# Differentiation

• Mar 21st 2007, 09:51 AM
Fnus
Differentiation
Hey, any help with this is appreciated, thanks. ^^

Question 1:
Find the coordinates of any points on this curve where the gradient is stated.

Y = x^2 + 3 / 2x^2, gradient = 3
(it's a fraction, I just don't know how to make those)

Question 2:
Given that the curve with equation y=Ax^2 + Bx has gradient 7 at the point (6,8), find the values of the constants A and B.

Question 3:
The two tangents to the curve y=x^2 at the points where y = 9 intersect at the point P. Find the coordinates of P.

Question 4:
Find the values of x at which the gradient of the curve
y = x^3 - 6x^2 + 9x + 2 = 0

Thanks again ^^
• Mar 21st 2007, 11:55 AM
Jhevon
Quote:

Originally Posted by Fnus
Question 1:
Find the coordinates of any points on this curve where the gradient is stated.

Y = x^2 + 3 / 2x^2, gradient = 3
(it's a fraction, I just don't know how to make those)

y = (x^2 + 3)/(2x^2)
=> y = (x^2)/(2x^2) + 3/(2x^2)
=> y = 1/2 + (3/2)x^-2
=> y' = -3x^-3

we want y' = 3

=> 3 = -3x^-3
=> -1 = x^-3
=> -1 = x^3 ..........i took the inverse
=> x = -1

when x = -1, y = [(-1)^2 + 3]/[2(-1)^2] = 4/2 = 2

so the coordinate where the gradient is 3 is (-1,2)

Quote:

Question 2:
Given that the curve with equation y=Ax^2 + Bx has gradient 7 at the point (6,8), find the values of the constants A and B.
y = Ax^2 + Bx
=> y' = 2Ax + B

when x = 6, y' = 7
=> 7 = 12A + B ..............(1)

also (6,8) is a point on the curve, so going back to the original equation and plugging these in we get:

8 = 36A + 6B
=> 4 = 18A + 3B ..............(2)

so we have the simultaneous equations:

7 = 12A + B ..............(1)
4 = 18A + 3B ............(2)

=> 4/3 = 6A + B .........(3) = (2)/3

=> 17/3 = 6A
=> A = 34

but 7 = 12A + B
=> B = 7 - 12(34) = -401

so A = 34, B = -401 ..........wierd numbers I know

Quote:

Question 3:
The two tangents to the curve y=x^2 at the points where y = 9 intersect at the point P. Find the coordinates of P.
when y = 9
=> 9 = x^2
=> x = +/- 3

so the tangents cut the curve at (3,9) and (-3,9). what are the equations of these tangent lines? why are we interested in that. if we find the equations of the tangent lines, we can equate them and find where they intersect. we can find the gradient from the derivative

y' = 2x

so at (3,9), gradient is 2(3) = 6
at (-3,9), gradient is 2(-3) = -6

for the line passing through (3,9)
y - y1 = m(x - x1)
=> y - 9 = 6(x - 3) = 6x - 18
=> y = 6x - 9

for the line passing through (-3,9)
y - y1 = m(x - x1)
=> y - 9 = -6(x + 3) = -6x - 18
=> y = -6x - 9

so we have y = 6x - 9 and y = -6x - 9

so they intersect when 6x - 9 = -6x - 9
=> 12x = 0
=> x = 0

so they intersect at (0,-9), which is the point P

Quote:

Question 4:
Find the values of x at which the gradient of the curve
y = x^3 - 6x^2 + 9x + 2 = 0
the gradient is given by the derivative
y = x^3 - 6x^2 + 9x + 2
=> y' = 3x^2 - 12x + 9
set y'=0
=> 0 = 3x^2 - 12x + 9
now we solve for x

=> 0 = x^2 - 4x + 3
=> (x - 3)(x - 1) = 0
so x = 3, x = 1

these are the x-values for which the condition holds

Disclaimer: the methods are correct, but you might want to double check my computations, i've been making stupid mistakes all week