   Chapter 10.3, Problem 15E

Chapter
Section
Textbook Problem

# Finding Equations of Tangent Lines In Exercises 19-22, find an equation of the tangent line to the curve at each given point. x = 2 cot θ , y = 2 sin 2 θ , ( − 2 3 , 3 2 ) , ( 0 , 2 ) , ( 2 3 , 1 2 )

To determine

To Calculate: The equation of tangent line to the provided curve x=2cosθ, y=2sin2θ at the points (23,32),(23,12),(0,2).

Explanation

Given:

The parametric equation,

x=2cotθy=2sin2θ

And, the points (23,32),(23,12),(0,2).

Formula used:

Point-slope form,

(yy1)=m(xx1).

Calculation:

Consider the parametric equations,

x=2cotθy=2sin2θ

Differentiate x=2cotθ with respect to θ, to get,

dxdθ=2cosec2θ ...... (1)

Differentiate y=2sin2θ with respect to θ, to get,

dydθ=4sinθcosθ ...... (2)

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

dydx=dydθdxdθ=4sinθcosθ2cosec2θ=2sin3θcosθ ….. (3)

Find value of θ at (23,32).

Substitute x=23 in the parametric equation x=2cotθ.

That is.,

23=2cotθcotθ=13 …… (4)

And, substitute y=32 in y=2sin2θ.

That is.,

32=2sin2θsinθ=32 …… (5)

From equation (4) and (5),

θ=2π3

So, slope of equation at θ=2π3 is,

Slope=(dydx)θ=2π3=2sin32π3cos2π3=2(32)3(12)=(338)

Now, use point slope form, (yy1)=m(xx1).

Therefore,

Equation of tangent line at (23,32) is,

(y32)=338(x+23)2y3=334(x+23)8y12=33x+633x8y+18=0

Therefore, equation of tangent line at (23,32) is 33x8y+18=0

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