Calculate the slope of the path dy/dx at the point (1/sqrt 2, sqrt 2) given x = cos t and y= 2 sin t.

To calculate the slope m =
dy/dx,

we'll have to re-write the function y, with respect
to x. we'll have to re-write the function y, with respect to
x.

y = 2sin t

x = cos t
=> t = arccos x

We'll substitute t in the expression
of y:

y = 2 sin (arccos x)

y =
2sqrt(1 - x^2)

We'll differentiate
dy/dx:

dy/dx

y = 2sin
t

x = cos t => t = arccos
x

We'll substitute t in the expression of
y:

y = 2 sin (arccos x)

y =
2sqrt(1 - x^2)

We'll differentiate
dy/dx:

dy/dx = 2(1 - x^2)'/2sqrt(1 -
x^2)

We'll simplify:

dy/dx =
-2x/sqrt(1 - x^2)

dy/dx = -2/sqrt2/sqrt(1 -
1/2)

dy/dx =
-2/sqrt2/1/sqrt2

The slope is: dy/dx =
-2

y - sqrt2 = (-2x/sqrt(1 - x^2))(x -
1/sqrt2)

y - sqrt2 = -2(x -
1/sqrt2)