# OCR Core 1 Differentiation

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Thread starter 6 years ago
#1
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
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6 years ago
#2
(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? 0
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6 years ago
#3
What have you tried?

Do you know what the normal means and how to find its gradient?
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6 years ago
#4
(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

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6 years ago
#5
Differentiate the equation and sub in the x value.
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Thread starter 6 years ago
#6
(Original post by usycool1)
Well, what have you tried? 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
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6 years ago
#7
(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.
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6 years ago
#8
the normal is perpendicular to the tangent
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6 years ago
#9
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).
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Thread starter 6 years ago
#10
(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
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Thread starter 6 years ago
#11
(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?
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6 years ago
#12
(Original post by danad2341)
would the gradient of the normal be 1/2 then?
Yup.
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Thread starter 6 years ago
#13
(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 0
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6 years ago
#14
(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 That would be a good idea.

Nice work 1
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