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.
To show that two curves intersect at right angles at a certain point, we need to demonstrate that their slopes are negative reciprocals of each other at that point.
Let's start by finding the slopes of the two curves. The first curve is given by y = √2sin(x). To find its slope, we can differentiate it with respect to x:
dy/dx = d/dx(√2sin(x))
= √2 * d/dx(sin(x))
= √2 * cos(x)
So the slope of the first curve is √2cos(x).
Next, let's find the slope of the second curve. The second curve is given by y = √2cos(x). We can differentiate it using the same steps we used for the first curve:
dy/dx = d/dx(√2cos(x))
= √2 * d/dx(cos(x))
= √2 * -sin(x)
So the slope of the second curve is -√2sin(x).
Now, we can check if the slopes are negative reciprocals of each other at a certain point. Let's consider the point of intersection, which is denoted by (x, y) on both curves.
At the point of intersection, the slope of the first curve (√2cos(x)) and the slope of the second curve (-√2sin(x)) should be negative reciprocals of each other. Mathematically, this can be written as:
(√2cos(x)) * (-√2sin(x)) = -1
Simplifying this equation, we have:
-2cos(x)sin(x) = -1
Dividing both sides by -2, we get:
cos(x)sin(x) = 1/2
Now, we can make use of the trigonometric identity cos(x)sin(x) = 1/2 to find the angle x between 0 and π/2 that satisfies this equation.
Using this identity, we know that one of the possible solutions is x = π/4. However, we need to check if this value satisfies the condition of 0 < x < π/2.
So, let's substitute x = π/4 into the original equations of the curves:
For the first curve, y = √2sin(π/4) = √2(1/√2) = 1
For the second curve, y = √2cos(π/4) = √2(1/√2) = 1
Therefore, at x = π/4, both curves intersect at the point (π/4, 1).
To conclude, the curves y = √2sin(x) and y = √2cos(x) intersect at right angles at the point (π/4, 1) in the interval 0 < x < π/2.