curve : \(\displaystyle xy^{2} - x^{3}y = 6\)
derivative : \(\displaystyle y'=\frac{3x^{2}y -y^{2}}{2xy - x^{3}}\)
question : find the x coordinate of each point on the curve where the tangent line is vertical.
I can find derivatives and stuff but I dont know how to answer this question. you're a genius if you can figure this one out!. Thanks a lot in advanceClick to expand...
If the tangent line is vertical, then the slope heads toward infinity.
If we differentiate \(\displaystyle xy^{2}-x^{3}y=6\)
we get \(\displaystyle 2xyy'+y^{2}-x^{3}y'-3x^{2}y=0\)
If we divide by y', then the terms without a y' will tend to 0 as \(\displaystyle y'\to {\infty}\).
So, eliminate the terms without a y' and we're left with:
\(\displaystyle 2xyy'-x^{3}y'=0\)
Divide out the y' and we are left with:
\(\displaystyle 2xy-x^{3}=0\)
Solve for y:
\(\displaystyle y=\frac{x^{2}}{2}\)
Now, sub this back into the original function and solve for x.
That will be your points of vertical tangency.
You could also solve the equation for y and graph it. Then, you can see any vertical tangents.
Be careful, though, when doing so. Vertical tangents and vertical asymptotes are two different critters.
The vertical tangent to a curve occurs at a point where the slope is undefined (infinite). This can also be explained in terms of calculus when the derivative at a point is undefined. There are many ways to find these problematic points ranging from simple graph observation to advanced calculus and beyond, spanning multiple coordinate systems. The method used depends on the skill level and the mathematic application. The first step to any method is to analyze the given information and find any values that may cause an undefined slope.
Graphically
Observe the graph of the curve and look for any point where the curve arcs drastically up and down for a moment.
Note the approximate "x" coordinate at these points. Use a straight edge to verify that the tangent line points straight up and down at that point.
Test the point by plugging it into the formula (if given). If the right-hand side of the equation differs from the left-hand side (or becomes zero), then there is a vertical tangent line at that point.
Using Calculus
- Pencil
- Paper
- Graphing calculator (optional)
Take the derivative (implicitly or explicitly) of the formula with respect to x. Solve for y' (or dy/dx). Factor out the right-hand side.
Set the denominator of any fractions to zero. The values at these points correspond to vertical tangents.
Plug the point back into the original formula. If the right-hand side differs (or is zero) from the left-hand side, then a vertical tangent is confirmed.
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References
- SOS Mathematics: Vertical Tangents and Cusps
About the Author
Residing in Pontiac, Mich., Hank MacLeod began writing professionally in 2010. He writes for various websites, tutors students of all levels and has experience in open-source software development. MacLeod is pursuing a Bachelor of Science in mathematics at Oakland University.
The second option can be very time consuming; Strong algebra skills (like knowing when an equation might result in division by zero) will help you to avoid having to make a table.
Tips:
- Some curves will have more than one vertical tangent. Always make sure you have found all the values of x that make the gradient infinite. You can use graph-plotting software to check by eye for places where the gradient becomes vertical.
- For more tips on where functions might return zero, see the .
CITE THIS AS:
Stephanie Glen. "Vertical Tangent: Definition, How to Find" From StatisticsHowTo.com: Elementary Statistics for the rest of us! //www.statisticshowto.com/calculus-definitions/vertical-tangent-definition-find/
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