Course:MATH102/Question Challenge/2002 December Q2(v)

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Question

In the triangle shown here, if ${\frac {d\theta }{dt}}=1$, the rate of change of side length $c$ at the instant that $c=1$ is given by

(a) $8-8\sin(\theta )$

(b) $8\sin(\theta )$

(c) $4\cos(\theta )$

(d) $8-8\cos(\theta )$

(e) $4\sin(\theta )$

Triangle

Hints

[show]Hint 1
Remember the Law of cosines: $c^{2}=a^{2}+b^{2}-2ab\cos(\theta )$ where $\theta$ is the angle opposite side $c$

[show]Hint 2
Use related rates.

Solutions

[show]Solution
$d\theta /dt=1$ $c=1$ To solve for $dc/dt$, we write an expression for this formula using the law of cosines, then derive to find $dc/dt$: $c^{2}=a^{2}+b^{2}-2ab\cos \theta$ $2c\cdot dc/dt=-8(-\sin \theta \cdot (1))$ $dc/dt=8\sin \theta /2c$ $dc/dt=8\sin \theta /2(1)$ $dc/dt=4\sin \theta$