Asked by Imad
Show that the curves (y= √2sinx) and (y=√2cosx) intersect at right angles at a certain point with 0<x<π/2
Answers
Answered by
Reiny
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.
√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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.