First slide

Rate Measure

Question

$Let x be the length of one of the equal sides of an isosceles triangle$ $and let θ be the angle between them. If x is increasing at the rate of$ $(1 / 12) m / hr and θ is increasing at the rate of \frac{π}{180} radians / hr, then$ $the rate in m^{2} / hr at which the area of the triangle is increasing when$ $x = 12 m and θ = \frac{π}{4}$

Moderate

A

$\sqrt{2} (\frac{1}{2} + \frac{π}{5})$
B

$\sqrt{2}$
C

$\frac{1}{2} + \frac{π}{5}$
D

$\frac{π}{2}$

Solution

$\begin{array}{l} Δ A B C = \frac{1}{2} x^{2} \sin θ \\ \frac{d Δ}{d t} = \frac{1}{2} (2 x \frac{d x}{d t} \sin θ + x^{2} \cos θ \frac{d θ}{d t}) \\ = x \frac{d x}{d t} \sin θ + \frac{x^{2}}{2} \cos θ \frac{d θ}{d t} \\ = (12) \frac{1}{12} \times \frac{1}{\sqrt{2}} + \frac{144}{2} \times \frac{1}{\sqrt{2}} \frac{π}{180} \\ = (\frac{1}{2} + \frac{π}{5}) \sqrt{2} \end{array}$

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