Question

Friends , plse explain its answer
a point on the curve 2y^3 + x^2 = 12y at which the tangent to the curve is vertical is

Answer 1

vertical line has a slope of ....?

Answer 2

infinity

Answer 3

good.
now, tangent slope is obtained by....?

Answer 4

dy/dx

Answer 5

perfect.
so, lets find it

Answer 6

implicit diff

Answer 7

dy/dx=x/(6-3y^2)

Answer 8

yep..
now, when does this slope become infinite?

Answer 9

hint: vertical asymptote

Answer 10

@msingh well?

Answer 11

i have done wrong

Answer 12

where?

Answer 13

is this answer

Answer 14

\[
6-3y^2=0\implies y=\pm\sqrt{2}
\]
for those value "y", you have a tangent to the graph that is vertical. -> two of them

Answer 15

we have to put coeff of y = 0 when tangent is vertical

Answer 16

how? and where?

Answer 17

when we have calculate dy/dx, what after that

Answer 18

ok, then you have x^2=0...

Answer 19

but, explain your statement geometrically

Answer 20

x or x^2=0

Answer 21

so, what is your answer.?

Answer 22

nahi pata

Answer 23

what?

Answer 24

i don't know

Answer 25

Then follow the steps.. math does not lie

Answer 26

k

Answer 27

dy/dx=x/(6-3y^2)
then 6-3y^2=0
y=+-under root 2
because, for those value "y", you have a tangent to the graph that is vertical. -> two of them

Answer 28

till here ,is it correct

Answer 29

right.

Answer 30

so, you know the ordinates at which the tangent goes vertical. now use f(x,y) and find "x" values

those two (may be 4, but some outside the domain) points will give you the required points

Answer 31

x^2=4 underrot 2, when y=+underroot 2

Answer 32

x^2=-4 underrot 2, when y=-underroot 2

Answer 33

no check again.. 8sqrt(2)

Answer 34

we have to put the value of y in the given equation
2y^3+x^2=12y

Answer 35

yup u r right
8 underroot 2

Answer 36

right.. so, now, take sq-root, you'd get one +ve and one -ve,
that is your answer..
the two (x,y) points

Answer 37

(2under root 2 under root 2,under root 2)

Answer 38

@electrokid

Answer 39

iguess.. one +ve x and one -ve "x" should be symmetric/

Answer 40

infinite slope? oy vey!!

Answer 41

assuming that y is a function of x ... implicitly.

that values of x that produce a tangent that is "undefined" will indicate vertical slopes

Answer 42

2y^3 + x^2 = 12y

6y^2 y' + 2x x' = 12; dx/dx=x' = 1

6y^2 y' + 2x = 12

6y^2 y' = 12 - 2x

y' = (12 - 2x)/6y^2

Answer 43

opps, i made an error :)

2y^3 + x^2 = 12y

6y^2 y' + 2x x' = 12 y'; dx/dx=x' = 1
^^
forgot to pop out a y' :)

2x = 12 y' - 6y^2 y'

x = 6 y' - 3y^2 y'

x = (6 - 3y^2) y'

x/(6-3y^2) = y'

thats better

Answer 44

so, your derivative is correct ... thats good

so this y' becomes undefined when we divide by 0

6-3y^2 = 0, when y=??

Answer 45

is say when y=+- sqrt(2)

Answer 46

so far youve done good, then you plugged that into the original setup to find teh x values

Answer 47

2y^3 + x^2 = 12y

2sqrt(2)^3 + x^2 = 12sqrt(2)

x^2 = 12sqrt(2) - 4sqrt(2)

x^2 = 8sqrt(2)

x = +- sqrt(8sqrt(2))

x = +- 2sqrt(2sqrt(2))

looks good to me

Answer 48

@amistre64
thankk you

Answer 49

youre welcome

Answer 50

@amistre64 so, what is the difference?
about using the words "infinite" vs "undefined"?

Answer 51

just semantics :) one method ignores mathing principles, the other applies them.

\[infinite ~slope:\frac{\infty}{1}\]\[undefined~slope:~\frac{k}{0}\]

Answer 52

of course you could have undefined slopes that use other undefinable function :)

ln(0)/3 is undefined along the reals

2/(2n)rt(-6) is undefined along the reals

etc.....

Answer 53

@even semantically, you are infinitesimally closer to "y" by not there, technically, you the ratio is still defined. Also, in Algebra, you say it is undefined. In geometry, it is "usually" safe to take it to define infinity.

Answer 54

Reference-> "1,2,3.. Infinity by George Gamow"

Answer 55

Gamow is a hack!! lol

Answer 56

@amistre64 lol. and who doesnt wish to be a membor of the whacky group!

Answer 57

... bob dole, or so ive been led to spread that rumor about

Answer 58

@amistre64 lmao

Answer 59

@amistre64 Drink to that brother.