1 / 11

Higher Revision WUQ

Higher Revision WUQ. 1. Find Equation of tangent to the curve y = 4cos(2x – π / 4 ) at x = π / 2 2. For equation tx 2 + (t + 10)x + 18 = 0, find the values of t where equation has equal roots. 3. A(2, 5, 9) B(4, 8, 12) C(5, 7, 14) are vertices of triangle ABC. Find size of L BAC.

viveka
Download Presentation

Higher Revision WUQ

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Higher Revision WUQ

  2. 1. Find Equation of tangent to the curve y = 4cos(2x – π/4) at x = π/2 2. For equation tx2 + (t + 10)x + 18 = 0, find the values of t where equation has equal roots. 3. A(2, 5, 9) B(4, 8, 12) C(5, 7, 14) are vertices of triangle ABC. Find size of LBAC

  3. 1. Find Equation of tangent to the curve y = 4cos(2x – π/4) at x = π/2 Need point and gradient gradient Let y = 4cosU where U = 2x – π/4 dy/du = 4sinU dU/dx = 2 dy/dx = 4sinU 2 = 8sinU = 8sin(2x – π/4) For gradient sub x = π/2 Find derivative m = 8sin(2 (π/2) – π/4) m = 8sin (3π/4) = 8(1/√2) = 8/√2

  4. 1. Find Equation of tangent to the curve y = 4cos(2x – π/4) at x = π/2 Need point and gradient gradient Let y = 4cosU where U = 2x – π/4 dy/du = 4sinU dU/dx = 2 dy/dx = 4sinU 2 = 8sinU = 8sin(2x – π/4) For gradient sub x = π/2 Find derivative m = 8sin(2 (π/2) – π/4) m = 8sin (3π/4) = 8(1/√2) = 8/√2

  5. 1. Find Equation of tangent to the curve y = 4cos(2x – π/4) at x = π/2 Need point and gradient sub x = π/2 into equation y = 4cos(2(π/2) – π/4) = 4cos(3π/4) = -4/√2 Use y – b = m(x – a), where m = 8/√2 and (a , b) =(π/2 , -4/√2)

  6. 2. For equation tx2 + (t + 10)x + 18 = 0, find the values of t where equation has equal roots. For equal roots b2 – 4ac = 0 a = t, b = t + 10, c = 18 (t + 10)2 – 4(t)(18) = 0 t2 + 20t + 100 – 72t = 0 t2 – 52t + 100 = 0 (t – 2)(t – 50) = 0 t = 2, t = 50

  7. B (4,8,12) 3. A(2, 5, 9) B(4, 8, 12) C(5, 7, 14) are vertices of triangle ABC. Find size of LBAC = b – a AB A ( ) ( ) 2 5 9 4 8 12 – = (2,5,9) C ( ) (5,7,14) 2 3 3 = = c – a AC a.b Cosϴ = ( ) ( ) 2 5 9 5 7 14 |a| |b| – = ( ) Need AB and AC 3 2 5 =

  8. 3. A(2, 5, 9) B(4, 8, 12) C(5, 7, 14) are vertices of triangle ABC. Find size of LBAC = b – a AB AB.AC = 2(3) + 3(2) + 3(5) ( ) ( ) 2 5 9 4 8 12 = 27 – = ( ) AB = √(22 + 32 + 32) = √22 2 3 3 = AC = √(32 + 22 + 52) = √38 = c – a AC Cosϴ = 27 ( ) =0.9338 ( ) 2 5 9 5 7 14 – = √22√38 ( ) ϴ = 21.00 3 2 5 =

More Related