Question

Suppose y is implicitly defined a function of x+y^3=16y
(a) Find all points on the curve with x-coordinate 0
I found them all (0,-4) (0,0) and (0,4)
(b) Find the slope of the curve at each point in part(a) ??
There are three values
(d) Find the points on the curve where the slope becomes vertical (vertical tangent line ) there are two points ?

Answer 1

(a) How did you find those? Please demonstrate.

Answer 2

x = y^3 - 16y

factorizing,

x = y(y^2 - 16)

x = y(y - 4)(y + 4)

The solutions:

y = 0, y = 4 and y = -4

Answer 3

Okay, there is some notational difficulty in there. Factoring to find solutions is not very meaningful unless the other side of the equation is zero. Actually substituting x = 0 in the beginning would have been a little better form. Also, you should have gotten -x = y^3 - 16x. Anyway, you lucked out and got it because -0 happens the be the same as +0.

Okay, now we need the derivative, dy/dx. Start from the original equation and let's see it.

Answer 4

x' = (3y^2)y' - 16y'
x' = y'((3y^2) - 16 )
y' = (x') / ( (3y^2) -16 )

Answer 5

Two things wrong with that.

x+y^3=16y ==> y^3 - 16y = -x

You're still not getting that -x. Be more careful.

You're also hanging onto that 'x' for dear life. If f(x) = x, what is f'(x)?

Answer 6

-x' = (3y^2)y' -16y'
x'= -(3y^2) y' + 16y'

Answer 7

Okay, now answer my question. If f(x) = x, what is f'(x)?

Answer 8

You mean I should write it like this?
f'(x) = (--3y^2) y' + 16y'

Answer 9

No. Look at it carefully. There is no f(x) in your problem statement. I invented f(x) to help you see something. Forget about your problem for a moment and work on this one...

If f(x) = x, what is f'(x)?

Answer 10

f'(x) = 1 ?

Answer 11

Perfect. Back to your problem. What is another expression for x'?

Answer 12

In words, "x'" is "the derivative of x with respect to x". Think about my f(x)? There is a MUCH simple way to express x'.

Answer 13

1= -(3y^2) y' + 16y'

Answer 14

THAT'S it!!! Next time you are tempted to write x', just write 1.

Now, solving for y', which you had almost correct a while back: \(y' = \dfrac{1}{16-3y^{2}}\).

Agreed?

Answer 15

Yees! Now we should equal y' to 1 and find y
Now I have a class , I will come later

Answer 16

Well, just remember the three points for part b).
For part c), just set that denominator to zero and solve. You'll be done in no time.

Answer 17

thank you I answered part (b correctly , but why should I set the denominator to zero ? ( they said vertical tangent line)

Answer 18

The slope is vertical. In algebra 1 parlance, that's "No Slope". This means a singularity in the derivative, since slope is defined as dy/dx. If y changes arbitrarily with NO change in x, that is what you seek.

When we do minimization (or maximization) problems, we usually look for dy/dx = 0. This is an horizontal line. To find a vertical line, we need dy/dx does not exist.

Answer 19

I didn't get it :(
can you explain more or write the answer with explanation, please ?

Answer 20

@bella__1 : Please be specific about what you want to know. You've posted a 3-part question; which part do you mean when you say, "I didn't get it"? Perhaps it's time for you to step back and assess what you do understand and what you don't, before you request more explanations after having received many explanations.

Answer 21

I didn't understand part (c) , I tried a lot but it always shows that the answer is wrong !

Answer 22

Please type in at least some of the work you have done on Part C. Then i could give you more meaningful feedback.

Answer 23

I set that denominator to one and I got y= \[\pm \sqrt{5}\] and I put x = 1 but here in the question they want two different x values ! I also tried to draw the function and I got approximately (24.634, 2.31) this is the largest x value :( , you can see the attachment above.

Answer 24

One? "1"? Why would you do that? There is no singularity with 1 in the denominator. You need ZERO.

\(16-3y^{2} = 0\) -- Solve for \(y\).

Answer 25

ohh sorry , now \[y =\frac{ 4\sqrt{3} }{ 3 }\] and \[y = -\frac{ 4\sqrt{3} }{ 3}\]
how can I find the x values ? from the origin function \[x + y^3 = 16y \] ?
x= 24.6336
x= -24.6336

Any thing wrong ?