Wednesday, February 13, 2008

Applications of differentiation

Finding the equations of tangents to curves

To obtain the equation of the tangent to a curve with equation y = f(x), we need to be able to differentiate f(x).

The following steps show how to find the equation of the tangent to the curve with given equation at the point where x is given.

Step 1:-
Differentiate y or find dy/dx
This would give the gradient function of the curve.
Substituting the given value of x would give the value of the gradient of the tangent to the curve at the point.
Hence the equation of the tangent would begin with y = m x + c where m is the value of the gradient found.

Step 2:-
Find the y-coordinate of the point with the given value of x. The values of x and y are needed to be substituted into the equation y = m x + c to calculate the value of c.
Hence the equation of the tangent would be obtained.

Worked example:-
Find the equation of the tangent to the curve with equation
y = 3x^2 + 2x - 5 at the point where x = - 2.

Step 1:-
dy/dx = 6x + 2
At x = -2, dy/dx = - 10
Hence y = - 10 x + c ---------equation [1]

Step 2:-
When x = - 2, y = 3
Substitute these values into equation [1]
Hence c = - 17
And the equation of the tangent is y = - 10 x – 17

5 comments:

Anonymous said...

Thanks,now I have a clear mind on doing dy/dx and finding the unkown points in the questions.

Anonymous said...

Mrs goh ,could you print some more worksheets about this topic , because from chapter 14 to 17 are all related to differentiation.and sometimes i feel this chapter quite complicated.

Anonymous said...

Hi Jiaying - ok, I will upload more of my explanation to the topics taught in class as and when I have finished teaching them.

Anonymous said...

dis topic hor..okay la quite easy to understand..mayb later topic cant understand ba..

Anonymous said...

Differentiation is okay until . . . the rates of change. I think i can't score in that @.@