Connected Rates on a Curve | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

A point can move right while moving down

For a point moving on $y=f(x)$ , $dx/dt$ measures horizontal motion and $dy/dt$ measures vertical motion. The curve gradient $dy/dx$ connects them:

\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}

A positive $dx/dt$ means moving right, not necessarily up. The sign of the product determines whether $y$ increases or decreases.

Worked example

On y = x + √(2x + 3), x increases at 0.06 units/s. Find the y-rate at x = 3.

\frac{dy}{dx}=1+\frac1{\sqrt{2x+3}}

At $x=3$ , the gradient is $4/3$ , so $dy/dt=(4/3)(0.06)=0.08$ units/s, increasing.

Your turnPractice

A point on $y=3x-2x^3$ has $dx/dt=0.015$ units/s. Find and interpret $dy/dt$ at $x=2$ .

Show worked answer

$dy/dx=3-6x^2=-21$ there. Thus $dy/dt=-0.315$ units/s. The point moves right and down: $y$ decreases at 0.315 units/s.

Use two rates to find the position

If both time rates are given and $dx/dt\ne0$ , divide them to find the gradient. Then solve the derivative equation for the position. Keep every allowed root.

Worked example

On y = 8/(7 − 2x), dx/dt = 0.125 and dy/dt = 0.08. Find the possible x-values.

\begin{gathered}\frac{dy}{dx}=\frac{0.08}{0.125}=0.64\\\frac{16}{(7-2x)^2}=0.64\\(7-2x)^2=25\\7-2x=\pm5\end{gathered}

Hence $x=1$ or $x=6$ . Both are in the domain, which excludes only $x=3.5$ . Substituting either into the derivative recovers 0.64.

Your turnPractice

A moving point on $y=x^3-3x$ has non-zero $dx/dt$ . Its $y$ -coordinate changes at twice its $x$ -coordinate’s rate. Find the possible $x$ -coordinates.

Show worked answer

$dy/dt=2\,dx/dt$ , so the chain rule gives $dy/dx=2$ . Then $3x^2-3=2$ , hence $x=\pm\sqrt{5/3}$ . Dividing by the given non-zero rate preserves both branches.

A point can move right while moving down

For a point moving on $y=f(x)$ , $dx/dt$ measures horizontal motion and $dy/dt$ measures vertical motion. The curve gradient $dy/dx$ connects them:

\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}

A positive $dx/dt$ means moving right, not necessarily up. The sign of the product determines whether $y$ increases or decreases.

Worked example

On y = x + √(2x + 3), x increases at 0.06 units/s. Find the y-rate at x = 3.

\frac{dy}{dx}=1+\frac1{\sqrt{2x+3}}

At $x=3$ , the gradient is $4/3$ , so $dy/dt=(4/3)(0.06)=0.08$ units/s, increasing.

Your turnPractice

A point on $y=3x-2x^3$ has $dx/dt=0.015$ units/s. Find and interpret $dy/dt$ at $x=2$ .

Show worked answer

$dy/dx=3-6x^2=-21$ there. Thus $dy/dt=-0.315$ units/s. The point moves right and down: $y$ decreases at 0.315 units/s.

Use two rates to find the position

If both time rates are given and $dx/dt\ne0$ , divide them to find the gradient. Then solve the derivative equation for the position. Keep every allowed root.

Worked example

On y = 8/(7 − 2x), dx/dt = 0.125 and dy/dt = 0.08. Find the possible x-values.

\begin{gathered}\frac{dy}{dx}=\frac{0.08}{0.125}=0.64\\\frac{16}{(7-2x)^2}=0.64\\(7-2x)^2=25\\7-2x=\pm5\end{gathered}

Hence $x=1$ or $x=6$ . Both are in the domain, which excludes only $x=3.5$ . Substituting either into the derivative recovers 0.64.

Your turnPractice

A moving point on $y=x^3-3x$ has non-zero $dx/dt$ . Its $y$ -coordinate changes at twice its $x$ -coordinate’s rate. Find the possible $x$ -coordinates.

Show worked answer

$dy/dt=2\,dx/dt$ , so the chain rule gives $dy/dx=2$ . Then $3x^2-3=2$ , hence $x=\pm\sqrt{5/3}$ . Dividing by the given non-zero rate preserves both branches.