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OCR Core 1 Differentiation

I am stuck on this question, I have checked the answers page but it does not show any working out on how they got the answer.

Find the equation of the normal to the curve at the point with the given x-coordinate.

y=-x^2 where x=1
Original post by danad2341
I am stuck on this question, I have checked the answers page but it does not show any working out on how they got the answer.

Find the equation of the normal to the curve at the point with the given x-coordinate.

y=-x^2 where x=1


Well, what have you tried? :smile:
Reply 2
Avatar for james22
james22
16
What have you tried?

Do you know what the normal means and how to find its gradient?
Original post by danad2341
I am stuck on this question, I have checked the answers page but it does not show any working out on how they got the answer.

Find the equation of the normal to the curve at the point with the given x-coordinate.

y=-x^2 where x=1


Differentiate with respect to x. This will give you the gradient function. Sub the x =1 into the gradient function to get the value of the gradient. From here use dat normal rule to find the gradient of the normal. Sub x=1 into y=-x^(2) to get y1.

Then do y-y1 = m(x-x1) where m is the gradient of the normal

Posted from TSR Mobile
(edited 9 years ago)
Differentiate the equation and sub in the x value.
Reply 5
Avatar for danad2341
danad2341
OP
Original post by usycool1
Well, what have you tried? :smile:



y=(-1)^2
y=1
dy/dx -x2 = -2x
-2(1) = -2
m=-2
y-1=-2(x-1)
y=-2x+3


The answer in the textbook is 2y=x-3
Reply 6
Avatar for james22
james22
16
Original post by danad2341
y=(-1)^2
y=1
dy/dx -x2 = -2x
-2(1) = -2
m=-2
y-1=-2(x-1)
y=-2x+3


The answer in the textbook is 2y=x-3


Why have you done (-1)^2? You should have done -(1)^2 to get -1 instead of 1.

You have used the gradient of the tangent, you need to find the gradient of the normal.
the normal is perpendicular to the tangent
Two things are incorrect here:

Original post by danad2341
y=(-1)^2
y=1


y = -x^2, not y = (-x)^2. So you should actually get y = - (1)^2 = -1.


-2(1) = -2
m=-2


That's the gradient of the tangent at the point, but you want the gradient of the normal (i.e. at right angles to the tangent).
Reply 9
Avatar for danad2341
danad2341
OP
Original post by james22
What have you tried?

Do you know what the normal means and how to find its gradient?


-1/m the normal is perpendicular to the tangent
Reply 10
Avatar for danad2341
danad2341
OP
Original post by usycool1
Two things are incorrect here:



y = -x^2, not y = (-x)^2. So you should actually get y = - (1)^2 = -1.




That's the gradient of the tangent at the point, but you want the gradient of the normal (i.e. at right angles to the tangent).


would the gradient of the normal be 1/2 then?
Original post by danad2341
would the gradient of the normal be 1/2 then?


Yup.
Reply 12
Avatar for danad2341
danad2341
OP
Original post by usycool1
Yup.


so.. y-1=1/2(x-1)

would you times the other side by two so you don't have to expand?

so
2y-2=x-1
2y=x-3


:smile:
Original post by danad2341
so.. y-1=1/2(x-1)

would you times the other side by two so you don't have to expand?

so
2y-2=x-1
2y=x-3


:smile:


That would be a good idea.

Nice work :smile:

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