# Thread: How to solve this problem

1. ## How to solve this problem

If the derivative dx/dy exists at a point and dx/dy = 0, then the tangent line is vertical. Use this information and the given equation to answer the following questions-
y4 + 5 = y2 + x2

1) Find dx/dy

Ans:
y4 + 5 = y2 + x2
4y3 =2 y + 2x dx/dy
(4y3 -2 y)/ 2x = dx/dy

2) Find the points on the graph where the tangent line is vertical. Arrange the values from smallest to largest.

(P.S.-please show the step so I will understand)

2. Originally Posted by skboss
If the derivative dx/dy exists at a point and dx/dy = 0, then the tangent line is vertical. Use this information and the given equation to answer the following questions-
y4 + 5 = y2 + x2

1) Find dx/dy

Ans:
y4 + 5 = y2 + x2
4y3 =2 y + 2x dx/dy
(4y3 -2 y)/ 2x = dx/dy

2) Find the points on the graph where the tangent line is vertical. Arrange the values from smallest to largest.

(P.S.-please show the step so I will understand)
If the derivative is zero, than the numerator of that fraction needs to be zero (and also $x\neq 0$).

So you need to solve:

$4y^3-2y=0$

Then plug the points back into the function to find the $x$ coordinates.

3. Originally Posted by redsoxfan325
If the derivative is zero, than the numerator of that fraction needs to be zero (and also $x\neq 0$).

So you need to solve:

$4y^3-2y=0$

Then plug the points back into the function to find the $x$ coordinates.

Thanks, that gave me two of the points which are (+-2 and 0) But there are 4 other values. How can I find those?

4. Originally Posted by skboss
Thanks, that gave me two of the points which are (+-2 and 0) But there are 4 other values. How can I find those?
Well the three points here are $y=\pm\frac{\sqrt{2}}{2}$ and $y=0$. Subbing $y=0$ into the first equation gives $x=\pm\sqrt{5}$.

Subbing in $y=\pm\frac{\sqrt{2}}{2}$ gives $x=\pm\frac{\sqrt{19}}{2}$

So the points are:

$\left\{\left(\pm\frac{\sqrt{19}}{2},\pm\frac{\sqrt {2}}{2}\right),\left(\pm\sqrt{5},0\right)\right\}$

for a total of six points. Is that what you meant by four more?