   Chapter 10.3, Problem 16E

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 − 3 cos θ , y = 3 + 2 sin θ , ( − 1 , 3 ) , ( 2 , 5 ) , ( 4 + 3 3 2 , 2 )

To determine

To Calculate: The equation of tangent to the provided curve x=23cosθ, y=3+2sinθ at the points (23,32),(0,2),(23,12).

Explanation

Given:

The parametric equation,

x=23cosθy=3+2sinθ

And, the points (1,3), (2,5), (4+332,2).

Formula used:

Point-slope form,

(yy1)=m(xx1).

Calculation:

Consider the parametric equations,

x=23cosθy=3+2sinθ

Differentiate x=23cosθ with respect to θ, to get,

dxdθ=3sinθ ...... (1)

Differentiate y=3+2sinθ with respect to θ, to get,

dydθ=2cosθ ...... (2)

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

dydx=dydθdxdθ=2cosθ3sinθ=23cotθ=23tanθ ...... (3)

Find value of θ at (1,3).

Substitute x=1 in the parametric equation x=23cosθ.

That is.,

1=23cosθ3=3cosθcosθ=1 …… (4)

And, substitute y=3 in y=3+2sinθ.

That is.,

3=3+2sinθsinθ=0 …… (5)

From equation (4) and (5), θ=0.

So, slope of equation at θ=0 is,

Slope=(dydx)θ=0=231tan0=20=

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

Therefore, equation of tangent line at (1,3) is.

(y3)=(x(1))y3=(x+1)x+1=0x=1

Therefore, equation of tangent at (1,3) is x=1.

Now, find value of θ at (2,5).

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

That is.,

2=23cosθcosθ=0 …… (6)

And, substitute y=5 in y=3+2sinθ.

That is.,

5=3+2sinθsinθ=1 …… (7)

From equation (6) and (7), θ=π2

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