# OCR Core 1 Differentiation

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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

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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(Original post by

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

**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

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#3

What have you tried?

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

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

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**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

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

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Well, what have you tried?

**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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#7

(Original post by

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

**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

You have used the gradient of the tangent, you need to find the gradient of the normal.

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#9

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).

-2(1) = -2

m=-2

m=-2

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(Original post by

What have you tried?

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

**james22**)What have you tried?

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

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(Original post by

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).

**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).

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(Original post by

Yup.

**usycool1**)Yup.

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

so

2y-2=x-1

2y=x-3

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#14

(Original post by

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

**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

Nice work

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