# Maths Question

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#1
This question has two parts. Hope someone can help!
Find the equation of the straight line that is normal to the curve y=x^2+x^(4/3) at x=1.
Determine area of region bounded by the curve, normal and the x axis.
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1 month ago
#2
what have you tried so far, where are you stuck?
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#3
(Original post by flaurie)
what have you tried so far, where are you stuck?
So I took the derivative of the curve. Evaluated at x=1 for the gradient. Got my pt (1,2) and then solved for the equation of line. Although not sure how to get from the tangent to normal.
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1 month ago
#4
(Original post by 121rp)
So I took the derivative of the curve. Evaluated at x=1 for the gradient. Got my pt (1,2) and then solved for the equation of line. Although not sure how to get from the tangent to normal.
(i.e. so if the gradient of the tangent was -7/2 then the gradient of the normal would be 2/7)
then you can sub in the (1,2) for the equation of the line 0
#5
(Original post by flaurie)
(i.e. so if the gradient of the tangent was -7/2 then the gradient of the normal would be 2/7)
then you can sub in the (1,2) for the equation of the line So that'd get me 10y=-3x+23.
For the 2nd part, would I add 2 separate definite integrals together? One for the straight line and another for the curve? Or could I combine them as normal into one integral to solve
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1 month ago
#6
(Original post by 121rp)
So that'd get me 10y=-3x+23. For the 2nd part, would I add 2 separate definite integrals together? One for the straight line and another for the curve? Or could I combine them as normal into one integral to solve
i find it easiest to split it up into an area under a curve (use integration) and a triangle (1/2*b*h) and add these together
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#7
(Original post by flaurie) i find it easiest to split it up into an area under a curve (use integration) and a triangle (1/2*b*h) and add these together
Right, that makes a lot more sense. I've followed that through and got a value of 52/7 units^2? Would that be right?
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