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April 21, 2014

April 21, 2014

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

- calculus -
**Reiny**, Thursday, April 8, 2010 at 12:26amlet'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.

- calculus -
**Reiny**, Thursday, April 8, 2010 at 12:27amlet'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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