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C1 - Differentiation

Can someone help me with this question :smile:.
It's question 9C January 2005 Edexcel paper.

Part c). Using dydx=(3x−1)2\frac{dy}{dx}=(3x-1)^2 show that there is no point on C at which the tangent if parallel to the line 1−2x1-2x.

I honestly don't know what it is asking :frown:

Any help would be appreciated.
Reply 1
Avatar for TenOfThem
TenOfThem
18
Original post by Super199
Can someone help me with this question :smile:.
It's question 9C January 2005 Edexcel paper.

Part c). Using dydx=(3x−1)2\frac{dy}{dx}=(3x-1)^2 show that there is no point on C at which the tangent if parallel to the line 1−2x1-2x.

I honestly don't know what it is asking :frown:

Any help would be appreciated.


What is the gradient of y=1-2x

Can the gradient of the curve be the same

If it can then it is a tangent, the question suggests that it cannot
1.) Expand out (3x -1)^2 to get the quadratic 9x^2 - 6x +1
2.) Since you know dy/dx gives the gradient of the tangent to the curve, we can compare this with the equation y = -2x + 1. The gradient for this line is -2.

For two lines to be parallel they must have the same gradient.

Assume 9x^2 - 6x +1 = -2, rearrange to get 9x^2 - 6x +3 = 0. This new quadratic has no real roots, so this shows that there's no point on C at which the tangent is parallel to the line y = 1-2x
Reply 3
Avatar for Super199
Super199
OP
20
Original post by TenOfThem
What is the gradient of y=1-2x

Can the gradient of the curve be the same

If it can then it is a tangent, the question suggests that it cannot


The gradient of the line is -2. But dy/dx cannot = -2 because we are squaring it? So the gradient has to be greater than 0 to be tangent?? :redface:
Reply 4
Avatar for davros
davros
16
Original post by StudyScience98
1.) Expand out (3x -1)^2 to get the quadratic 9x^2 - 6x +1
2.) Since you know dy/dx gives the gradient of the tangent to the curve, we can compare this with the equation y = -2x + 1. The gradient for this line is -2.

For two lines to be parallel they must have the same gradient.

Assume 9x^2 - 6x +1 = -2, rearrange to get 9x^2 - 6x +3 = 0. This new quadratic has no real roots, so this shows that there's no point on C at which the tangent is parallel to the line y = 1-2x


There is no need to do any expansion with this question!

The expression for dy/dx is something squared, so it can never be negative.

The gradient of the line is ???

Therefore ...
Reply 5
Avatar for Super199
Super199
OP
20
Original post by StudyScience98
1.) Expand out (3x -1)^2 to get the quadratic 9x^2 - 6x +1
2.) Since you know dy/dx gives the gradient of the tangent to the curve, we can compare this with the equation y = -2x + 1. The gradient for this line is -2.

For two lines to be parallel they must have the same gradient.

Assume 9x^2 - 6x +1 = -2, rearrange to get 9x^2 - 6x +3 = 0. This new quadratic has no real roots, so this shows that there's no point on C at which the tangent is parallel to the line y = 1-2x

I was thinking of doing that. For some stupid reason I put 9x^2-6x+1=1-2x. Right need to brush up a bit I guess. :s-smilie: Thanks :smile:
Reply 6
Avatar for TenOfThem
TenOfThem
18
Original post by Super199
The gradient of the line is -2. But dy/dx cannot = -2 because we are squaring it? So the gradient has to be greater than 0 to be tangent?? :redface:


:biggrin:

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