Question

Calculus is hard. T-T

Answer 1

The chord length x is growing at a rate of 1 m/min after 2π minutes.

Explanation:

The formula for chord length x in terms of radius r is:

`\boxed{x &= 2r\sin\left((\theta)/(2)\right)}`

where θ is the central angle subtended by the chord.

The formula for arc length s in terms of radius r is:

`\boxed{s=r\theta}`

where θ is the central angle subtended by the chord.

Rearrange the formula for arc length to isolate θ, then substitute this into the the formula for chord length to create an equation for x in terms of s:

`\begin{aligned}\theta=(s)/(r) \implies x &= 2r\sin\left(((s)/(r))/(2)\right)\\x&= 2r\sin\left((s)/(2r)\right)\end{aligned}`

Given the radius is 6 m, substitute r = 6 into the equation for x:

`\implies x &= 12\sin\left((s)/(12)\right)`

Differentiate with respect to s:

`\begin{aligned}\implies \frac{\text{d}x}{\text{d}s} &= (1)/(12) \cdot 12\cos\left((s)/(12)\right)\\\\&=\cos\left((s)/(12)\right)\end{aligned}`

Given the arc length s is growing at a rate of 2 m/min:

`\implies \frac{\text{d}s}{\text{d}t}=2`

Use the chain rule to create an equation for the rate of change of the chord x:

`\begin{aligned}\implies \frac{\text{d}x}{\text{d}t}&=\frac{\text{d}x}{\text{d}s}* \frac{\text{d}s}{\text{d}t}\\\\&=\cos\left((s)/(12)\right) * 2\\\\&=2\cos\left((s)/(12)\right) \end{aligned}`

Given the arc length is growing at a rate of 2 m/min:

`\implies \frac{\text{d}s}{\text{d}t}=2`

Integrate to create an equation for s in terms of t:

`\begin{aligned}\displaystyle \int 1 \; \text{d}s&=\int 2\; \text{d}t}\\\\ s&=2t+\text{C}\end{aligned}`

Given s = 0 when t = 0, this means that C = 0. Therefore:

`\implies s=2t`

If t = 2π then:

`\implies s=2 \cdot 2\pi = 4\pi`

Finally, substitute s = 4π into the equation to calculate the rate of change of the chord at 2π minutes:

`\begin{aligned}\implies \frac{\text{d}x}{\text{d}t}&=2\cos\left((4\pi)/(12)\right) \\\\&=2 \cos \left((\pi)/(3)\right)\\\\&=2 \cdot (1)/(2)\\\\&=1\end{aligned}`

Therefore, the chord length x is growing at a rate of 1 m/min after 2π minutes.