Implicit Curve Gradients | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Collect dy/dx

After differentiating, collect every term containing $dy/dx$ on one side, factor it out, and divide by its coefficient.

Worked example

A product of x and y

x^2+y^2=xy+7

2x+2y\frac{dy}{dx}=x\frac{dy}{dx}+y

(2y-x)\frac{dy}{dx}=y-2x

\frac{dy}{dx}=\frac{y-2x}{2y-x}

Common mistake

xy

has two changing factors. Its derivative is

x\,dy/dx+y

, not 1 and not just

y

Differentiate mixed terms

Worked example

A current-paper implicit product

\begin{aligned}6e^{-x}y^2+e^{2x}\\{}-12y+7&=0\end{aligned}

\begin{aligned}0={}&-6e^{-x}y^2\\&+12e^{-x}y\frac{dy}{dx}\\&+2e^{2x}-12\frac{dy}{dx}\end{aligned}

\frac{dy}{dx}=\frac{6e^{-x}y^2-2e^{2x}}{12e^{-x}y-12}

At $(x,y)=(\ln3,2)$ , use $e^{-\ln3}=1/3$ and $e^{2\ln3}=9$ . The exact gradient is $5/2$ . (9709/22/O/N/24 Q3)

Use the gradient

A horizontal tangent needs $dy/dx=0$ : the numerator is zero while the denominator is non-zero.
A vertical tangent occurs when the denominator is zero while the numerator is non-zero, so the gradient is not finite.
If both are zero, the fraction gives $0/0$ and does not decide the tangent.

Your turnIndependent check [6 marks]

The curve $x^2+xy+y^2=7$ passes through $(1,2)$ . Find the equation of its tangent there.

Show worked answer

2x+x\frac{dy}{dx}+y+2y\frac{dy}{dx}=0

\frac{dy}{dx}=-\frac{2x+y}{x+2y}

Answer

4x+5y-14=0

Collect dy/dx

After differentiating, collect every term containing $dy/dx$ on one side, factor it out, and divide by its coefficient.

Worked example

A product of x and y

x^2+y^2=xy+7

2x+2y\frac{dy}{dx}=x\frac{dy}{dx}+y

(2y-x)\frac{dy}{dx}=y-2x

\frac{dy}{dx}=\frac{y-2x}{2y-x}

Common mistake

xy

has two changing factors. Its derivative is

x\,dy/dx+y

, not 1 and not just

y

Differentiate mixed terms

Worked example

A current-paper implicit product

\begin{aligned}6e^{-x}y^2+e^{2x}\\{}-12y+7&=0\end{aligned}

\begin{aligned}0={}&-6e^{-x}y^2\\&+12e^{-x}y\frac{dy}{dx}\\&+2e^{2x}-12\frac{dy}{dx}\end{aligned}

\frac{dy}{dx}=\frac{6e^{-x}y^2-2e^{2x}}{12e^{-x}y-12}

At $(x,y)=(\ln3,2)$ , use $e^{-\ln3}=1/3$ and $e^{2\ln3}=9$ . The exact gradient is $5/2$ . (9709/22/O/N/24 Q3)

Use the gradient

A horizontal tangent needs $dy/dx=0$ : the numerator is zero while the denominator is non-zero.
A vertical tangent occurs when the denominator is zero while the numerator is non-zero, so the gradient is not finite.
If both are zero, the fraction gives $0/0$ and does not decide the tangent.

Your turnIndependent check [6 marks]

The curve $x^2+xy+y^2=7$ passes through $(1,2)$ . Find the equation of its tangent there.

Show worked answer

2x+x\frac{dy}{dx}+y+2y\frac{dy}{dx}=0

\frac{dy}{dx}=-\frac{2x+y}{x+2y}

Answer

4x+5y-14=0