Theory question 109
DifferentiationA curve is given by y = 8x + 27/(2x2). Find: (a) an expression for dy/dx; (b) the co-ordinates of the stationary point on the curve and the nature of the stationary points; (c) the equation of normal to the curve at (2, 2).
Show the model answer
Model answer
(a) y = 8x + (27/2)x⁻², so dy/dx = 8 − 27/x³.
(b) 8 − 27/x³ = 0 gives x³ = 27/8, x = 3/2; y = 12 + 27/(2 × 9/4) = 12 + 6 = 18. Stationary point (3/2, 18).
d²y/dx² = 81/x⁴ > 0, so it is a minimum point.
(c) At x = 2, dy/dx = 8 − 27/8 = 37/8; normal gradient = −8/37.
Through (2, 2) as stated: y − 2 = −(8/37)(x − 2), i.e. 8x + 37y − 90 = 0.
(Note: the curve point at x = 2 is (2, 155/8); the normal there is 64x + 296y − 5863 = 0.)
Explanation
Set dy/dx = 0 for stationary points and use the second derivative for their nature.