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 $\pi/3$.

Answer 1

Toolbox:

• 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}$

http://clay6.com/mpaimg/tn%205-1-7.JPG

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 $0.3 m^2 /sec$