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C1 help

Not sure about how to do question 10.Any assistance would be welcomed.
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Because the gradients are parallel, the gradients will be the same.

The straight line is already in the form y=mx+c, so you can see the gradient for this line is simply 8

That means also that dy/dx=8

Then just differentiate it, solve for x and sub into the equation to get y

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Original post by Kadak


What would the differential have to equal?

Posted from TSR Mobile
Original post by dont ban me :(
Because the gradients are parallel, the gradients will be the same.

The straight line is already in the form y=mx+c, so you can see the gradient for this line is simply 8

That means also that dy/dx=8

Then just differentiate it, solve for x and sub into the equation to get y

Posted from TSR Mobile




Posted from TSR Mobile
I love you quantum man,no homo.
Original post by Kadak
Not sure about how to do question 10.Any assistance would be welcomed.


When I did this question, I was told to change the dy/dx equation back into fraction form as it would be easier to work with later on.
You set the dy/dx equation to equal the gradient and then multiply through by x^2 to get rid of the fraction.
You then rearrange this equation so all terms are on the LHS and the RHS = 0.
Now minimise into a quadratic that you can work out easily to factorise by letting y = x^2
Factorise.
Solve for x and then sub in to original equation to find y and these are the coordinates.

I hope this is right and helps you a bit :smile:

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