Calculus Examples

$r (t) = (3 \sin (t)) i - 5 \cos (3 t) j + e^{- 7 t} k$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $(3 \sin (t)) i - 5 \cos (3 t) j + e^{- 7 t} k$ with respect to $t$ is $\frac{d}{d t} [(3 \sin (t)) i] + \frac{d}{d t} [- 5 \cos (3 t) j] + \frac{d}{d t} [e^{- 7 t} k]$ .

$\frac{d}{d t} [(3 \sin (t)) i] + \frac{d}{d t} [- 5 \cos (3 t) j] + \frac{d}{d t} [e^{- 7 t} k]$

Step 1.2

Evaluate $\frac{d}{d t} [(3 \sin (t)) i]$ .

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Step 1.2.1

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

$3 i \frac{d}{d t} [\sin (t)] + \frac{d}{d t} [- 5 \cos (3 t) j] + \frac{d}{d t} [e^{- 7 t} k]$

Step 1.2.2

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

$3 i \cos (t) + \frac{d}{d t} [- 5 \cos (3 t) j] + \frac{d}{d t} [e^{- 7 t} k]$

Step 1.3

Evaluate $\frac{d}{d t} [- 5 \cos (3 t) j]$ .

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Step 1.3.1

Since $- 5 j$ is constant with respect to $t$ , the derivative of $- 5 \cos (3 t) j$ with respect to $t$ is $- 5 j \frac{d}{d t} [\cos (3 t)]$ .

$3 i \cos (t) - 5 j \frac{d}{d t} [\cos (3 t)] + \frac{d}{d t} [e^{- 7 t} k]$

Step 1.3.2

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 $g (t) = 3 t$ .

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Step 1.3.2.1

To apply the Chain Rule, set $u_{1}$ as $3 t$ .

$3 i \cos (t) - 5 j (\frac{d}{d u_{1}} [\cos (u_{1})] \frac{d}{d t} [3 t]) + \frac{d}{d t} [e^{- 7 t} k]$

Step 1.3.2.2

The derivative of $\cos (u_{1})$ with respect to $u_{1}$ is $- \sin (u_{1})$ .

$3 i \cos (t) - 5 j (- \sin (u_{1}) \frac{d}{d t} [3 t]) + \frac{d}{d t} [e^{- 7 t} k]$

Step 1.3.2.3

Replace all occurrences of $u_{1}$ with $3 t$ .

$3 i \cos (t) - 5 j (- \sin (3 t) \frac{d}{d t} [3 t]) + \frac{d}{d t} [e^{- 7 t} k]$

Step 1.3.3

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

$3 i \cos (t) - 5 j (- \sin (3 t) (3 \frac{d}{d t} [t])) + \frac{d}{d t} [e^{- 7 t} k]$

Step 1.3.4

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

$3 i \cos (t) - 5 j (- \sin (3 t) (3 \cdot 1)) + \frac{d}{d t} [e^{- 7 t} k]$

Step 1.3.5

Multiply $3$ by $1$ .

$3 i \cos (t) - 5 j (- \sin (3 t) \cdot 3) + \frac{d}{d t} [e^{- 7 t} k]$

Step 1.3.6

Multiply $3$ by $- 1$ .

$3 i \cos (t) - 5 j (- 3 \sin (3 t)) + \frac{d}{d t} [e^{- 7 t} k]$

Step 1.3.7

Multiply $- 3$ by $- 5$ .

$3 i \cos (t) + 15 j \sin (3 t) + \frac{d}{d t} [e^{- 7 t} k]$

Step 1.4

Evaluate $\frac{d}{d t} [e^{- 7 t} k]$ .

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Step 1.4.1

Since $k$ is constant with respect to $t$ , the derivative of $e^{- 7 t} k$ with respect to $t$ is $k \frac{d}{d t} [e^{- 7 t}]$ .

$3 i \cos (t) + 15 j \sin (3 t) + k \frac{d}{d t} [e^{- 7 t}]$

Step 1.4.2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = - 7 t$ .

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Step 1.4.2.1

To apply the Chain Rule, set $u_{2}$ as $- 7 t$ .

$3 i \cos (t) + 15 j \sin (3 t) + k (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d t} [- 7 t])$

Step 1.4.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$3 i \cos (t) + 15 j \sin (3 t) + k (e^{u_{2}} \frac{d}{d t} [- 7 t])$

Step 1.4.2.3

Replace all occurrences of $u_{2}$ with $- 7 t$ .

$3 i \cos (t) + 15 j \sin (3 t) + k (e^{- 7 t} \frac{d}{d t} [- 7 t])$

Step 1.4.3

Since $- 7$ is constant with respect to $t$ , the derivative of $- 7 t$ with respect to $t$ is $- 7 \frac{d}{d t} [t]$ .

