Posted by Imad on Thursday, April 8, 2010 at 12:17am.
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.
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.
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