Your SlideShare is downloading. ×
Gradient of a curve and equation of a
Gradient of a curve and equation of a
Gradient of a curve and equation of a
Gradient of a curve and equation of a
Gradient of a curve and equation of a
Gradient of a curve and equation of a
Gradient of a curve and equation of a
Gradient of a curve and equation of a
Gradient of a curve and equation of a
Gradient of a curve and equation of a
Gradient of a curve and equation of a
Gradient of a curve and equation of a
Gradient of a curve and equation of a
Gradient of a curve and equation of a
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Text the download link to your phone
Standard text messaging rates apply

Gradient of a curve and equation of a

5,891

Published on

Published in: Education
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total Views
5,891
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
16
Comments
0
Likes
0
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide

Transcript

  • 1. Gradient of a Curve andEquation of a Tangent to a Curve Math Studies 2 – Topic 7
  • 2. Find the gradient of the tangent to the curve whose 4equation is y = + 2 at the point where x=2. x −1 y = 4x + 2 substitute f ( x ) = −4 x + 0 −2 −4 f ( x) = 2 = −1 2 4 f ( x ) = −4 x −2 = 2 x
  • 3. A power boat moves in a straight line such that at time tseconds its distance s from a fixed point O on that line isgiven by s = 2t − 3t + 1 2 Find the speed after 3 seconds. ds substitute = 4t − 3 t =3 dt speed = 4(3) − 3 = 9m / s
  • 4. Find the value of the gradient of the curve whoseequation is y=(x-3)(x+2) at the point where it crosses thepositive x-axis (x=3). multiply substitute y = x − x−6 2 x=3 f ( x) = 2(3) − 1 = 5 f ( x) = 2 x − 1
  • 5. Think Back What are thesteps I used to solve the problems?
  • 6. Steps to finding the gradient of any curve• Differentiate to find the gradient• Substitute the particular value of x to find the gradient of the curve at that point
  • 7. Find the equation of the tangent to the curve whose 2equation is y = 3x 2 + − 5 at the point P where x=2. x dy 2 2 −2 = 6x − 2x = 6x − 2 y = 3(2) 2 + −5 = 8 dx x 2 y =8 P(2,8) substitute dy 2 1 y = mx + c = 12 − = 11 dx 4 2 1 1 8 = 11 (2) + c m = 11 2 2 c = 15
  • 8. equation 1y = 11 x − 15 2
  • 9. Think Back What are thesteps I used to solve the problems?
  • 10. Equation of the Tangent at a Given Point• Find the gradient at the point P and call it m• Find the y-coordinate of P as well as the x- coordinate (sometimes it will be given)• Use y=mx+c and substitute for m, x, and y at the point P• Write the equation with values of m and c
  • 11. Independent Practice
  • 12. 1) Find the equation of the tangent to the curve whose 4 equation is y=x²-2x +3 at the point where x=1. Show all working.2) Find the equation of a tangent to the curve whose equation is y=2x³-4x at the point where x=2.3) The curve whose equation is y=x²-4x+15 has gradient 6 when x=a. Find the value of a.4) The tangent to the parabola given by y=x²+3x-8 has gradient 7 at the point P. Find the coordinates of P.
  • 13. Solution1)y=2x+42)y=20x-323)54)(2,2)
  • 14. Resourceswww.info.oxford-consulting.comwww.wclipart.comwww.clipartspot.netMathematical Studies Course Companion (Oxford)

×