MATH 221 Lecture 19

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 9 August 2012

MATH 221 Lecture 19

The tangent line to a curve f(x) at the point (a,b) is the line through (a,b) with the same slope as f(x) at the point (a,b).

The normal line is the line through (a,b) which is perpendicular to the tangent line.

x y a b y=f(x) tangent line normal line

The slope of the tangent line (a,b) is

df dx | x=a

If a line has slope 25

x y -2 5 2 5

then the perpendicular line has slope 5-2

Example:

Find the equations of the tangent and normal to the curve y=x4-6x3 +13x2-10x+5 at the point where x=1.

The slope of the tangent line at x=1 is

df dx | x=1 = 4x3-18x2 +26x-10 | x=1 = 4-18+26-10=2

The equation of a line is y=mx+b where m is the slope. So, for our line

m=2 and 3=m·1+b=2·1+b

So b=1.So the tangent line is

y=2x+1.

The slope of the normal line is 1-2=-12.

The equation of the normal line is y=mx+c with m=-12 and 3=m·1+b.

So b=72 and y=-12+72 is the normal line.

Example:

Find the equation of the tangent and normal lines to the curve

x=acosθ, y=bsinθ at θ=π4 .

First graph this:

xa=cosθ, yb=sinθ. So xa2 + yb2=1.

x y -a a b -b π4 ( 22a , 22b )

Whenθ=π4, x=acosπ4 =22a y=bsinπ4 =22b

The slope of the tangent line is

dy dx | x=22a y=22b = dy/dθ dx/dθ | θ=π4 = dbsinθ dθ dacosθ dθ | θ=π4 = bcosθ -asinθ | θ=π4 = b22 -a22 =-ba .

So the equation of the tangent line is y=mx+y0 with m=-ba and 22=m2 2a+y0=-ba 22a+y0.

So y0=22b +22b=2b.

So the equation of the tangent line is

y=-b ax+2b.

The equation of the normal line is y=mx+y0 with m=ab and 22b=m 22a+y0= ab22a+y0

So y0=22b- a2b=22 (b2-a2b).

So the equation of the normal line is

y=abx+22 (b2-a2b)

Example:

Find the equations of the normal to 2x2-y2=14

The line x+3y=4 is the same as

y=-13x+43 .

So it has slope -13.

So the slope of the normal line is -13.

So the slope of the tangent line is 3.

So

dy dx | x=3 =3 .

Now 4x-2dydx=0. So we want 2xy=3 and 2x2-y2=14.

So 2x2-49x2=14.

So 149x2=14. So x2=9. So x=±3.

So x=3 and y=23·3=2 or x=-3 and y=23(-3)=-2.

In the first case:

The normal has slope -13 and goes through (3,2).
So m=-13 and 2=m·3+y0
So y0=3 and the equation of the normal line is

y=-13x+3.

In the second case:

The normal has slope -13 and goes through (-3,-2).

So m=-13 and -2=m(-3) +y0=(-13) (-3)+y0.

So y0=-3 and the equation of the normal line is

y=-13x-3.

The graph should explain how there can be two normal lines parallel to x+3y=4

x y 4 -3 -2 -1 1 2 3 7 -7 -2 2 43 x+3y=4 (-3,-2) (3,2) first normal line second normal line

Notes:2x2-y2=14
(a) If y=0,x=±7
(b) 2- (yx) 2 =14x2 . So, as x, this becomes 2- (yx) 2 =0 .
(yx)2, (yx)=±2 , y=±2x .

Notes and References

These are a typed copy of lecture notes given by Arun Ram on October 23, 2000.

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