1. The document discusses concepts in differential calculus including envelopes, asymptotes, and finding vertical, horizontal, and oblique asymptotes of curves. 2. It provides examples and steps for finding the envelope of a family of curves, and the oblique asymptotes of an algebraic curve using the highest degree terms. 3. Key aspects covered are parametrizing lines, using partial derivatives to eliminate parameters for envelopes, and determining asymptotes by examining the limiting behavior of curves as they tend toward infinity.

1. Differential Calculus
K.Muthulakshmi M.Sc.,M.Phil.,

2. Content
• Envelope
• Asymptote

3. Finding the Envelope
• Family of curves given by F(x,y,a) = 0
• For each a the equation defines a curve
• Take the partial derivative with respect to a
• Use the equations of F and Fa to eliminate the parameter
a
• Resulting equation in x and y is the envelope

4. Parametrize Lines
• L is the length of ladder
• Parameter is angle a
• Note x and y intercepts
1sincos  aa L
y
L
x
Lyx
 aa sincos

5. Finding Envelope

6. Finding Envelope

7. Example: No intersections
• Start with given ellipse
• At each point construct the osculating circle (radius =
radius of curvature)
• Original ellipse is the envelope of this family of circles
• Neighboring ellipses are disjoint!

10. ASYMPTOTES

11. An asymptote of a curve is a line such that
the distance between the curve and the line
approaches zero as they tend to infinity

12. X
Y

13. TYPES OF ASYMPTOTES

14. ASYMPTOTES
VERTICAL
ASYMPTOTES
HORZONTAL
ASYMPTOTES
OBLIQUE
ASYMPTOTES

15. VERTICAL ASYMPTOTES
The line x = a is a vertical asymptote of the graph of the
function
y = ƒ (x) if at least one of the following statements is true:
Lim f (x) =+∞ or -∞ or Lim f (x)=+∞ or -∞
X → a+ X → a-

16. CURVE
Y
X
ASYMPTOTES

17. HORIZONTAL ASYMPTOTES
Horizontal asymptotes are horizontal lines that the graph
of the function approaches as x → ±∞. The horizontal
line y = c is a horizontal asymptote of the function
y = ƒ(x)
if Lim f(x)=c
or
Lim f(x)=c
In the first case, ƒ(x) has y = c as asymptote when x
tends to −∞, and in the second that ƒ(x) has y = c as an
asymptote as x tends to +∞
x→∞
x→ -∞

18. curve
asymptotes
Y
X

19. OBLIQUE ASYMPTOTES
When a linear asymptote is not parallel to the x- or y-axis, it is called an oblique asymptote or
slant asymptote. A function f(x) is asymptotic to the straight line y = mx + n (m ≠ 0)
if Lim [f (x) - (mx+n) ]=0
Or
Lim [f (x) - (mx+n) ]=0
X → -∞
x→∞

20. m x +b
curve
asymptotes
Y
X

21. WORKING RULETO FIND OBLIQUE
ASYMPTOTESOF AN ALGEBRAIC CURVES
1. In the highest degree terms,put x=1,y=m and obtain Φ
(m).Then in the next highest degree term again put
x=1,y=m to obtain Φ (m) and so on.
2. Put Φ (m)=0,then its roots say m1,m2…. are the slopes of
the asymptotes.
n
n-1
n

22. 3.For the non repeated roots of Φ (m)=0,find c from the relation
c=-Φ (m)/Φ (m) for each value
of m.
The required asymptotes are
y=m1x+c1
y=m2x+c2
y=m3x+c3 …………
n
n-1 n

23. 4.If Φ (m)=0 for some value of m but
Φ (m)≠0 ,then there is no asymptotes corresponding to that
value of m.
5.If Φ "'(m)=0 and also Φ (m)=0 which is the case when two of
the asymptotes are parallel, then find c from the equation
(c²/2!)Φ (m)+ cΦ (m)+Φ (m)=0
which gives two values of c.Thus there are two parallel
asymptotes corresponding to this value of m.
n
n-1
n
‘
n-1
n n-1 n-2
‘
′
″
′

24. 6.If Φ (m)=Φ (m)=Φ (m)=0,then the values of c corresponding
to this value of m are determined from the equation
(c³/3!)Φ (m)+(c²/2!)Φ (m)+cΦ’ (m)=0.
n
n-1 n-2
n n-1 n-1
″
″′
″
′

25. EXAMPLE
TO FIND ASYMPTOTESOF
ALGEBRAIC CURVE

26. Find the asymptotes of the curve
x³+3x²y-4y³-x+y+3=0?
SOLUTION:
The given curve is
x³+3x²y-4y³-x+y+3=0
To find the oblique asymptotes
Putting x=1,y=m in the third, second and first degree
terms one by one, we get

27. Φ (m)=1+3-4m³
Φ (m)=0
Φ (m)=-1+m
Now Φ (m)=0
1+3m-4m³=0
(1-m)(1+4m+4m²)=0
m=1,m=-½,m=-½
Also Φ (m)=3-12m²
and Φ (m)=-24m
3
2
1
3
3
′
3
″

28. therefore, c=-Φ (m)/Φ (m)
=-Φ (m)/Φ (m)
=-(0/(3-12m²))
When m=1, c=-(0/(3-12))=0
When m=-1/2
c=-(0/(3-3))=0
Therefore in this case, c is given by
(c²/2!)Φ (m)+cΦ (m)+Φ(m)=0
(c²/2!)(-24m)+c.0+(m-1)=0
n-1 n
′
2 3
′
3
″
2
′
1

29. (c²/2)(12)-(1/2)-1=0
c²=1/4
c=±1/2, for m=-1/2
Now the asymptotes are obtained by putting the values of m
and c in y=mx+c,
i.e. y=x+0 ,
y=(-1/2)x+1/2
and y=(-1/2)x+(-1/2)