# 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.

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Here we have to find the slope of dy/dx at the point ((1/ sqrt 2), sqrt 2). We are given that x = cos t and y= 2 sin t.

Now from x = cos t , we get dx/ dt = -sin t = -y/2.

From y = 2 sin t we get dy/dt = 2 cos t = 2x

Divide dy/dt by dx/ dt

=> (dy/dt)/ (dx/ dt) = 2x / (-y/2)

=> dy/dx = -4x / y

dy/dx = -4x / y

At the point ((1 / sqrt 2), sqrt 2), dy/dx = -4x/y

=> (-4/sqrt 2) / sqrt 2

=> -4 / sqrt 2* sqrt 2

=> -4 / 2

=> -2

**Therefore the required slope is -2.**

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)**