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Applications of Chain Rule (DP IB Maths: AA HL)
Revision Note
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
- 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
- A positive rate of change would indicate an increase
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
- 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.
STEP 2
Use an appropriate differentiation rule to set up an equation linking ‘rates of change’
e.g. Chain rule:
STEP 3
Substitute in known values
e.g. If, when , and , then
STEP 4
Solve the problem and interpret the answer in context if required
e.g. ‘when , changes at a rate of 4, with respect to ’
Examiner 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
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 seconds the triangle has base cm and height 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.
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Differentiating Inverse Functions
What is meant by an inverse function?
- Some functions are easier to process with (rather than ) as the subject
- i.e. in the form
- This is particularly true when dealing with inverse functions
- e.g. If the inverse would be written as
- finding can be awkward
- so write instead
How do I differentiate inverse functions?
- Since it is easier to differentiate “ with respect to ” rather than “ with respect to ”
- i.e. find rather than
- Note that will be in terms of but can be substituted
STEP 1
For the function , the inverse will be
Rewrite this as
STEP 2
From find
STEP 3
Find using - this will usually be in terms of
- If an algebraic solution in terms of is required substitute for in
- If a numerical derivative (e.g. a gradient) is required then use the -coordinate
- If the -coordinate is not given, you should be able to work it out from the orginal function and -coordinate
Examiner Tip
- With 's and 's everywhere this can soon get confusing!
- Be clear of the key information and steps - and set your wokring out accordingly
- The orginal function,
- Its inverse,
- Rewriting the inverse,
- Finding first, then finding its reciprocal for
- Be clear of the key information and steps - and set your wokring out accordingly
- Your GDC can help when numerical derivatives (gradients) are required
Worked example
b) Given that show that the derivative of is .
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