   Chapter 10.3, Problem 11E

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 = 2 + sec θ , y = 1 + 2 tan θ θ = − π 3

To determine

To calculate: The value of dydx, d2ydx2 slope and concavity of parametric equations x=2+secθ,y=1+2tanθ, At parameter θ=π3.

Explanation

Given:

The parametric equations are,

x=2+secθy=1+2tanθ

And, parameter is θ=π3.

Formula used:

Differentiation of some standard functions are

d(constant)dx=0, d(secθ)dθ=secθtanθ, d(tanθ)dθ=2sec2θ

Chain rule:

dudv=dudx×dxdy............×dzdv

Calculation:

Consider the equations,

x=2+secθy=1+2tanθ

Differentiate the equation x=2+secθ with respect to θ, to obtain,

dxdθ=secθtanθ ….. (1)

Differentiate the equation y=1+2tanθ with respect to θ, to obtain,

dydθ=2sec2θ …… (2)

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

dydx=2sec2θsecθtanθ=2cosecθ ….. (3)

For the slope.

Put θ=π3 in equation (3).

That is,

(dydx)θ=π3=2cosec(π3)=433

Therefore,

At θ=π3, slope is 433

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