The diagram shows the curve $y = 2x^{2} + 3$, where $P(x, y)$ is a point moving along the c

Question

The diagram shows the curve $y = 2x^{2} + 3$, where $P(x, y)$ is a point moving along the curve for $0 < x < 6$. $Q$ is a point on the $x$-axis such that $PQ$ is parallel to the $y$-axis and $R(6, 0)$ is a fixed point on the $x$-axis. Express the area of triangle $PQR$, $T$ $\text{units}^{2}$, in terms of $x$. Hence find $\frac{\mathrm{d}T}{\mathrm{d}x}$. Given that when $x = 2$, $T$ is increasing at the rate of $0.8$ $\text{units}^{2}/\text{s}$, find the corresponding rate of change of $x$ at this instant.

Accepted Answer

(i) $T=\dfrac12(6-x)(2x^2+3)$ (ii) $\dfrac{\mathrm{d}T}{\mathrm{d}x}=-\dfrac32(2x^2-8x+1)$ (iii) $\dfrac{8}{105}\text{ unit s}^{-1}\approx0.0762\text{ unit s}^{-1}$

Differentiation - Connected Rates of Change

Differentiation - Connected Rates of Change

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