$3 i \cos (t) + 15 j \sin (3 t) + k (e^{- 7 t} (- 7 \frac{d}{d t} [t]))$

Step 1.4.4

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

$3 i \cos (t) + 15 j \sin (3 t) + k (e^{- 7 t} (- 7 \cdot 1))$

Step 1.4.5

Multiply $- 7$ by $1$ .

$3 i \cos (t) + 15 j \sin (3 t) + k (e^{- 7 t} \cdot - 7)$

Step 1.4.6

Move $- 7$ to the left of $e^{- 7 t}$ .

$3 i \cos (t) + 15 j \sin (3 t) + k (- 7 e^{- 7 t})$

Step 1.5

Simplify.

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Step 1.5.1

Reorder terms.

$3 i \cos (t) + 15 j \sin (3 t) - 7 e^{- 7 t} k$

Step 1.5.2

Reorder factors in $3 i \cos (t) + 15 j \sin (3 t) - 7 e^{- 7 t} k$ .

$3 i \cos (t) + 15 j \sin (3 t) - 7 k e^{- 7 t}$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $3 i \cos (t) + 15 j \sin (3 t) - 7 k e^{- 7 t}$ with respect to $t$ is $\frac{d}{d t} [3 i \cos (t)] + \frac{d}{d t} [15 j \sin (3 t)] + \frac{d}{d t} [- 7 k e^{- 7 t}]$ .

$r'' (t) = \frac{d}{d t} (3 i \cos (t)) + \frac{d}{d t} (15 j \sin (3 t)) + \frac{d}{d t} (- 7 k e^{- 7 t})$

Step 2.2

Evaluate $\frac{d}{d t} [3 i \cos (t)]$ .

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Step 2.2.1

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

$r'' (t) = 3 i \frac{d}{d t} (\cos (t)) + \frac{d}{d t} (15 j \sin (3 t)) + \frac{d}{d t} (- 7 k e^{- 7 t})$

Step 2.2.2

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

$r'' (t) = 3 i (- \sin (t)) + \frac{d}{d t} (15 j \sin (3 t)) + \frac{d}{d t} (- 7 k e^{- 7 t})$

Step 2.2.3

Multiply $- 1$ by $3$ .

$r'' (t) = - 3 i \sin (t) + \frac{d}{d t} (15 j \sin (3 t)) + \frac{d}{d t} (- 7 k e^{- 7 t})$

Step 2.3

Evaluate $\frac{d}{d t} [15 j \sin (3 t)]$ .

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Step 2.3.1

Since $15 j$ is constant with respect to $t$ , the derivative of $15 j \sin (3 t)$ with respect to $t$ is $15 j \frac{d}{d t} [\sin (3 t)]$ .

$r'' (t) = - 3 i \sin (t) + 15 j \frac{d}{d t} (\sin (3 t)) + \frac{d}{d t} (- 7 k e^{- 7 t})$

Step 2.3.2

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 $g (t) = 3 t$ .

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Step 2.3.2.1

To apply the Chain Rule, set $u_{1}$ as $3 t$ .

$r'' (t) = - 3 i \sin (t) + 15 j (\frac{d}{d u (1)} (\sin (u_{1})) \frac{d}{d t} (3 t)) + \frac{d}{d t} (- 7 k e^{- 7 t})$

Step 2.3.2.2

The derivative of $\sin (u_{1})$ with respect to $u_{1}$ is $\cos (u_{1})$ .

$r'' (t) = - 3 i \sin (t) + 15 j (\cos (u_{1}) \frac{d}{d t} (3 t)) + \frac{d}{d t} (- 7 k e^{- 7 t})$

Step 2.3.2.3

Replace all occurrences of $u_{1}$ with $3 t$ .

$r'' (t) = - 3 i \sin (t) + 15 j (\cos (3 t) \frac{d}{d t} (3 t)) + \frac{d}{d t} (- 7 k e^{- 7 t})$

Step 2.3.3

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

$r'' (t) = - 3 i \sin (t) + 15 j (\cos (3 t) (3 \frac{d}{d t} (t))) + \frac{d}{d t} (- 7 k e^{- 7 t})$

Step 2.3.4

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

$r'' (t) = - 3 i \sin (t) + 15 j (\cos (3 t) (3 \cdot 1)) + \frac{d}{d t} (- 7 k e^{- 7 t})$

Step 2.3.5

Multiply $3$ by $1$ .

