A point P(p,0) moves along the x-axis and a point Q lies on the curve y = x^2 such that PQ is vertical. The fixed point S is (2,0). The area of triangle PQS is A square units. Show that A = p^2 - \dfrac{1}{2}p^3, and hence find \dfrac{dA}{dt} when p = -2 and \dfrac{dp}{dt} = -0.5 units per second.

-5\text{ units}^2\text{ s}^{-1}

10\text{ units}^2\text{ s}^{-1}

-10\text{ units}^2\text{ s}^{-1}

AM-1948

Connected rates of change

A Math · Secondary 4 · Hard · Connected rates of change

A point

P(p,0)

moves along the

x

-axis and a point

Q

lies on the curve

y = x^2

such that

PQ

is vertical. The fixed point

S

(2,0)

. The area of triangle

PQS

A

square units. Show that

A = p^2 - \dfrac{1}{2}p^3

, and hence find

\dfrac{dA}{dt}

when

p = -2

and

\dfrac{dp}{dt} = -0.5

units per second.

Loading problem…