Rates and Integration

Expert reviewed • 21 July 2024 • 4 minute read

HSC Maths Advanced Syllabus

use any of the functions covered in the scope of this syllabus and their derivatives to solve practical and abstract problems
solve optimisation problems for any of the functions covered in the scope of this syllabus, in a wide variety of contexts including displacement, velocity, acceleration, area, volume, business, finance and growth and decay

Note:

Video coming soon!

Tips to help with Integrating Rates

Before we look at problems to do with integrating rates. it is important to know a few things.

Integrating rates is done following the same processes of integration relating to any other equation. we must simply apply the specific rules to suit the context and equation.
We should never remove the constant from the expression once integrated.
Substitute in points to find unknown variables/rates

Practice Question 1

A reservoir contains 50,000 litres of water. When the draining valve is opened, the volume $V$ (in litres of water) in the reservoir decreases at a variable rate given by: $\frac{dV}{dt}=−1200+25t$ , where $t$ is the time in seconds after opening the valve. Once the water stops flowing, the valve shuts off. At what time does the water stop flowing.

This occurs when $\frac{dV}{dt}=0$ :

\frac{dV}{dt}=0\\0=-1200+25t\\25t=1200\\t=48\;seconds

Practice Question 2

Use the information found in the previous question to answer the following. Integrate the rate of change of volume, to find the volume of water $V$ in the reservoir as a function of time $t$ .

V=\int(-1200+25t)dt\\=-1200t+\frac{25}{2}t^2+C

Given the initial condition that at $t=0$ , $V=50000$ we can find $C$ :

V(0)=50000\\50000=-1200(0)+\frac{25}{2}(0)^2+C\\C=50000

Thus, the function $V$ is:

V=-1200t+\frac{25}{2}t^2+50000

Rates and Differentiation

Return to Module 7: Motion and Rates