1. © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules 39: Connected Rates of Change

2. Connected Rates of Change "Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages" Module C3 AQA Edexcel Module C4 MEI/OCR OCR

3. Connected Rates of Change If water drips from a tap into a container, the volume of water, V, in the container increases and the height, h, also increases. The rates at which the volume and the height increase are connected. Both rates are given by derivatives and we can calculate one rate of increase from the other using the chain rule. V h

4. Connected Rates of Change Solution: Let h cm be the height at time t minutes. What is the rate of increase of the height of water in the glass? The question is asking us to find the “ rate of increase of the height ”. e.g. 1 Water drips from a tap at the rate of per minute. It is collected in a cylindrical glass with a base radius of 2 cm. V h 2 cm

5. Connected Rates of Change Solution: Let h cm be the height at time t minutes. What is the rate of increase of the height of water in the glass? The question is asking us to find the “ rate of increase of the height ”. e.g. 1 Water drips from a tap at the rate of per minute. It is collected in a cylindrical glass with a base radius of 2 cm. V h 2 cm We want. We are given that

6. Connected Rates of Change Tip: The units confirm that we are being given a rate of change of a volume. Solution: Let h cm be the height at time t minutes. What is the rate of increase of the height of water in the glass? The question is asking us to find the “ rate of increase of the height ”. e.g. 1 Water drips from a tap at the rate of per minute. It is collected in a cylindrical glass with a base radius of 2 cm. V h 2 cm The chain rule links the 2 quantities together. We want. We are given that

7. Connected Rates of Change We are given that We want. Using the chain rule: Since the derivatives behave like fractions, the missing expression needs to “cancel” dV and introduce dh. So, V h 2 cm

8. Connected Rates of Change  is a constant, so V h 2 cm We now need. For a cylinder, We want a link between h and V that we can differentiate. so There’s no need to rearrange this since we don’t mind finding instead of. So,

9. Connected Rates of Change  is a constant, so V h 2 cm We now need. For a cylinder, We want a link between h and V that we can differentiate. so So, cm

10. Connected Rates of Change  is a constant, so V h 2 cm We now need. For a cylinder, We want a link between h and V that we can differentiate. so So, cm /

11. Connected Rates of Change  is a constant, so V h 2 cm We now need. For a cylinder, We want a link between h and V that we can differentiate. so So, cm min. /

12. Connected Rates of Change  is a constant, so V h 2 cm We now need. For a cylinder, So, cm We want a link between h and V that we can differentiate. so min. /

13. Connected Rates of Change SUMMARY Method for Calculating connected rates of change Define appropriate letters for the variables. This can be done on a diagram. Write down the given rate of change as a derivative Write down the required rate of change and link it to the given one by the chain rule. Write down an expression linking the 2 variables in the 2 nd part of the chain. Differentiate the new expression. Substitute and evaluate if necessary.

14. Connected Rates of Change e.g. 2 A block of ice is melting at the rate of per minute. Assuming it remains in the shape of a cube, find the rate at which the edge, x, is decreasing when x = 10 mm. Solution: We want given x x x V Subst. in (1) : Be careful, the volume is decreasing so the gradient of the graph of volume against time is negative.

15. Connected Rates of Change When x = 10, mm. per minute So, the edge is decreasing at a rate of 0.01 mm. per minute.

16. Connected Rates of Change Exercises 1.A circular stain on a floor is increasing at. 2.A large snowball is melting at a rate of. Find the rate of decrease of the radius in cm. per hour when the radius is 50 cm. Assume the snowball is spherical. Volume Find the rate of increase of the radius when the radius is 5 cm.

17. Connected Rates of Change 1.A circular stain on a floor is increasing at 10. Find the rate of increase of the radius when the radius is 5 Solution: r A We want given Area of a circle, Subst. in ( 1 ): - - - - ( 1 )

18. Connected Rates of Change 2.A large snowball is melting at a rate of. Find the rate of decrease of the radius in cm. per hour when the radius is 50 cm. Solution: We want given Volume of ball, Subst. in ( 1 ): - - - - ( 1 ) I’ve chosen metres as my unit of length. If you chose cm. you need The rate of decrease is 3 cm. per hour.

19. Connected Rates of Change The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied. For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

20. Connected Rates of Change SUMMARY Method for Calculating Connected Rates of Change Define appropriate letters for the variables. This can be done on a diagram. Write down the given rate of change as a derivative Write down the required rate of change and link it to the given one by the chain rule. Write down an expression linking the 2 variables in the 2 nd part of the chain. Differentiate the new expression. Substitute and evaluate if necessary.

21. Connected Rates of Change Tip: The units confirm that we are being given a rate of change of a volume. Solution: Let h cm be the height at time t minutes. What is the rate of increase of the height of water in the glass? The question is asking us to find the “ rate of increase of the height ”. e.g. 1 Water drips from a tap at the rate of per minute. It is collected in a cylindrical glass with a base radius of 2 cm. The chain rule links the 2 quantities together. We want. We are given that V h 2 cm

22. Connected Rates of Change We are given that We want. Using the chain rule: Since the derivatives behave like fractions, the missing expression needs to “cancel” dV and introduce dh. So, We now need.

23. Connected Rates of Change  is a constant, so For a cylinder, So, cm. per min. We want a link between h and V that we can differentiate. so There’s no need to rearrange this since we don’t mind finding instead of.

24. Connected Rates of Change e.g. 2 A block of ice in the shape of a cube is melting at the rate of per minute. Find the rate at which the edge is decreasing when x = 10 mm. Solution: We want given x x x V Be careful, the volume is decreasing so the gradient of the graph of volume against time is negative.

25. Connected Rates of Change When x = 10, mm. per minute So, the edge is decreasing at a rate of 0.01 mm. per minute. Subst. in (1) :