• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Mathematics: Post your doubts here!

salvatore

salvatore

Messages
382
Reaction score
315
Points
73
Obsidian Fl1ght said:
Hey,, no worries. Glad 2 help.

So, 9(ii):
I guess you know what we gotta do in this - we treat the gradient of normal as a line. To find the stationary (max) point on it, we find the dy/dx of this 'line'. (Technically we're findin the d^2y/dx^2)

From (i), we know dy/dx is (1+x).[(1-x^2)]^(1/2)___________ [Here, (1+x) is u and v is [(1-x^2)]^(1/2)]
Apply (u.v)' = u'.v+u.v'
d^2y/dx^2 = (1).[(1-x^2)]^(1/2) + (1+x).(1/2)[(1-x^2)]^(-1/2)(-2x)
= [(1-x^2)]^(1/2) + (-x)(1+x)[(1-x^2)]^(-1/2)
Now equate this to zero:
[(1-x^2)]^(1/2) + (-x)(1+x)[(1-x^2)]^(-1/2) = 0
I then got: (1-x^2) - x(1+x) = 0 ___________ (Coz [(1-x^2)]^(1/2).[(1-x^2)]^(1/2) = (1-x^2) )
Then: 1 - x^2 - x - x^2 = 0
-2x^2 - x + 1 = 0 (multiply with -1 in both sides)
2x^2 + x - 1 = 0
Solving this you get two sol., x = 1/2 (ANS.) or x = -1
From graph, we see dat point P ain't on x = -1
Ans. is thus x = 1/2
Click to expand...
damn.. I had made a small error in differentiating.
Thanks a lot man!
 
Suchal Riaz

Suchal Riaz

Messages
1,983
Reaction score
3,044
Points
273
ahmed abdulla said:
question
7 (ii)
Click to expand...
I am just guessing btw. For y to be zero, either sin or cos have to be zero. Graph starts at zero as sin is zero. Then graph comes to zero when cos is zero. Which is at 2X=pi/2. Therefore x=pi/4. For one area the interval is of pi/4. For 40 A it must be 40/4=10pi. Is my answer correct?
 
You must log in or register to reply here.
Top