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Differentiation II

Connected rates of change

Connected rates of change

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Maxima and minima problems

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Vertical motion under gravity

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Integration II

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Differentiation II

Differentiating sin x and cos x

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The chain rule

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Differentiating inverse functions

Differentiating trigonometric functions

Parametric differentiation

Implicit differentiation

Second derivatives: Concave and convex functions

Connected rates of change

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Proof II

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Integration I

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Differentiation I

Finding the gradient of a curve

Differentiation from first principles

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Explainer Video

Tutor: Toby

Summary

Connected rates of change

​​In a nutshell

One of the many uses of the chain rule is to relate multiple rates of change. Recall that a derivative gives a rate of change.


Example 1

The volume VVV​of a cylinder with a base circle with radius rrr (in centimetres) is given by the formula V=πr2hV=\pi r^2hV=πr2h where hhh is the cylinder's height (also in centimetres). Given that the radius of this cylinder grows by 222 centimetres per second, find the rate of change of the volume when the radius is 555 centimetres. Assume that the height stays constant, at 10 cm10\text{ cm}10 cm​.


Start by giving the rate of change of the radius with differentiation notation:

drdt=2\frac{\text dr}{\text dt}=2dtdr​=2​​


where ttt is the time passed in seconds. Next find the rate of change of the volume with respect to the radius. This can be found by differentiating the equation given for the volume:

V=πr2hdVdr=2πrh\begin{aligned}V&=\pi r^2h\\\frac{\text dV}{\text dr}&=2\pi rh\end{aligned}VdrdV​​=πr2h=2πrh​​​


The question asks for the rate of change of the volume. Implicitly, this is with respect to time, thus you seek dVdt\frac{\text dV}{\text dt}dtdV​. The chain rule says that

dVdt=dVdr×drdt\frac{\text dV}{\text dt}=\frac{\text dV}{\text dr}\times\frac{\text dr}{\text dt}dtdV​=drdV​×dtdr​​​


Thus

dVdt=2πrh×2=4πrh\frac{\text dV}{\text dt}=2\pi rh\times 2=4\pi rhdtdV​=2πrh×2=4πrh​​


You have that ​h=10h=10h=10 and the question asks about when r=5r=5r=5:

dVdt=4π×5×10=200π≈628.32\frac{\text dV}{\text dt}=4\pi \times 5\times10=200\pi\approx628.32dtdV​=4π×5×10=200π≈628.32​​


Therefore, when the radius reaches 555 centimetres, the rate of change of the volume of the cylinder is about 628 cm3s−1‾(3s.f.)\underline{628\text{ cm}^3\text{s}^{-1}} (3{ s.f.})628 cm3s−1​(3s.f.). In other words, the volume increases by nearly 630 cm3630\text{ cm}^3630 cm3 per second at the point when the radius is 5 cm5\text{ cm}5 cm​. 



Differential equations

Equations involving rates of change are called differential equations. 


Example 2

The rate of change of the population PPP is directly proportional to the size of the population itself. Form a differential equation from this information.


You have that "the rate of change" refers to ddt\frac{\text d}{\text dt}dtd​, unless otherwise stated. Hence this information says

dPdt∝P\frac{\text dP}{\text dt}\propto PdtdP​∝P​​


To turn this into an equation, you change the "∝\propto∝" to "===" and multiply one side by a constant kkk:

‾\underline{}​dPdt=kP‾\underline{\frac{\text dP}{\text dt}=kP}dtdP​=kP​​​​



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Connected rates of change

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FAQs - Frequently Asked Questions

What is a differential equation?

Equations involving rates of change are called differential equations.

How do you relate multiple rates of change?

Using the chain rule.

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