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Differentiation Gradient problems.

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Presentation on theme: "Differentiation Gradient problems."β€” Presentation transcript:

1 Differentiation Gradient problems

2 Differentiation Starter: KUS objectives
BAT differentiate polynomials and find the gradient of curves Starter: 𝑦= 3 π‘₯ βˆ’ π‘₯ 𝑦=3 π‘₯ 5 +2 π‘₯ 3 βˆ’π‘₯ 𝑦=βˆ’3 π‘₯ βˆ’2 βˆ’ π‘₯ βˆ’4 βˆ’6 Find the y value of each function when x = 2 Calculate the gradient when x = 2

3 Practice 1

4 Practice 1 solutions

5 a) Given that 𝑦= 5π‘₯ 3 +6+ 5 π‘₯ 2 find 𝑑𝑦 𝑑π‘₯ in its simplest form
WB10 a) Given that 𝑦= 5π‘₯ π‘₯ find 𝑑𝑦 𝑑π‘₯ in its simplest form b) Given that 𝑦= 9π‘₯ 3 βˆ’8 π‘₯ + 9 π‘₯ 2 +4 π‘₯ find 𝑑𝑦 𝑑π‘₯ in its simplest form

6 a) Find the coordinate of the point on the curve 𝑦= 4π‘₯ 2 βˆ’10π‘₯+6
WB11 a) Find the coordinate of the point on the curve 𝑦= 4π‘₯ 2 βˆ’10π‘₯+6 Where the gradient is -2 b) Sketch this graph and point on a diagram Now find the point by going back to the original equation for y When x = 1 … y = 0

7 We need to be able to find these points using algebra
Points with a Given Gradient WB12a Determine the points on the curve 𝑦= π‘₯ 3 βˆ’8π‘₯+7 Where the gradient is equal to 4 Find the coordinates of the points on the curve where the gradient equals 4 Gradient of curve = gradient of tangent = 4 We need to be able to find these points using algebra

8 Points with a Given Gradient
WB112b Determine the points on the curve 𝑦= π‘₯ 3 βˆ’8π‘₯+7 Where the gradient is equal to 4 𝑑𝑦 𝑑π‘₯ =3 π‘₯ 2 βˆ’8 𝑑𝑦 𝑑π‘₯ =3 π‘₯ 2 βˆ’8=4 3 π‘₯ 2 =12 π‘₯=Β±2 When π‘₯ = 2, 𝑦= (2) 3 βˆ’8 2 +7=βˆ’1 When π‘₯ =βˆ’2, 𝑦= (βˆ’2) 3 βˆ’8 βˆ’2 +7=15 Points where the gradient is four are (2, βˆ’1) and (βˆ’2, 15)

9 Determine the points on the curve 𝑦= π‘₯ 3 +5π‘₯+4
WB13 Determine the points on the curve 𝑦= π‘₯ 3 +5π‘₯+4 Where the gradient is equal to 17 𝑑𝑦 𝑑π‘₯ =3 π‘₯ 2 +5 𝑑𝑦 𝑑π‘₯ =3 π‘₯ 2 +5=17 3 π‘₯ 2 =12 π‘₯=Β±2 When π‘₯ = 2, 𝑦= (2) =27 When π‘₯ =βˆ’2, 𝑦= βˆ’ βˆ’2 +4=βˆ’14 Points where the gradient is 17 are (2, 27) and (βˆ’2, βˆ’14)

10 Determine the points on the curve 𝑦= 1 4 π‘₯ 4 βˆ’ π‘₯ 2 +π‘₯+3
WB14 Determine the points on the curve 𝑦= 1 4 π‘₯ 4 βˆ’ π‘₯ 2 +π‘₯+3 Where the gradient is equal to 1 𝑑𝑦 𝑑π‘₯ = π‘₯ 3 βˆ’2π‘₯+1 𝑑𝑦 𝑑π‘₯ = π‘₯ 3 βˆ’2π‘₯+1=1 π‘₯ 3 βˆ’2π‘₯=0 π‘₯(π‘₯+2)(π‘₯βˆ’2)=0 π‘₯=βˆ’2, 0 2 When π‘₯ = 2, 𝑦= 1 4 (2) 4 βˆ’ =5 When π‘₯ =βˆ’2, 𝑦= βˆ’ βˆ’2 +4=βˆ’14 When π‘₯ =βˆ’2, 𝑦= βˆ’ βˆ’2 +4=βˆ’14 Points where the gradient is 1 are (2, 5). 0 and (βˆ’2, βˆ’14)

11 How can we use the gradient?
WB15 Sketch the graph of 𝑦= π‘₯ 2 βˆ’4π‘₯βˆ’21 showing the minimum point and the places where the graph intersects the axes. Find the minimum point At points and And At point How can we use the gradient? the gradient = 0 at the minimum point At point

12 One thing to improve is –
KUS objectives BAT differentiate polynomials and find the gradient of curves self-assess One thing learned is – One thing to improve is –

13 END


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