Add Maths

[Spidey21]

Hello Everybody...

I will be really happy if someone helped me out on a question of Add maths...

The question is of June 2000, Paper 1 and the Q2...its from Differentiation and its applications....

I just want to know that how did we take out the area at the end...

Thank you in Advance!!

 

[mohdumar]

reproduce the question here because we dont have online past papers before 2001

otherwise those with past paper books can help

 

[Spidey21]

Q.Find the gradient of the curve y=4x^2 -20x+27 at the point P(2,3).
The tangent to the curve at P meets the x-axis at A. Calculate the area of the triangle AOP, where O is the origin..

Please answer ASAP!!

 

[Anon]

Here is how you do it:

1. Find derivative of the equation.
2. substitute 2 in place of x.
3.You get the gradient.

4.Write the equation, "y-y1 = m(x-x1)"
5. Substitute as:
x1=2
y1=3
m=(the gradient you found at step 3)

6.You now have equation of the tangent
7.substitute, (x,0) in equation.
8.You now have the coordinate of point A.

9.Now apply area of triangle formula:

(1/2) * (value of x at point A) * (value of y at point P)

simplifying:
(1/2)*(3)*(x coordinate of A)

ta daaa, you get the answer...

 

[Spidey21]

i did the same thing but the answer on the past papers answer and mine didnt match!! mind doing just the last part???

 

[Silent Hunter]

what is your answer spidey21 ?

 

[Silent Hunter]

0.5 x 2.75 x 3 = 4.125 unit^2

correct or not? what does the answer say?

 

[Anon]

i did the same thing but the answer on the past papers answer and mine didnt match!! mind doing just the last part???

Lets see:

Co ordinate are:
A = (11/4 , 0)
P= (2,3)

height= 3
base= 11/4

So,

11/4 * 3 *0.5

= 4.125cm^2

 

[Spidey21]

4.125 it is!! =P good job!!