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O lvl A Maths: Applications of Differentiation (Rates of Change)

The figure shows the curve y² = 2x. The points O (0,0) and P(x, y) lie on the curve. Given that the coordinates of Q is (x, 0),

(a) express the area, A cm², of the triangle OPQ in terms of x.
(b) If x is increasing at a rate of 4 units per seconds, calculate the rate of increase of A when x = 2.



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Answer:

(a) Area of triangle OPQ, A = ½(OQ)(PQ) = ½xy
y² = 2x
\Rightarrow y=\pm \sqrt{2x}

Since area of triangle is positive, we take the positive value of y

Hence,
A=\frac{1}{2}x\left( \sqrt{2x}\right)


(b) Using product rule,
\frac{dA}{dx}=\left(\sqrt{2x} \right)\left(\frac{1}{2} \right)+\frac{1}{2}x\left[\frac{1}{2}\left(2x \right)^{-\frac{1}{2}} \right]\left(2 \right)
=\frac{1}{2}\sqrt{2x}+\frac{x}{2\sqrt{2x}}
=\frac{2x+x}{2\sqrt{2x}}
=\frac{3x}{2\sqrt{2x}}

If x is increasing at a rate of 4 units per seconds, then dx/dt = 4.
\frac{dA}{dt}=\frac{dA}{dx}\times \frac{dx}{dt} (chain rule)
=\frac{3x}{2\sqrt{2x}}\times 4
=\frac{6x}{\sqrt{2x}}

When x = 2, dA/dt = 6(2) / √(2)(2)
= 6



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