   Chapter 10.3, Problem 14E

Chapter
Section
Textbook Problem

# Finding Slope and Concavity In Exercises 9-18, find d y / d x and d 2 y / d x 2 , and find the slope and concavity (if possible) at the given value of the parameter.Parametric Equations Parameter x = θ − sin θ , y = 1 − cos θ θ = π

To determine

To calculate: The value of dydx, d2ydx2, and slope and concavity of parametric equations

x=θsinθ,y=1cosθ, at Parameter θ=π.

Explanation

Given:

The parametric equations,

x=θsinθy=1cosθ

And, parameter θ=π.

Formula used:

If a smooth curve is given by the parametric equations x(t) and y(t), then the slope of the curve is,

dydx=dy/dtdx/dt, dxdt0

Calculation:

Consider the equations,

x=θsinθy=1cosθ

Differentiate x=θsinθ with respect to t, to get,

dxdθ=1cosθ ….. (1)

Differentiate y=1cosθ with respect to t, to get,

dydθ=(sinθ)=sinθ …… (2)

Divide equation (1) by (2), to get,

dydx=dydθdxdθ=sinθ1cosθ ….. (3)

Now, in order to find slope at θ=π, substitute θ=π in equation (3).

That is.,

(dydx)θ=π=sinπ1cosπ=01(1)=0

Therefore, at θ=π, slope is 0

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