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Maths Differentiation question - HELP!

Ok so the q is: find the equation of the NORMAL to y=x^4 - 4x^3 at the point for which x=1/2

So I differentiated and got 4x^3 and 12x^2. I got the gradient of the normal as 2/5 and y as - 7/16. I can't get the right answer! :confused:
Please Help! :smile:
wat. you have that the gradient of the tangent at that point is 1/2+3=7/2 so the gradient of the normal is -2/7...
Reply 2
Avatar for Palladium
Palladium
OP
The gradient of the tangent is 1/2 ^ 3 - 12 X (1/2^2) = 1/2 - 3??? not +!
Palladium
The gradient of the tangent is 1/2 ^ 3 - 12 X (1/2^2) = 1/2 - 3??? not +!

what crack are you smoking?
Reply 4
Avatar for Palladium
Palladium
OP
Sorry I wrote the Q wrong, my fault it is x^4 - 4x^3 soz
Palladium
Sorry I wrote the Q wrong, my fault it is x^4 - 4x^3 soz

you've got the right answer there. book is wrong.
maxfire
[br]dydx=4x3+12x2[br]\frac{dy}{dx} = 4x^3 + 12x^2
as you've worked out.

[br]y=124+4×123=916[br]y = \frac{1}{2}^4 + 4\times\frac{1}{2}^3 = \frac{9}{16}
I don't know how you got -7/16

[br]dydx=4×123+12×122=72[br]\frac{dy}{dx} = 4\times\frac{1}{2}^3 + 12\times\frac{1}{2}^2 = \frac{7}{2}
gradient = 7/2

so you just put them into the equation
[br]dydx=4x3+12x2[br]\frac{dy}{dx} = 4x^3 + 12x^2
as you've worked out.

[br]y=124+4×123=916[br]y = \frac{1}{2}^4 + 4\times\frac{1}{2}^3 = \frac{9}{16}
I don't know how you got -7/16

[br]y−y1=m(x−x1)[br]y - y_1 = m(x - x_1)

i think. Sorry if I'm wrong, it was a lot of confusing LaTeX

lolwaste of time.
Reply 7
Avatar for maxfire
maxfire
14
Totally Tom
lolwaste of time.


i know. latex makes me sad. I'm gunna stop using it ;(
Reply 8
Avatar for Palladium
Palladium
OP
what answers do you guys get then?
Palladium
what answers do you guys get then?

the same as you...
Reply 10
Avatar for Palladium
Palladium
OP
I have another horrid question, find the point on the curve y=2x^3-3x+1 where the gradient of the tangent = 1
differentiate it. you'll get a quadratic, equate this to 1 and solve for x.
Reply 12
Avatar for Adam92
Adam92
13
Palladium
I have another horrid question, find the point on the curve y=2x^3-3x+1 where the gradient of the tangent = 1


That's not horrid, I miss these easy questions :rolleyes:

Differentiate, and make dy/dx equal to one. Then solve for x and substitute it into the original curves equation
Reply 13
Avatar for Palladium
Palladium
OP
I did that! :'( I got dy/dx= 6x^2 -3, and the I got x = 1/3 and y = 2/27! and the answer is something completely different :/
you want
6x2−3=16x^2-3=1
x2=2/3x^2=2/3
x=...x=...
no work it out knob.
then sub x into the original y= equation.
Reply 16
Avatar for rbnphlp
rbnphlp
11
Palladium
x=0?

I think you have missed the basic stuff in maths
Reply 17
Avatar for Palladium
Palladium
OP
the answer is (1,0) so x=1! not root 2/3...

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