Original post by skent6
I’m struggling on this question (I’m pretty bad at maths just to clarify), can anyone give me some pointers please?
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Working out? Find the derivative, put x=-3 in and then get the gradient of the normal. That has the same gradient as the line given
Original post by Maximus 190
Working out? Find the derivative, put x=-3 in and then get the gradient of the normal. That has the same gradient as the line given
Tried that but I couldn’t get it cos I don’t like the k being there
Original post by skent6
Tried that but I couldn’t get it cos I don’t like the k being there
It’s just y=k * x^-2 , try differentiate that
use the equation of the line to find the gradient of that line (6), the gradient of this line is the negative reciprocal of the tangent to the curve at that point (as they are perpendicular) (i.e the gradient of the curve at this point is -1/6), so the derivative of k/x^2 when x=-3 is equal to -1/6 if that helps
The curve isn’t in indice form it should be kx^-2, you can then differentiate that and substitute in the -3 to find the gradient. The normal to the curve is that gradient with -1/m where m is the gradient of the curve you found by differentiating. You have the k value which is unknown but you know the gradient of -1/m is equal to 6x as you have to double the equation of the line to make ‘y’ rather than 1/2 of y.
Lol I think this is the method but I tried doing it but the unknown k value threw me off sorry
Lol I think this is the method but I tried doing it but the unknown k value threw me off sorry
Original post by izzychloe247
use the equation of the line to find the gradient of that line (6), the gradient of this line is the negative reciprocal of the tangent to the curve at that point (as they are perpendicular) (i.e the gradient of the curve at this point is -1/6), so the derivative of k/x^2 when x=-3 is equal to -1/6 if that helps
I don’t think it’s equal to the negative reciprocal of the line because it just says there normal of the curve is parallel to the line
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