$r'' (t) = - 3 i \sin (t) + 15 j (\cos (3 t) \cdot 3) + \frac{d}{d t} (- 7 k e^{- 7 t})$

Step 2.3.6

Move $3$ to the left of $\cos (3 t)$ .

$r'' (t) = - 3 i \sin (t) + 15 j (3 \cdot \cos (3 t)) + \frac{d}{d t} (- 7 k e^{- 7 t})$

Step 2.3.7

Multiply $3$ by $15$ .

$r'' (t) = - 3 i \sin (t) + 45 j \cos (3 t) + \frac{d}{d t} (- 7 k e^{- 7 t})$

Step 2.4

Evaluate $\frac{d}{d t} [- 7 k e^{- 7 t}]$ .

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Step 2.4.1

Since $- 7 k$ is constant with respect to $t$ , the derivative of $- 7 k e^{- 7 t}$ with respect to $t$ is $- 7 k \frac{d}{d t} [e^{- 7 t}]$ .

$r'' (t) = - 3 i \sin (t) + 45 j \cos (3 t) - 7 k \frac{d}{d t} e^{- 7 t}$

Step 2.4.2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = - 7 t$ .

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Step 2.4.2.1

To apply the Chain Rule, set $u_{2}$ as $- 7 t$ .

$r'' (t) = - 3 i \sin (t) + 45 j \cos (3 t) - 7 k (\frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d t} (- 7 t))$

Step 2.4.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$r'' (t) = - 3 i \sin (t) + 45 j \cos (3 t) - 7 k (e^{u_{2}} \frac{d}{d t} (- 7 t))$

Step 2.4.2.3

Replace all occurrences of $u_{2}$ with $- 7 t$ .

$r'' (t) = - 3 i \sin (t) + 45 j \cos (3 t) - 7 k (e^{- 7 t} \frac{d}{d t} (- 7 t))$

Step 2.4.3

Since $- 7$ is constant with respect to $t$ , the derivative of $- 7 t$ with respect to $t$ is $- 7 \frac{d}{d t} [t]$ .

$r'' (t) = - 3 i \sin (t) + 45 j \cos (3 t) - 7 k (e^{- 7 t} (- 7 \frac{d}{d t} t))$

Step 2.4.4

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

$r'' (t) = - 3 i \sin (t) + 45 j \cos (3 t) - 7 k (e^{- 7 t} (- 7 \cdot 1))$

Step 2.4.5

Multiply $- 7$ by $1$ .

$r'' (t) = - 3 i \sin (t) + 45 j \cos (3 t) - 7 k (e^{- 7 t} \cdot - 7)$

Step 2.4.6

Move $- 7$ to the left of $e^{- 7 t}$ .

$r'' (t) = - 3 i \sin (t) + 45 j \cos (3 t) - 7 k (- 7 \cdot e^{- 7 t})$

Step 2.4.7

Multiply $- 7$ by $- 7$ .

$r'' (t) = - 3 i \sin (t) + 45 j \cos (3 t) + 49 k e^{- 7 t}$

Step 3

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

$3 i \cos (t) + 15 j \sin (3 t) - 7 k e^{- 7 t} = 0$

Step 4

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.

$- 3 i \sin (0) + 45 j \cos (3 (0)) + 49 k e^{- 7 \cdot 0}$

Step 5

Evaluate the second derivative.

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Step 5.1

Simplify each term.

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Step 5.1.1

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

$- 3 i \cdot 0 + 45 j \cos (3 (0)) + 49 k e^{- 7 \cdot 0}$

Step 5.1.2

Multiply $0$ by $- 3$ .

$0 i + 45 j \cos (3 (0)) + 49 k e^{- 7 \cdot 0}$

Step 5.1.3

Multiply $0$ by $i$ .

$0 + 45 j \cos (3 (0)) + 49 k e^{- 7 \cdot 0}$

Step 5.1.4

Multiply $3$ by $0$ .

$0 + 45 j \cos (0) + 49 k e^{- 7 \cdot 0}$

Step 5.1.5

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

$0 + 45 j \cdot 1 + 49 k e^{- 7 \cdot 0}$

Step 5.1.6

Multiply $45$ by $1$ .

$0 + 45 j + 49 k e^{- 7 \cdot 0}$

Step 5.1.7

Multiply $- 7$ by $0$ .

$0 + 45 j + 49 k e^{0}$

Step 5.1.8

Anything raised to $0$ is $1$ .

$0 + 45 j + 49 k \cdot 1$

Step 5.1.9

Multiply $49$ by $1$ .

$0 + 45 j + 49 k$

Step 5.2

Add $0$ and $45 j$ .

$45 j + 49 k$

Step 6

Since the first derivative test failed, there are no local extrema.

No Local Extrema

Step 7