# Integration help needed C2Watch

#1
The question
The curve touches the x-axis at point A. B is the point (2,2)
on the curve and N is the foot of the perpendicular from B to the x-axis. Prove that the tangent B divides the area between the arc AB, BN and AN in the ratio 11:24

My problem I have found the area of Area B as shown in my working but the problem is area A I could a different answer please have a look at my working and inform me where I went wrong

I have attached my working below
Only my calculations for area B was correct not my area A

thank you
0
3 years ago
#2
(Original post by bigmansouf)
Only my calculations for area B was correct not my area A

thank you

Area above the tangent is 11/60

Area below is 2/5

Both of which you have.

Hence ratio 11/60 : 2/5
or 11/60 : 24/60...

You just put them together incorrectly.
#3
(Original post by ghostwalker)
Area above the tangent is 11/60

Area below is 2/5

Both of which you have.

Hence ratio 11/60 : 2/5
or 11/60 : 24/60...

You just put them together incorrectly.
no I'm sorry I didnt find the area for Area A thats the problem I found
I wrote in the picture that I expected but could
Also I proceeded to write
I am sorry I didnt make it clear but I wanted help in finding Area A as I got the wrong answer.
0
3 years ago
#4
(Original post by bigmansouf)
no I'm sorry I didnt find the area for Area A thats the problem I found
I wrote in the picture that I expected but could
Also I proceeded to write
I am sorry I didnt make it clear but I wanted help in finding Area A as I got the wrong answer.
OK, I just saw you had the correct value.

In your working for finding the area for A you subtracted the eqn of the tangent from the equation of the curve. This gives you the area between the curve and the tangent. This corresponds with your desired area as long as the tangent is above the x-axis. But for x<8/5 this isn't true.

You want to work out the area under the curve. Then the area in the triangle (area B). And subtract one from the other to get the area A.
#5

Attachment 473551473553

I have atteched the workings I did but I still got it wrong the book says the equation of the curve is
Where did I go wrong?
0
3 years ago
#6
(Original post by bigmansouf)

Attachment 473551473553

I have atteched the workings I did but I still got it wrong the book says the equation of the curve is
Where did I go wrong?
Don't understand - what's the original question. Are you sure you've on the right thread?
#7
(Original post by ghostwalker)
Don't understand - what's the original question. Are you sure you've on the right thread?
0
#8
(Original post by ghostwalker)
OK, I just saw you had the correct value.

In your working for finding the area for A you subtracted the eqn of the tangent from the equation of the curve. This gives you the area between the curve and the tangent. This corresponds with your desired area as long as the tangent is above the x-axis. But for x<8/5 this isn't true.

You want to work out the area under the curve. Then the area in the triangle (area B). And subtract one from the other to get the area A.
thank you very much i will do this and report back after work
0
#9
SOLVED

find the tangent at B (2,2)

b(2,2) m = 5 thus tangent at B is
Area of BND sq units

Area of integral from 2 to 1
thus area of ABD

ratio:

thus; tangent at B(2,2) divides AB, BN, and AN into ratio 11:24
Last edited by bigmansouf; 1 week ago
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