   Chapter 2.R, Problem 50E

Chapter
Section
Textbook Problem

# 49–50 Find an equation of the tangent line and normal line to the curve at the given point. x 2 + 4 x y + y 2 = 13 , ( 2 , 1 )

To determine

To find: Equation of tangent line and normal line to the given curve at given point.

Explanation

Differentiate the given curve and then plug the given point in it to get slope of tangent line. Then use point slope form of line to get equation of tangent line. Then calculate the negative reciprocal of slope of tangent line. It is slope of normal line. Again with this new slope find the equation of normal line using the same slope point form.

1) Formula:

i. Power rule combined with chain rule:

ddxf(x)n=n.fxn-1.f'(x)

ii. Sum rule:

ddxfx+gx=ddxfx+ddxg(x)

iii. Constant multiple rule:

ddxk.fx=k.ddxfx where k is constant.

iv.

ddxsinx=cosx

v. Constant function rule:

ddxk=0 where k is constant.

vi. Normal line and tangent lines are perpendicular to each other.

vii. Slopes of perpendicular lines are negative reciprocal of each other.

viii. Equation of line passing through point (a,f(a)) with slope m is, (y-f(a))=m(x-a)

2) Given:

x2+4xy+y2=13,      (2 ,1)

3) Calculation:

Differentiate given equation with respect to x,

ddx(x2+4xy+y2)=ddx13

Using power rule, product rule and constant function rule,

2x+4.xddxy+y.ddxx+2y.dydx=0

Using power rule,

2x+4xdydx+4y+2y

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