calculus
posted by Imad on .
Show that the curves (y= √2sinx) and (y=√2cosx) intersect at right angles at a certain point with 0<x<π/2

let's find their intersection points
√2sinx = √2cosx
sinx/cosx = 1
tanx = 1
x = π/4 in the given interval
for y = √2sinx
dy/dx = √2cosx
for x= π/4
dy/dx = √2sinπ/4 = √2(1/√2) = 1
for y = √2cosx
dy/dx = √2sinx
at x=π/4
dy/dx = √2(1/√2) =  1
since the slopes are negative reciprocals of each other when x = π/4, the tangents are perpendicular at that point.