Given that , find the approximate percentage change in when increases from 2 by 2%.

Cultivating Champions, Moulding Success

B6

let

Sub to

Thus, it is a min point.

C7

let

(NA).

There are no stationary points for this curve.

C8

Let

Evaluate with a calculator…

When , it implies we have a stationary point.

To determine the nature of the stationary point, we can do either the first derivative test **or** the second derivative.

The first derivative test:

Students should write the actual values of and in the table.

We use this under these two situations:

1. is difficult to solve for, that is, is tough to be differentiated

2.

The second derivative test:

Other things students should take note is concavity and drawing of the derivative graph.

VJC P1 Q6

A curve has equation and a line has equation as shown above.

is a fixed point on C and A is an arbitrary point on . State the geometrical relationship between the line segment AB and is the distance from B to A is the least.

Taking the coordinates of A as , find an equation relating and for which AB is the least.

Deduce that when AB is the least, where is a constant to be found. Hence or otherwise, find the coordinates of the point on C that is nearest on .

(a)(i)

Consider is a circle with centre Origin and radius .

Thus, required area is the area of a quadrant with radius

(ii)

(b)

(a)(i)

Let

(a)(ii)

*Students are expected to prove that gives the maximum area.

(b)(i)

latex xlatex \Rightarrow \frac{dy}{dx}=0latex y=-x^2latex y=-x^2latex x^3 + 2y^3 +3xy=klatex x^3 + 2(-x^2)^3 +3x(-x^2)=klatex 2x^6 + 2x^3 + k=0latex y=-1latex -1=-x^2latex x =\pm xlatex x = 1, k=-4latex x = -1, k=0$