Applications of Chain Rule | DP IB Maths: AA HL Revision Notes 2021

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Related Rates of Change

What is meant by rates of change?

A rate of change is a measure of how a quantity is changing with respect to another quantity
Mathematically rates of change are derivatives
- $\frac{d V}{d r}$ could be the rate at which the volume of a sphere changes relative to how its radius is changing
Context is important when interpreting positive and negative rates of change
- A positive rate of change would indicate an increase
  - e.g. the change in volume of water as a bathtub fills
- A negative rate of change would indicate a decrease
  - e.g. the change in volume of water in a leaking bucket

What is meant by related rates of change?

Related rates of change are connected by a linking variable or parameter

this is often time, represented by $t$
seconds is the standard unit for time but this will depend on context

e.g. Water running into a large hemi-spherical bowl

both the height and volume of water in the bowl are changing with time

time is the linking parameter between the rate of change of height and the rate of change of volume

How do I solve problems involving related rates of change?

Use of chain rule and product rule are common in such problems
Be clear about which variables are representing which quantities

STEP 1
Write down any variables and derivatives involved in the problem
e.g. $x, y, t, \frac{d y}{d x}, \frac{d x}{d t}, \frac{d y}{d t}$

STEP 2
Use an appropriate differentiation rule to set up an equation linking ‘rates of change’
e.g. Chain rule: $\frac{d y}{d t} = \frac{d y}{d x} \times \frac{d x}{d t}$

STEP 3
Substitute in known values
e.g. If, when $t = 3$ , $\frac{d x}{d t} = 2$ and $\frac{d y}{d t} = 8$ , then $8 = \frac{d y}{d x} \times 2$

STEP 4
Solve the problem and interpret the answer in context if required
e.g. $\frac{d y}{d x} = \frac{8}{2} = 4$ ‘when $t = 3$ , $y$ changes at a rate of 4, with respect to $x$ ’

Exam Tip

If you struggle to determine which rate to use then you can look at the units to help
- e.g. A rate of 5 cm³ per second implies volume per time so the rate would be $\frac{d V}{d t}$

Worked example

In a manufacturing process a metal component is heated such that it’s cross-sectional area expands but always retains the shape of a right-angled triangle. At time $t$ seconds the triangle has base $b$ cm and height $h$ cm.

At the time when the component’s cross-sectional area is changing at 4 cm s^-1, the base of the triangle is 3 cm and its height is 6 cm. Also at this time, the rate of change of the height is twice the rate of change of the base.

Find the rate of change of the base at this point of time.

MR5QgbAY_5-8-1-ib-hl-aa-only-rel-roc-we1-soln-

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Differentiating Inverse Functions

What is meant by an inverse function?

Some functions are easier to process with $x$ (rather than $y$ ) as the subject

i.e. in the form $x = f (y)$

This is particularly true when dealing with inverse functions

e.g. If $y = f (x)$ the inverse would be written as $y = f^{- 1} (x)$

finding $f^{- 1} (x)$ can be awkward
so write $x = f (y)$ instead

How do I differentiate inverse functions?

Since $x = f (y)$ it is easier to differentiate “ $x$ with respect to $y$ ” rather than “ $y$ with respect to $x$ ”
- i.e. find $\frac{d x}{d y}$ rather than $\frac{d y}{d x}$
- Note that $\frac{d x}{d y}$ will be in terms of $y$ but can be substituted

STEP 1
For the function $y = f (x)$ , the inverse will be $y = f^{- 1} (x)$

Rewrite this as $x = f (y)$

STEP 2
From $x = f (y)$ find $\frac{d x}{d y}$

STEP 3
Find $\frac{d y}{d x}$ using $\frac{d y}{d x} = \frac{1}{\frac{d x}{d y}}$ - this will usually be in terms of $y$

If an algebraic solution in terms of $x$ is required substitute $f (x)$ for $y$ in $\frac{d y}{d x}$
If a numerical derivative (e.g. a gradient) is required then use the $y$ -coordinate
- If the $y$ -coordinate is not given, you should be able to work it out from the orginal function and $x$ -coordinate

Exam Tip

With $x$ 's and $y$ 's everywhere this can soon get confusing!
- Be clear of the key information and steps - and set your wokring out accordingly
  - The orginal function, $y = f (x)$
  - Its inverse, $y = f^{- 1} (x)$
  - Rewriting the inverse, $x = f (y)$
  - Finding $\frac{d x}{d y}$ first, then finding its reciprocal for $\frac{d y}{d x}$
Your GDC can help when numerical derivatives (gradients) are required

Worked example

Find the gradient of the curve at the point where

y = 3

on the graph of

y = f^{- 1} (x)

where

f (x) = \sqrt{{(5 x + 1)}^{3}}

5-8-1-ib-hl-aa-only-we2a-soltn

b) Given that $y = e^{x}$ show that the derivative of $y = \ln x$ is $\frac{1}{x}$ .

5-8-1-ib-hl-aa-only-we2b-soltn-

Applications of Chain Rule (DP IB Maths: AA HL)

Revision Note

Related Rates of Change

What is meant by rates of change?

What is meant by related rates of change?

How do I solve problems involving related rates of change?

Exam Tip

Worked example

Differentiating Inverse Functions

What is meant by an inverse function?

How do I differentiate inverse functions?

Exam Tip

Worked example

Author: Paul