Horizontal tangent line to the curve

[Mooch22]

The curve is defined by 2(y^3)+6(x^2)y-12(x^2)+6y = 1

a.) Show that (dy/dx) = (4x-2xy)/(x^2+y^2+1)

b.) Write an equation of each horizontal tangent line to the curve.

c.) The line through the origin with slope -1 is tangent to the curve at point P. Find the x- and y-coordinates of point P.

I think I'm ok with part A, but part B and C.... I haven't the slightest idea! Some help, please? It would be greatly appreciated!

 

[stapel]

a) They're asking you to differentiate implicitly, and then isolate "dy/dx". (In essense, they've told you "differentiate, and, by the way, here's the answer.)

b) Plug zero in for "dy/dx", and recall that a fraction is zero when its numerator is zero. Factor, and solve for the x-value and y-value for which dy/dx is zero. Plug the x-value into the original equation and solve for any corresponding y-value(s). Do the same with the y-value.

Now you have the x,y-points at which the tangent lines are horizontal. An horizontal line is of the form "x = a" for some number "a". Use this fact to write the equations of the tangent lines.

c) If the line is tangent to the curve, then that point on the curve has a slope of -1. So plug "-1" in for "dy/dx", and solve as above.

If you get stuck, please reply showing what steps you have tried. Thank you.

Eliz.

 

[soroban]

Hello, Mooch22!

Is there a typo in the problem?
I get equations that are unsolvable and/or baffling.

The curve is defined by: $\displaystyle 2y^3\,+\,6x^2y\,-\,12x^2\,+\,6y\;=\;1$

a) Show that: $\displaystyle \frac{dy}{dx}\:=\:\frac{4x\,-\,2xy}{x^2\,+\,y^2\,+\,1}$

b) Write an equation of each horizontal tangent line to the curve.

c) The line through the origin with slope -1 is tangent to the curve at point $\displaystyle P$.
Find the $\displaystyle x$- and $\displaystyle y$-coordinates of point $\displaystyle P$.

(b) As Eliz pointed out, that derivative equals zero when its numerator equals zero.

So we have: .$\displaystyle 4x\,-\,2xy\:=\:0\;\;\Rightarrow\;\;2x(2\,-\,y)\:=\:0$

. . Hence, there are horizontal tangents when $\displaystyle x=0$ or $\displaystyle y = 2$.


If $\displaystyle x = 0$, there is a horizontal tangent at the $\displaystyle y$-intercept(s).

. . . . . $\displaystyle 2y^3\,+\,6\cdot0^2y\,-\,12\cdot0^2\,+\,6y\:=\:1\;\;\Rightarrow\;\;2y^3\,+\,6y\,-\,1\:=\:0$

There could be up to three y-intercepts, but I can't solve for a single one.
. . This cubic equation has no rational roots.


If $\displaystyle y = 2$, we seem to have found a horizontal tangent.
. . Just curious, I tried to locate the point(s) of tangency.

Let $\displaystyle y=2$ and we have: .$\displaystyle 2\cdot2^3\,+\,6x^2\cdot2\,-\,12x^2\,+\,6\cdot2\:=\:1\;\;\Rightarrow\;\;28\,=\,1$ ??

Evidently, the curve never has a $\displaystyle y$-coordinate of 2.
. . . What's going on?


(c) The line through the origin with slope -1 is: $\displaystyle y\,=\,-x$

We know two things:
. . [1] $\displaystyle y\,=\,-x$ intersects the graph at least once.
. . [2] The slope at $\displaystyle P$ is -1.

[1] Substitute $\displaystyle y=-x$ into the equation: .$\displaystyle 2(-x)^3\,+\,6x^2(-x)\,-\,12x^2\,+\,6(-x)\:=\:1$
. . . and we get: .$\displaystyle 8x^3\,+\,12x^2\,+\,6x\,+\,1\:=\:0$ . . . another unsolvable cubic.

[2] Since the derivative equals -1 and $\displaystyle y = -x$, we have: .$\displaystyle \frac{4x\,-\,2x(-x)}{x^2\,+\,(-x)^2\,+\,1}\:=\:-1$

. . . This simplifies to: .$\displaystyle 4x^2\,+\,4x\,+\,1\:=\:0\;\;\Rightarrow\;\;(2x + 1)^2\:=\:0\;\;\Rightarrow\;\;x\,=\,-\frac{1}{2}$
. . . (Hey, finally got something reasonable!)

. . . Find the $\displaystyle y$-coordinate: .$\displaystyle y^3\,+\,6\left(-\frac{1}{2}\right)^2y\,-\,12\left(-\frac{1}{2}\right)^2\,+\,6y\:=\:1\;\;\Rightarrow\;\;4y^3\,+\,15y\,-\,8\:=\:0$

. . . Glory be! . . . I found a root! . . . $\displaystyle y\,=\,\frac{1}{2}$

Therefore, the coordinates of $\displaystyle P$ are: .$\displaystyle \left(-\frac{1}{2},\,\frac{1}{2}\right)$


. . . a most baffling, frustrating, and unsatisfying problem.

 

[Mooch22]

THANK YOU!

THANKS GUYS... YA'LL ARE GREAT!