   Chapter 3.3, Problem 62E

Chapter
Section
Textbook Problem

# For what values of a and b is (2, 2.5) an inflection point of the curve x 2 y + a x + b y = 0 ? What additional inflection points does the curve have?

To determine

To find:

(i) The values of a and b where the inflection point is (2,2.5) on the curve

x2y+ax+by=0

(ii) Additional inflection points of the given curve.

Explanation

1) Concept:

Inflection point is a point where the concavity of function changes from upward to downward or downward to upward, that is f"(x)=0

2) Given:

i) x2y+ax+by=0

ii) Inflection point is 2, 2.5

3) Calculation:

(i)

The inflection point 2, 2.5 which is same as 2,52 lies on the curve,

x2y+ax+by=0 (1)

Therefore,

2252+2a+52b=0

4+b52+2a=0

20+5b+4a=0

4a+5b=-20(I)

Rearranging (1),

(x2+b)y+ax=0 (1)

Differentiating (1),

x2+by'+2xy+a=0(2)

Again differentiating (2),

x2+by"+2xy'+2xy'+2y=0

Simplifying,

x2+by"+4xy'+2y=0 (3)

The inflection point 2, 2.5=2,52 lies on the curve x2y+ax+by=0, then y"=0

Therefore,

4xy'+2y=0(4)

Since the inflection point 2, 2.5=2,52 lies on the curve x2y+ax+by=0

Substitute x=2, y=52 in 4 and (2)

In equation (4),

4(2)y'+252=0

8y'+5=0

y'=-58

In equation (2),

4+by'+2(2)52+a=0

4+by'+10+a=0

Substitute y'=-58

4+b-58+10+a=0

-584+b+10+a=0

-54+b+80+8a=0

-20-5b+80+8a=0

-5b+60+8a=0

8a-5b=-60(II)

4a+5b=-20(I)

$\underset{_}{8a-5b=-60\dots \left(II\right)}$

12a+0=-80

a=-8012

Factoring 80=4·20 and 12=4·3

=-4·204·3

Cancelling out common factor,

a=-203

Substitute a=-203 in (I)

4-203+5b=-20

-80+15b=-60

15b=-60+80=20

15b=20

b=2015

Factoring 20=5·4 and 15=5·3

b=2015=5·45·3

Cancelling out common factor,

b=43

Therefore,a=-203 and b=43

Substitute a=-203 and b=43 in the given equation,

x2y+ax+by=0

x2y+-203x+43y=0

Rearranging,

x2+43y=203x is the given curve

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