Calculating Chord Gradients | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Keep the two points distinct

To find a chord gradient from two x-coordinates, find both y-coordinates from the curve, then use the straight-line gradient formula. The letter $h$ can represent their horizontal separation.

On $y=x^2$ , take $P=(1,1)$ and $Q=(1+h,(1+h)^2)$ . Here $h\ne0$ , because the points must be distinct. The horizontal change is $(1+h)-1=h$ .

m_{PQ}=\frac{(1+h)^2-1}{h}.

Simplify before taking the limit

Expand the whole numerator. Since $(1+h)^2=1+2h+h^2$ ,

\begin{aligned}m_{PQ}&=\frac{2h+h^2}{h}\\&=\frac{h(2+h)}{h}\\&=2+h,\qquad h\ne0.\end{aligned}

The common factor $h$ cancels because it is non-zero. Only now consider $h\to0$ : the gradient approaches $2$ . At $h=0.1$ , the chord gradient is $2.1$ , still not the exact tangent gradient.

Using this limit to find a derivative is called differentiation from first principles. Pure 1 requires the chord idea and calculations such as this, not a formal general proof of every differentiation rule.

Your turn9709/12 Oct/Nov 2024 Q3 [5 marks]

On $y=2x^2-3$ , points $A,B$ have x-coordinates $2,2+h$ . Find the chord gradient in terms of $h$ . Explain how it gives the gradient at $A$ .

Show worked answer

$A=(2,5)$ . At $B$ , expand $2(2+h)^2-3=5+8h+2h^2$ .

\begin{aligned}m_{AB}&=\frac{8h+2h^2}{h}\\&=8+2h.\end{aligned}

As $h\to0$ , the chord approaches the tangent at $A$ and its gradient approaches $8$ .

Use the same idea on a cubic

Your turnIndependent practice

For $y=x^3$ , use the points with x-coordinates $1$ and $1+h$ to find the chord gradient and then the tangent gradient at $x=1$ .

Show worked answer

Expand $(1+h)^3=1+3h+3h^2+h^3$ . Subtract the first y-coordinate, $1$ , then divide by $h\ne0$ .

\begin{aligned}m&=\frac{3h+3h^2+h^3}{h}\\&=3+3h+h^2.\end{aligned}

As $h\to0$ , both remaining terms in $h$ approach zero. The tangent gradient is $3$ .

Keep the two points distinct

To find a chord gradient from two x-coordinates, find both y-coordinates from the curve, then use the straight-line gradient formula. The letter $h$ can represent their horizontal separation.

On $y=x^2$ , take $P=(1,1)$ and $Q=(1+h,(1+h)^2)$ . Here $h\ne0$ , because the points must be distinct. The horizontal change is $(1+h)-1=h$ .

m_{PQ}=\frac{(1+h)^2-1}{h}.

Simplify before taking the limit

Expand the whole numerator. Since $(1+h)^2=1+2h+h^2$ ,

\begin{aligned}m_{PQ}&=\frac{2h+h^2}{h}\\&=\frac{h(2+h)}{h}\\&=2+h,\qquad h\ne0.\end{aligned}

The common factor $h$ cancels because it is non-zero. Only now consider $h\to0$ : the gradient approaches $2$ . At $h=0.1$ , the chord gradient is $2.1$ , still not the exact tangent gradient.

Your turn9709/12 Oct/Nov 2024 Q3 [5 marks]

On $y=2x^2-3$ , points $A,B$ have x-coordinates $2,2+h$ . Find the chord gradient in terms of $h$ . Explain how it gives the gradient at $A$ .

Show worked answer

$A=(2,5)$ . At $B$ , expand $2(2+h)^2-3=5+8h+2h^2$ .

\begin{aligned}m_{AB}&=\frac{8h+2h^2}{h}\\&=8+2h.\end{aligned}

As $h\to0$ , the chord approaches the tangent at $A$ and its gradient approaches $8$ .

Use the same idea on a cubic

Your turnIndependent practice

For $y=x^3$ , use the points with x-coordinates $1$ and $1+h$ to find the chord gradient and then the tangent gradient at $x=1$ .

Show worked answer

Expand $(1+h)^3=1+3h+3h^2+h^3$ . Subtract the first y-coordinate, $1$ , then divide by $h\ne0$ .

\begin{aligned}m&=\frac{3h+3h^2+h^3}{h}\\&=3+3h+h^2.\end{aligned}

As $h\to0$ , both remaining terms in $h$ approach zero. The tangent gradient is $3$ .