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CaR

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What would you do to find the angle between the curves y=e^x and y=e^-x at the point of intersection?

'elp! ><"
 
withoutaface

withoutaface

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find the point of intersection e<sup>x</sup>=1/e<sup>x</sup>
e<sup>2x</sup>=1
2x=ln1
x=0

then find dy/dx for each at the point of intersection

d/dx(e<sup>x</sup>)=e<sup>x</sup>
m=e<sup>0</sup>=1

d/dx(e<sup>-x</sup>)=-e<sup>-x</sup>

m=-e<sup>0</sup>=-1

as can be seen, one gradient is the negative reciprocal of the other
:. intersect at perpendiculars(90')
 
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