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Two sides of a tringle are $4$m and $5$m in length and the angle between them is increasing at a rate of $0.06$ rad/sec. Find the rate at which the area of tringle is increasing when the angle between the sides of fixed length is $\pi/3$.

1 Answer

  • If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x)$ is the rate of change of w.r.t x
  • $\large\frac{dy}{dx_{(x_1,y_1}}$ is the slope of the tangent to the curve at the point $(x_1,y_1) $ on the curve. It is the slope of the curve at that point.
  • The normal at a point $(x_1,y_1)$ on $y=f(x)$ is perpendicular to the tangent at $(x_1,y_1)$
  • It $y=f(x)$ and both depend on 't' then $\large\frac{dy}{dt}$$=f'(x).\large\frac{dx}{dt}$
Step 1:
Let $OA,OB$ be the sides of $\Delta\; OA$ and $OB$ where $OA=4m,OB=5m \;and \;\angle AOB=\theta$
The area of the triangle at time 't' is
$A=\large\frac{1}{2}$$OA \times OB \sin \theta$
$\quad=\large\frac{1}{2}$$ \times 4 \times 5 \sin \theta$
$\quad=10 \sin \theta$
Step 2:
The angle $\theta$ is increasing at the rate of $0.06^c/sec$
(ie) $\large\frac{d \theta}{dt}$$=0.06^c/sec$
Step 3:
The rate of change of the area is
$\large\frac{dA}{dt}=$$10 \cos \theta \large\frac{d \theta}{dt}$
Step 4:
When $\theta=\large\frac{\pi}{3} $ and $\large\frac{d\theta}{dt}$$=0.06$
$\large\frac{dA}{dt}$$=10 \cos \large\frac{\pi}{3} $$\times 0.06$
$\quad=10 \times \large\frac{1}{2}$$ \times 0.06$
$\quad=0.3 m^2 /sec$
The area is increasing at $0.3 m^2 /sec$
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