)$ is a local minima

$(10, 115)$ is a local maxima

Step 15