Starter. Differentiating Parametric Equations Aims: To be able to differentiate equations defined parametrically.

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Presentation transcript:

Starter

Differentiating Parametric Equations Aims: To be able to differentiate equations defined parametrically.

Differentiating Parametric Equations We can differentiate parametric equations using the chain rule. We can differentiate parametric equations using the chain rule. Differentiate y with respect to t. Differentiate y with respect to t. Differentiate x with respect to t. Differentiate x with respect to t. Flip dx/dt to get dt/dx and plug into the chain rule. Flip dx/dt to get dt/dx and plug into the chain rule.

Example Find the gradient at t = 3 on the parabola x = t 2, y = 2t

Matching exercise Each set contains 5 cards one of each colour. Each set contains 5 cards one of each colour. Match up the cards that definitely go together first. Match up the cards that definitely go together first. To fully complete the puzzle you will need to consider which combinations are possible and check which will match with the final cards. To fully complete the puzzle you will need to consider which combinations are possible and check which will match with the final cards.

Questions

Tangents and Normals Aims: To find the equations of tangents and normals to parametric equations To answer hard/exam questions including questions with proofs

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

So, at We can now substitute in to find c. To find c we also need x and y at

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

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

Find x and y at t = 2: Using to find c : Substitute in : Equation of the normal is Substitute in

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

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.

Group Quiz Group Quiz Working in groups of 3 or 4 Working in groups of 3 or 4 Collect a mini white boards, pen and cloth each. Collect a mini white boards, pen and cloth each. Show all workings. Show all workings. Decide on an answer in your group Decide on an answer in your group Don’t hold up your answer cards until asked. Don’t hold up your answer cards until asked. If someone in your group does not understand the question make sure you explain it to each other. If someone in your group does not understand the question make sure you explain it to each other.

Relay Race in groups 4 x hard/exam/show that questions

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

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

Differentiating Parametric Equations At Substituting in for m and c : Find c :

Differentiating Parametric Equations At Substituting in for m and c : Find c :

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

Differentiating Parametric Equations At Substitute in to find c : Substitute for c in :

notes

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.

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

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

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

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

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.