Define connected rates of change .

A situation involving more than two variables, where the chain rule links several rates of change into a single equation. The rate you actually want is usually built out of rates you can find, rather than being available directly.

Why must you be clear which letters are variables and which are constants?

Because the two behave completely differently under differentiation: a constant disappears, while a variable does not. Treating a fixed radius as a variable, or the other way round, changes the derivative entirely.

What is the role of connected rates of change in a differential equation problem?

They are how the equation gets set up from the information given. Solving it afterwards is a separate matter of integration, and assembling the right chain of rates is often the harder half.

International A Level (IAL)MathsEdexcelPure 4FlashcardsDifferentiationFurther Applications of Differentiation

Further Applications of Differentiation (Edexcel International A Level (IAL) Maths: Pure 4): Flashcards

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FrontConnected Rates of Change
Define connected rates of change.

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Define connected rates of change.
A situation involving more than two variables, where the chain rule links several rates of change into a single equation.
The rate you actually want is usually built out of rates you can find, rather than being available directly.
Complete the chain rule linking three rates of change:
$\frac{d V}{d t} = \frac{d V}{d r} \times______$
The completed chain is:
$\frac{d V}{d t} = \frac{d V}{d r} \times \frac{d r}{d t}$
The two inner letters cancel in the way they would in ordinary fractions, which is a quick check that the right rates have been chained together.
Where does the middle rate of change in a chain usually come from?
Differentiating a formula, either one the question supplies or one you are expected to know.
A sphere's $V = \frac{4}{3} π r^{3}$ , for instance, differentiates to give $\frac{d V}{d r}$ .
Why must you be clear which letters are variables and which are constants?
Because the two behave completely differently under differentiation: a constant disappears, while a variable does not.
Treating a fixed radius as a variable, or the other way round, changes the derivative entirely.
What is the role of connected rates of change in a differential equation problem?
They are how the equation gets set up from the information given.
Solving it afterwards is a separate matter of integration, and assembling the right chain of rates is often the harder half.
True or False?
A rate of change is always a rate with respect to time.
False.
A rate of change is with respect to whatever variable sits underneath, so $\frac{d V}{d r}$ is a rate with respect to radius.
Rates with respect to time are simply the commonest kind in these particular problems.

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