Unknown Coefficients from Gradients | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Separate point information from gradient information

An unknown coefficient is still a constant when you differentiate with respect to $x$ . For $y=ax^2+bx$ , the derivative is $2ax+b$ . The letters $a,b$ do not change as $x$ changes.

Worked example

The curve $y=ax^2+bx$ passes through $(1,5)$ with gradient $7$ . Find $a,b$ .

Put the point into the curve and the gradient into the derivative:

\begin{aligned}a+b&=5,\\2a+b&=7.\end{aligned}

Subtract the first equation from the second to eliminate $b$ : $a=2$ . Then $b=3$ . Check both conditions: $2+3=5$ and $4+3=7$ .

Two gradient conditions instead give two equations from the derivative. Do not put a gradient into $y$ . Two conditions determine two constants only if they give enough independent information; duplicated conditions do not do that.

Use a known height inside a root

Worked example

The curve $y=\sqrt{ax+b}$ passes through $(12,4)$ with gradient $1/4$ . Find $a,b$ .

The point gives $\sqrt{12a+b}=4$ . Squaring gives $12a+b=16$ . The chain rule gives

\frac{dy}{dx}=\frac{a}{2\sqrt{ax+b}}.

At the given point the square root is already known to be $4$ . Therefore $1/4=a/8$ , so $a=2$ . Substitute into $12a+b=16$ to obtain $b=-8$ .

Check the original root equation after squaring: $\sqrt{24-8}=4$ . The expression inside the square root is positive here, so the derivative is defined.

Build the equations from two gradients

Your turnIndependent practice

The curve $y=ax^4+bx^2+x$ has gradient $3$ at $x=1$ and gradient $-51$ at $x=-2$ . Find $a,b$ .

Show worked answer

\frac{dy}{dx}=4ax^3+2bx+1.

The two conditions give $4a+2b+1=3$ and $-32a-4b+1=-51$ . Simplify:

\begin{aligned}2a+b&=1,\\8a+b&=13.\end{aligned}

Subtract to obtain $6a=12$ , so $a=2,b=-3$ . The derivative becomes $8x^3-6x+1$ , which gives both stated gradients.

Separate point information from gradient information

An unknown coefficient is still a constant when you differentiate with respect to $x$ . For $y=ax^2+bx$ , the derivative is $2ax+b$ . The letters $a,b$ do not change as $x$ changes.

Worked example

The curve $y=ax^2+bx$ passes through $(1,5)$ with gradient $7$ . Find $a,b$ .

Put the point into the curve and the gradient into the derivative:

\begin{aligned}a+b&=5,\\2a+b&=7.\end{aligned}

Subtract the first equation from the second to eliminate $b$ : $a=2$ . Then $b=3$ . Check both conditions: $2+3=5$ and $4+3=7$ .

Use a known height inside a root

Worked example

The curve $y=\sqrt{ax+b}$ passes through $(12,4)$ with gradient $1/4$ . Find $a,b$ .

The point gives $\sqrt{12a+b}=4$ . Squaring gives $12a+b=16$ . The chain rule gives

\frac{dy}{dx}=\frac{a}{2\sqrt{ax+b}}.

At the given point the square root is already known to be $4$ . Therefore $1/4=a/8$ , so $a=2$ . Substitute into $12a+b=16$ to obtain $b=-8$ .

Check the original root equation after squaring: $\sqrt{24-8}=4$ . The expression inside the square root is positive here, so the derivative is defined.

Build the equations from two gradients

Your turnIndependent practice

The curve $y=ax^4+bx^2+x$ has gradient $3$ at $x=1$ and gradient $-51$ at $x=-2$ . Find $a,b$ .

Show worked answer

\frac{dy}{dx}=4ax^3+2bx+1.

The two conditions give $4a+2b+1=3$ and $-32a-4b+1=-51$ . Simplify:

\begin{aligned}2a+b&=1,\\8a+b&=13.\end{aligned}

Subtract to obtain $6a=12$ , so $a=2,b=-3$ . The derivative becomes $8x^3-6x+1$ , which gives both stated gradients.