Question
$\mathrm{A}=\frac{1}{2} \mathrm{x}^{2} \sin \theta$$\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\mathrm{x} \sin \theta \frac{\mathrm{dx}}{\mathrm{dt}}+\frac{\mathrm{x}^{2}}{2} \cos \theta \frac{\mathrm{d} \theta}{\mathrm{dt}}$$\quad=12 \times \frac{1}{\sqrt{2}} \times \frac{1}{12}+\frac{144}{2} \times \frac{1}{\sqrt{2}} \times \frac{\pi}{180}$$=2^{1 / 2}\left(\frac{1}{2}+\frac{\pi}{5}\right)$
Step 1
Step 1: Given the area of the triangle as $A=\frac{1}{2} x^{2} \sin \theta$, we need to find the rate of change of the area with respect to time, which is $\frac{dA}{dt}$. Show more…
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