Can we find a percentage increase using the gradient?

Can we find a percentage increase using the gradient?

Question

The function $y$ is defined by $y=3x-2x^2$. Write down its derivative and show that, when $x=5$, a small increase in $x$ of $p\%$ causes an increase in the magnitude of $y$ of approximately $\dfrac{17p}{7}\%$.

Suggestion

The function $y$ is defined by $y=3x-2x^2$. Write down its derivative and show that, when $x=5$, a small increase in $x$ of $p\%$ causes an increase in the magnitude of $y$ of approximately $17p/7\%$.

Suppose $y = f(x)$, and that $(x,y)$ is on this curve. We now increase $x$ by a small amount $\delta x$, which increases $y$ by a small amount $\delta y$ ($y$ could decrease, in which case $\delta y$ would be negative).

Can you see why $y'$ evaluated at $x$ is roughly $\dfrac{\delta y}{\delta x}$? A diagram might help…

Solution

The function $y$ is defined by $y=3x-2x^2$. Write down its derivative and show that, when $x=5$, a small increase in $x$ of $p\%$ causes an increase in the magnitude of $y$ of approximately $17p/7\%$.

The derivative of $y=3x-2x^2$ is $y'=\dfrac{dy}{dx}=3-4x$, so when $x = 5, y = -35$ and $y'=-17$.

We now approximate the curve with the tangent to the curve at $x=5,$ which is justified if we are close to $(5,-35)$.

Suppose that $(x + \delta x, y + \delta y)$ is on the curve. Then we can express the percentage increase in $x$ by $\frac{\delta x}{x}\times 100=p,$ and similarly the percentage increase in $y$ by $\frac{\delta y}{y} \times 100 = q.$ Dividing these, we get $\dfrac{\delta y}{\delta x} \dfrac{x}{y} = \dfrac{q}{p}$. Now if $\delta x$ and $\delta y$ are small, $\dfrac{\delta y}{\delta x} \approx \dfrac{dy}{dx}$, so at $x = 5, \dfrac{q}{p} \approx -17 \times \dfrac{5}{-35}$.

This gives us that $q = \dfrac{17p}{7}$, and therefore the magnitude of $y$ increases by approximately $\dfrac{17p}{7}$ per cent.