Thread: 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.

How do I do this one? Some one please help me with this.
(P.S.-please show the step so I will understand)

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.

How do I do this one? Some one please help me with this.
(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 ).

So you need to solve:



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

Originally Posted by redsoxfan325

If the derivative is zero, than the numerator of that fraction needs to be zero (and also ).

So you need to solve:



Then plug the points back into the function to find the 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?

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 and . Subbing into the first equation gives .

Subbing in gives

So the points are:



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