Nyala is bungee jumping. The following function gives her elevation, in meters, $t$ seconds after jumping: $\newline$ $E(t)=\frac{80 \sin (0.6 t)}{t}+15$ $\newline$ What is the instantaneous rate of change of Nyala's elevation after $5$ seconds? $\newline$ Choose $1$ answer: $\newline$ (A) $-9$ . $96$ meters per second $\newline$ (B) $-9$ . $96$ seconds per meter $\newline$ (C) $\mathbf{1 7 . 2 6}$ meters per second $\newline$ (D) $\mathbf{1 7 . 2 6}$ seconds per meter

View step-by-step help

Home

Math Problems

Calculus

Find derivatives of sine and cosine functions

Full solution

Q. Nyala is bungee jumping. The following function gives her elevation, in meters,

t

seconds after jumping:

\newline

E(t)=\frac{80 \sin (0.6 t)}{t}+15

\newline

What is the instantaneous rate of change of Nyala's elevation after

5

seconds?

\newline

Choose

1

answer:

\newline

(A)

-9

96

meters per second

\newline

(B)

-9

96

seconds per meter

\newline

(C)

\mathbf{1 7 . 2 6}

meters per second

\newline

(D)

\mathbf{1 7 . 2 6}

seconds per meter

Take Derivative: To find the instantaneous rate of change, we need to take the derivative of $E(t)$ with respect to $t$ .
Apply Quotient Rule: Differentiate $E(t)$ using the quotient rule: $\frac{v(u') - u(v')}{v^2}$ , where $u = 80 \sin(0.6 t)$ and $v = t$ .
Find $u'$ and $v'$ : First, find the derivative of $u$ , which is $80 \cos(0.6 t) \times 0.6$ .
Simplify Expression: Then, find the derivative of $v$ , which is just $1$ since $v = t$ .
Plug in $t=5$ : Now apply the quotient rule: $E'(t) = \frac{t \times (80 \cos(0.6 t) \times 0.6) - 80 \sin(0.6 t) \times 1}{t^2}$ .
Calculate Values: Simplify the expression: $E'(t) = \frac{80t \cos(0.6 t) \cdot 0.6 - 80 \sin(0.6 t)}{t^2}$ .
Calculate Values: Simplify the expression: $E'(t) = \frac{80t \cos(0.6 t) \times 0.6 - 80 \sin(0.6 t)}{t^2}$ .Plug in $t = 5$ seconds into the derivative to find the instantaneous rate of change at $t = 5$ .
Calculate Values: Simplify the expression: $E'(t) = \frac{80t \cos(0.6 t) \cdot 0.6 - 80 \sin(0.6 t)}{t^2}$ .Plug in $t = 5$ seconds into the derivative to find the instantaneous rate of change at $t = 5$ . $E'(5) = \frac{80\cdot 5 \cos(0.6\cdot 5) \cdot 0.6 - 80 \sin(0.6\cdot 5)}{5^2}$ .
Calculate Values: Simplify the expression: $E'(t) = \frac{80t \cos(0.6 t) \times 0.6 - 80 \sin(0.6 t)}{t^2}$ .Plug in $t = 5$ seconds into the derivative to find the instantaneous rate of change at $t = 5$ . $E'(5) = \frac{80\times5 \cos(0.6\times5) \times 0.6 - 80 \sin(0.6\times5)}{5^2}$ .Calculate the values: $E'(5) = \frac{400 \cos(3) \times 0.6 - 80 \sin(3)}{25}$ .
Calculate Values: Simplify the expression: $E'(t) = \frac{80t \cos(0.6 t) \times 0.6 - 80 \sin(0.6 t)}{t^2}$ .Plug in $t = 5$ seconds into the derivative to find the instantaneous rate of change at $t = 5$ . $E'(5) = \frac{80\times5 \cos(0.6\times5) \times 0.6 - 80 \sin(0.6\times5)}{5^2}$ .Calculate the values: $E'(5) = \frac{400 \cos(3) \times 0.6 - 80 \sin(3)}{25}$ .Use a calculator to find the numerical values of $\cos(3)$ and $\sin(3)$ .
Calculate Values: Simplify the expression: $E'(t) = \frac{(80t \cos(0.6 t) \times 0.6 - 80 \sin(0.6 t))}{t^2}$ . Plug in $t = 5$ seconds into the derivative to find the instantaneous rate of change at $t = 5$ . $E'(5) = \frac{(80\times5 \cos(0.6\times5) \times 0.6 - 80 \sin(0.6\times5))}{5^2}$ . Calculate the values: $E'(5) = \frac{(400 \cos(3) \times 0.6 - 80 \sin(3))}{25}$ . Use a calculator to find the numerical values of $\cos(3)$ and $\sin(3)$ . $E'(5) = \frac{(400 \times 0.6 \times (-0.99) - 80 \times 0.14)}{25}$ .