Find the stationary points on the curve x 2 - x y + y 2 = 48 . By finding d 2 y d x 2 determine the nature of each of these stationary points.

Given equation of the curve is x 2 - x y + y 2 = 48 . . . . . . i First we need to find the coordinates of the stationary points on the given curve. We know that the stationary point is a point on the curve where the gradient is zero. So, differentiating both sides of equation i with respect to x , we have d d x x 2 - x y + y 2 = d d x 48 ⇒ d d x x 2 - d d x x y + d d x y 2 = 0 ⇒ 2 x - x d y d x + y d d x x + d d y y 2 d y d x = 0 ∵ d d x u v = u d v d x + v d u d x ⇒ 2 x - x d y d x - y + 2 y d y d x = 0 ⇒ d y d x 2 y - x = y - 2 x ⇒ d y d x = y - 2 x 2 y - x . . . . i i Now, for the stationary points we must have d y d x = 0 ⇒ y - 2 x = 0 [using i i ] ⇒ y = 2 x Substituting y = 2 x in the given equation i , we have x 2 - x 2 x + 2 x 2 = 48 ⇒ x 2 - 2 x 2 + 4 x 2 = 48 ⇒ 3 x 2 = 48 ⇒ x 2 = 16 ⇒ x = ± 4 When x = 4 , we have y = 8 , and when x = - 4 , we have y = - 8 Thus, the coordinates of the stationary points on the given curve are 4 , 8 , and - 4 , - 8 . Now from i i , we have d y d x = y - 2 x 2 y - x ⇒ d 2 y d x 2 = d d x d y d x = d d x y - 2 x 2 y - x = 2 y - x d d x y - 2 x - y - 2 x d d x 2 y - x 2 y - x 2 = 2 y - x × - 2 - y - 2 x × - 1 2 y - x 2 = - 4 y + 2 x + y - 2 x 2 y - x 2 = - 3 y 2 y - x 2 At 4 , 8 d 2 y d x 2 ] 4 , 8 = - 3 8 2 8 - 4 2 < 0 ⇒ 4 , 8 is a point of local maxima And, d 2 y d x 2 ] - 4 , - 8 = - 3 - 8 2 - 8 - 4 2 > 0 ⇒ - 4 , - 8 is a point of local minima.

For the following parametric equation, find d y d x in terms of the given parameter. x = 2 t 3 , y = t 2 - 5

Given parametric equations are x = 2 t 3 , and y = t 2 - 5 We know that when a curve is given in parametric form, in terms of the parameter t we can use the chain rule to find d y d x in terms of t . Since x = 2 t 3 ⇒ d x d t = d d t 2 t 3 = 2 d d t t 3 = 2 × 3 t 2 ∵ d d x x n = n x n - 1 , n ∈ R = 6 t 2 . . . . i And, y = t 2 - 5 ⇒ d y d t = d d t t 2 - 5 = 2 t . . . . i i ∵ d d x x n = n x n - 1 , n ∈ R Now, applying the chain rule, we have d y d x = d y d t × d t d x = 2 t × 1 6 t 2 [Using i , and i i ] = 1 3 t Hence, d y d x = 1 3 t

For the following parametric equation, find d y d x in terms of the given parameter. x = 2 + sin 2 θ , y = 4 θ + 2 cos 2 θ

Given parametric equations are x = 2 + sin 2 θ , and y = 4 θ + 2 cos 2 θ We know that when a curve is given in parametric form, in terms of the parameter θ we can use the chain rule to find d y d x in terms of θ . Since x = 2 + sin 2 θ ⇒ d x d θ = d d θ 2 + sin 2 θ = d d θ 2 + d d θ sin 2 θ = 0 + cos 2 θ × d d θ 2 θ ∵ d d x sin x = cos x = 2 cos 2 θ . . . . i ∵ d d x x n = n x n - 1 , n ∈ R And, y = 4 θ + 2 cos 2 θ ⇒ d y d θ = d d θ 4 θ + 2 cos 2 θ = 4 d d θ θ + 2 d d θ cos 2 θ = 4 + 2 × - sin 2 θ d d θ 2 θ = 4 - 4 sin 2 θ . . . . i i Now, applying the chain rule, we have d y d x = d y d θ × d θ d x = 4 - 4 sin 2 θ × 1 2 cos 2 θ [Using i , and i i ] = 2 - 2 sin 2 θ cos 2 θ Hence, d y d x = 2 - 2 sin 2 θ cos 2 θ

E M B I B E

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 4 - Differentiation>EXERCISE 4F>Q 9

MEDIUM

AS and A Level

IMPORTANT

Earn 100

The equation of a curve is $y^{3} - 12 x y + 16 = 0 .$

Show that the curve has no stationary points.

Important Questions on Differentiation

MEDIUM

AS and A Level

IMPORTANT

Cambridge International AS & A Level Mathematics : Pure Mathematics 2 & 3 Course Book>Chapter 4 - Differentiation>EXERCISE 4F>Q 9

The equation of a curve is $\Rightarrow \frac{d y}{d x} (3 y^{2} - 12 x) = 12 y$ Find the coordinates of the point on the curve where the tangent is parallel to the $y$ -axis.