1. 42: Differentiating Parametric Equations © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules

2. Differentiating Parametric Equations "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 C4

3. Differentiating Parametric Equations The main reason for using parametric equations is that they are easy to differentiate. Using the chain rule, Suppose the parameter is t.

4. Differentiating Parametric Equations e.g. 1 Find the gradient where t = 2 on the curve given by Solution: Where t = 2, the gradient Be careful, we want not

5. Differentiating Parametric Equations SUMMARY  The gradient of a curve given in terms of a parameter t is

6. Differentiating Parametric Equations Solution: e.g. 2 Find the equation of the tangent to the ellipse at the point where giving an exact answer. The equation of the tangent is given by We first find m at

7. Differentiating Parametric Equations So, at We can now substitute in y – y 1 =m(x – x 1 ) To find c we also need x and y at So, the tangent is

8. Differentiating Parametric Equations e.g. 3(a) Find for the curve given by the parametric equations Solution: Using, the grad. of the normal is 4 when t = 2 (b) Show that the equation of the normal at the point where t = 2 is

9. Differentiating Parametric Equations Find x and y at t = 2: Substitute in Using Equation of the normal is or

10. Differentiating Parametric Equations Tip: It is a common mistake to forget to substitute for the parameter so beware ! SUMMARY  To find the equation of a tangent at a point: Substitute to find m at the given point. Substitute to find x and y at the given point. Use to find c. Find the gradient function. Substitute for m and c in  To find the gradient of a normal use where m 1 is the grad. of the tangent and m 2 the grad. of the normal.

11. Differentiating Parametric Equations Exercise 1. Find the equation of the tangent at the point where t =  1 to the curve 2. Find the equation of the normal at the point where to the curve given by

12. Differentiating Parametric Equations 1. Find the equation of the tangent at the point where t =  1 to the curve Solution:

13. Differentiating Parametric Equations At

14. Differentiating Parametric Equations 2. Find the equation of the normal at the point where to the curve given by Solution: Gradient of normal:

15. Differentiating Parametric Equations At Substitute in

16. Differentiating Parametric Equations The next ( final ) section is only for those of you studying the EDEXCEL spec.

17. Differentiating Parametric Equations The formula for finding the area under a curve, between the curve, the x -axis and the lines x = a and x = b is If y is given in terms of a parameter, t, we need to integrate w.r.t. t. We substitute for y, AND for the limits, all in terms of t.

18. Differentiating Parametric Equations Solution: Limits: e.g. Find the area shaded in the diagram where the curve is given by 4  x ( We need t > 0 since y > 0 )

19. Differentiating Parametric Equations

20. The method for finding volumes when using parametric equations is similar to that for finding areas.

21. Differentiating Parametric Equations Solution: Limits: e.g. Find the volume when the area shaded in the diagram is rotated about the x -axis.

22. Differentiating Parametric Equations

24. 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.

25. Differentiating Parametric Equations The main reason for using parametric equations is that they are easy to differentiate. Using the chain rule, Suppose the parameter is t.

26. Differentiating Parametric Equations Solution: e.g. 1 Find the equation of the tangent to the ellipse at the point where giving an exact answer. The equation of the tangent is given by We first find m at

27. Differentiating Parametric Equations So, at We can now substitute in to find c. To find c we also need x and y at So, the tangent is

28. Differentiating Parametric Equations e.g. 2(a) Find for the curve given by the parametric equations Solution: Using, the grad. of the normal is 4 when t = 2 (b) Show that the equation of the normal at the point where t = 2 is

29. Differentiating Parametric Equations Find x and y at t = 2: Substitute in Using to find c : Substitute in : Equation of the normal is or

30. Differentiating Parametric Equations Tip: It is a common mistake to forget to substitute for the parameter so beware ! SUMMARY  To find the equation of a tangent at a point: Substitute to find m at the given point. Substitute to find x and y at the given point. Use to find c. Find the gradient function. Substitute for m and c in  To find the gradient of a normal use where m 1 is the grad. of the tangent and m 2 the grad. of the normal.