Question

I have to find the horizontal tangent and vertical tangent for the following function

Answer 1

\[x^3+y^3=4xy+1\]

Answer 2

for the horizontal tangent, I know I need to take the derivative and solve for 0

Answer 3

Not sure the approach for the vertical tangent

Answer 4

By horizontal and vertical, do we mean tangent and normal?

Answer 5

For what values of x will the curve x^3+y^3=4xy+1 have a horizontal tangent?

Answer 6

ok I thought I had the right approach, not sure anymore

Answer 7

horizontal tangent : dy/dx = 0
vertical tangent : dx/dy = 0

Answer 8

\[\frac{ dy }{ dx }=\frac{ 4xy-3x^2 }{ 3y^2-1 }\]

Answer 9

how do I solve this one for 0?

Answer 10

\[4xy-3x^2=0\]

Answer 11

Careful, if the denominator is also zero, you may not have what you think.

Answer 12

by horizontal tangent switch the equation

Answer 13

do you mean for the vertical tangent switch the equations?
\[\frac{ dx }{ dy }=\frac{ 3y^2-1 }{ 4xy-3x^2 }\]
I can solve this one,
\[3y^2-1=0\]
\[y=\pm \sqrt{\frac{ 1 }{ 3 }}\]

Answer 14

so you got the y co-ordinate.
get the x co-ordinate from the original equation.
because vertical lines have x value constant (hence their equation is x=c)

Answer 15

so do I sub this y value into the original to find the x values

Answer 16

yes

Answer 17

you might want to double check your derivative

Answer 18

ok still don't know what to do with my horizontal tangent
ok I will double check it

Answer 19

oh I dropped something, I will redo it

Answer 20

you're not given any point, at which the tangents are to be found ?

Answer 21

\[\frac{ dy }{ dx }=\frac{ 4xy-3x^2 }{ 3y^2-4x}\]

Answer 22

ok this is the correct derivative

Answer 23

No, this is what I was given.....
For what values of x will the curve \[x^3+y^3=4xy+1\] have a horizontal tangent? Show your work and explain your thinking.

Answer 24

My second part,
In terms of y, describe the values of x for which the curve \[x^3+y^3=4xy+1\] will have a vertical tangent? Show your work and explain your thinking.

Answer 25

so for this part (the second part) if I solve for y I get:
\[y=\pm \sqrt{\frac{ 4x }{ 3 }}\]

Answer 26

what values of x in terms of y
so,
x= 3y^2/4

Answer 27

are you sure that derivative is correct

Answer 28

omg am I that dumb???? ok I will triple check it.......my brain is fried

Answer 29

ok here is my last attempt......\[\frac{ dy }{ dx }=\frac{ 4y-3x^2 }{ 3y^2-4x }\]

Answer 30

you're learning, silly mistakes will happen

Answer 31

that is correct

Answer 32

ok did I get it correct this time? No, I am just tired (been doing math all day long) Maybe I have reached my limit

Answer 33

haha calculus joke

Answer 34

ok my second part is ok is that what they meant for the second part

Answer 35

1st part am I solving for the solution in terms of x?

Answer 36

so is the first part
\[x=\pm \sqrt{\frac{ 4y }{ 3 }}\]

Answer 37

not sure anymore......

Answer 38

plug that into original equation

Answer 39

Is that what I need to do for both of them?

Answer 40

you will get equation only in 'x'
solving that equation will give you x values for which you have horizontal tangent

Answer 41

for 2nd part, you CAN do the same thing, but it is not asked for
"In terms of y"
x =3y^2/4 is good enough

Answer 42

ok thanks, I am not sure this is what I need to do....I don't recall these equations being sooooooo bad or so complex.....