Question

The curve defined by x^3+xy-y^2=10 has a vertical tangent line when x=?

How does one go about solving this here problem?

Answer 1

implicit differentiation, solve for dy/dx = infinity (or the denominator is zero.)

Answer 2

3x^2+x(dy/dx)+y=2y(dy/dx)=10
?

Answer 3

* -2y(dy/dx)

Answer 4

good, except it should equal zero

Answer 5

right

Answer 6

so -( 3x^2 +y) /(x-2y)

Answer 7

wait a minute, how did you get the divided?

Answer 8

(Can also find dx/dy and set equal to 0.)

Answer 9

nvm, I get it, what do you do after that?

Answer 10

\[x ^{3}+xy -y ^{2}=10\]
Now implicitly differentiate to get\[3x ^{2}+y +x \frac{ dy }{ dx }+2y \frac{ dy }{ dx }=0\]Now solve this equation for dy/dx. Thus\[\frac{ dy }{ dx }=\frac{ -3x ^{2}-y }{ x +2y }\]For a vertical tangent, the derivative is undefined. This implies that\[x +2y =0\]
\[x =-2y\]
Substitute x into the given equation.
\[(-2y)^{2}+(-2y)(y)-y ^{2}=10\]
\[4y ^{2}-2y ^{2}-y ^{2}=10\]
\[y ^{2}=10\]
\[y =\pm \sqrt{10}\]
Therefore the vertical tangents are at \[x =\pm \sqrt{10}\]

Answer 11

I think you're a bit off @calculusfunctions

Answer 12

The given eauation is x^3, not 3x^2

Answer 13

that and x=2y

Answer 14

not x^2, rather

Answer 15

I know that! Look at my work. I said that the derivative is 3x²

Answer 16

Oh yes that is true! That was a careless mistake. The correct answer is \[x =\pm2\sqrt{10}\]

Answer 17

how would you get that answer if x is cubes,a nd cannot simplify with the y^2?

Answer 18

would you not use the quadratic formula?

Answer 19

@calculusfunctions,

my good man, that is not the correct answer.

Answer 20

web assign ? or solution key?

Answer 21

Worksheet,
A. 0 or -1/3
B. 1.037
C. 2.074
D. 2.096
E. 2.154

Answer 22

your teacher is a d***
it's 2.074

Answer 23

Many apologies, but I do not know how you would come to that answer. Would you care to explain?

Answer 24

just what he did, but correctly.

Answer 25

(haha) Yes, but I came up with -1/4

would you then use that to solve for x?

Answer 26

rather, -1/8

Answer 27

no. y is 1.0371
and x is 2y

Answer 28

how did you find y...

Answer 29

plug in y=2x (it's a cubic)
find the root.

Answer 30

wouldn't y=x/2?

Answer 31

whoops x=2y

plug in x=2y (it's a cubic)
find the root.

Answer 32

either way though should work.

Answer 33

8y^3+y^2=10, right?

Answer 34

yep

Answer 35

would you then use the quadratic formula?

Answer 36

...

Answer 37

it's not a quadratic

Answer 38

you use the cubatic formula aka Wolfram

Answer 39

I was solving it this way and it occurred to me there must be some simpler way your teacher wants you to do the problem... but I haven't figured that bit out yet.

Answer 40

OK! I see my mistake. So give me a few minutes to type it up again.

Answer 41

\[x ^{3}+xy -y ^{2}=10\]
\[3x ^{2}+y +x \frac{ dy }{ dx }-2y \frac{ dy }{ dx }=0\]
\[\frac{ dy }{ dx }=\frac{ 3x ^{2}+y }{ 2y -x }\]Now \[2y -x =0\]
\[x =2y\]
Substitute x = 2y into the given equation.
\[(2y)^{3}+(2y)(y)-y ^{2}=10\]
\[4y ^{2}+2y ^{2}-y ^{2}=10\]
\[5y ^{2}=10\]
\[y ^{2}=2\]
\[y =\pm \sqrt{2}\]
\[ x =\pm2\sqrt{2}\]

Answer 42

...

Answer 43

it should be 2.07424

Answer 44

you're 0-2 tonight @calculusfunctions

Answer 45

Thanks algebraic, I looked at the directions and I could use my Graphing calc, so I just did that. Thanks again, to both of you. :)

Answer 46

ah

Answer 47

that's why it was a goofy answer.

Answer 48

:3 sorry

Answer 49

np:)

Answer 50

@Algebraic! How is it 2.07424? Why don't you offer your solution?

Answer 51

lol

Answer 52

Um, y should be 1.0371208412063, and x=2y, sooo...

x=2.07424

Answer 53

Turns out you can't do this just by hand... ?

Answer 54

Again, my good man, it is x^3, not x^2

Answer 55

@Jack, that is x³, but when you type it using the equation feature at the bottom, it looks like x². If you don't believe me, why don't you try doing x³ yourself, using the equation feature at the bottom left.

Answer 56

4y2+2y2−y2=10

5y2=10

y2=2

y=±2√

x=±22√


this is what you typed, copy paste, good sir.

Answer 57

(2y)3+(2y)(y)−y2=10

4y2+2y2−y2=10

5y2=10

y2=2

y=±2√

x=±22√

This is all of it, so yes, I see there WAS an x^3, but you changed it to x^2

Answer 58

And I did get x = 2y. My good man, you need to look again to see that I have x = 2y the second time I typed the solution.

Answer 59

Of course it changes to 3x squared when you take the derivative of x cubed.

Answer 60

My second solution is therefore perfect. I'm sorry you don't see that.

Answer 61

Many apologies, but your answer may still be incorrect, my good man.

Answer 62

THIS is the second solution you allowed me:

Substitute x = 2y into the given equation.
(2y)3+(2y)(y)−y2=10

4y2+2y2−y2=10

5y2=10

y2=2

y=±2√

x=±22√

Answer 63

A simply typo, which resulted in an incorrect calculation.

Answer 64

Ah I see when I plugged x = 2y back into the original equation, I did (2y)²... instead of (2y)³. In that case you are right. I apologize.

Answer 65

No apologies are required, my good man. :)

Answer 66

Thanks, I appreciate that but I was wrong nonetheless.

Answer 67

Be more careful typing next time.

Answer 68

Good man! This was a good challenge, nonetheless. Quite fun, was it not? Lessons learned: always check work, and that lesson applies to the best of us, does it not?

Answer 69

Thank you for your time and effort, it is greatly appreciated.

Answer 70

Of course. I agree with everything you said. Thank you!