alevels2020
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Report Thread starter 10 months ago
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for part c, I thought that you’d sub in x=-5 to dy/dx to get the gradient (which is 20) and then get the negative reciprocal because it’s a tangent. apparently the gradient for that point is just 20 and not -1/20 ? please help. Attachment 787338
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BobbJo
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but dy/dx gives the gradient of tangent, not the gradient of the normal

the gradient of the normal is -1/m where m is the gradient of the tangent
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alevels2020
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oh, I thought dy/dx gave the gradient of the curve at that point so the tangent would be the negative reciprocal. thank you!
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