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# Need help with differentiation...yet again watch

1. the question is

the normals to the curve y=√x at the points where x=1 and x=4 meet at P, find the coordinates at P

ive tried loads of diff things but im jus not gettin the right answer

so to start of with when x=1 y=1 when x=4 y=2

and because its the normal the gradients will be the reciprocals(sp?) of the gradients of the tangent so

dy/dx= -1/2x^-3/2 so grad of tangent when x=1 = -1/2 so grad of normal is 2 and when x=4 grad of tangent is -1/16 so grad of normal is 16

now what...cause all i tried after this point didnt work

2. y = x^(1/2)

dy/dx = (1/2)x^(-1/2) = 1/(2√x)

x = 1, y = 1, dy/dx = 1/2
x = 4, y = 2, dy/dx = 1/4

Gradient of normal when x=1 ---> -1/(1/2) = -2
Gradient of normal when x=4 ---> -1/(1/4) = -4

Therefore y = -2x + c
Sub x = 1 & y = 1 in:
1 = -2 + c
--> c = 3
---> y = -2x + 3

Also, y = -4x + c
Sub x = 4 & y = 2 in:
2 = -16 + c
--> c = 18
---> y = -4x + 18

Equate two normals:
-2x + 3 = -4x + 18
2x = 15
x = 7.5

Sub x = 7.5 into either of the two normal equations:
y = -2(7.5) + 3
y = -15 + 3
y = 12 -12

---> (7.5,12-12)

This could be wrong, go through my working.

EDIT: Fixed errors
3. your differentiation seems to be wrong...dy/dx= -1/2x^-3/2 should be dy/dx= +1/2x^-1/2

you need the whole equation of the normal , not just the gradient.

the equation will be y = mx + c where m is the normal eg -2 and c is the intercept.
to find c use the fact that the point (1,1) is on the line...

1= -2*1 + c
c = 3

so y = -2x + 3

then you repeat for the second normal.
once you have both equations you solve them simultaneously.

and Robert is your mother's brother-in-law
4. (Original post by TomX)
y = x^(1/2)

dy/dx = (1/2)x^(-1/2) = 1/(2√x)

x = 1, y = 1, dy/dx = 1/2
x = 4, y = 2, dy/dx = 1/4

Gradient of normal when x=1 ---> -1/(1/2) = -2
Gradient of normal when x=4 ---> -1/(1/4) = -4

Therefore y = -2x + c
Sub x = 1 & y = 1 in:
1 = -2 + c
--> c = 3
---> y = -2x + 3

Also, y = -4x + c
Sub x = 4 & y = 2 in:
2 = -16 + c
--> c = 18
---> y = -4x + 18

Equate two normals:
-2x + 3 = -4x + 18
2x = 15
x = 7.5

Sub x = 7.5 into either of the two normal equations:
y = -2(7.5) + 3
y = -15 + 3
y = 12

---> (7.5,12)

This could be wrong, go through my working.
thank you so much except its 7.5.-12

but anyways its not even hard i was jus an idiot and though √x = x^-1/2 when its jus a half and so my dy/dx was completely wrong...cheers tho!
5. can anyone help me with this question?

f(x) = (c-1/c -x)(4 - 3x^2) where c is a positive constant and x varies over the real numbers.

i) Show that f(x) has one max and one min
ii) Show that the difference between these values of f(x) at it turning points is:

4/9(c+1/c)^3

What is the least value that the difference in (ii) can have for c > 0

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Updated: November 12, 2005
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