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belka [17]
3 years ago
11

Find dy/dx and d2y/dx2, and find the slope and concavity (if possible) at the given value of the parameter. (if an answer does n

ot exist, enter dne.) parametric equations point x = 6 cos θ, y = 6 sin θ θ = π 4
Mathematics
1 answer:
Reptile [31]3 years ago
6 0

We are given the parametric equations:

x = 6 cos θ

y = 6 sin θ

We know that the derivative of cos a = - sin a and the derivative of sin a = cos a, therefore taking the 1st and 2nd derivates of x and y:

d x = 6 (-sin θ) = - 6 sin θ

d^2 x = -6 (cos θ) = - 6 cos θ

d y = 6 (cos θ) = 6 cos θ

d^2 y = 6 (-sin θ) = - 6 sin θ

 

Therefore the values we are asked to find are:

dy / dx = 6 cos θ / - 6 sin θ = - cos θ / sin θ = - cot θ

d^2 y / d^2 x = - 6 sin θ / - 6 cos θ = sin θ / tan θ = tan θ

 

We can find the value of the slope at θ = π/4 by using the dy/dx:

dy/dx = slope = - cot θ

dy/dx = - cot (π/4) = - 1 / tan (π/4)

dy/dx = -1 = slope

 

We can find the concavity at θ = π/4 by using the d^2 y/d^2 x:

d^2 y / d^2 x = tan θ

d^2 y / d^2 x = tan (π/4)

d^2 y / d^2 x = 1

Since the value of the 2nd derivative is positive, hence the concavity is going up or the function is concaved upward.

 

Summary of Answers:

dy/dx = - cot θ

d^2 y/d^2 x = tan θ

slope = -1

concaved upward

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