# Change of axis in case of Horizontal tangent

Discussion in 'Homework Help' started by zulfi100, Dec 14, 2013.

Jun 7, 2012
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2. ### studiot AAC Fanatic!

Nov 9, 2007
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It tells you the answer in 8.36, just before 8.37.

What does it say about dy/dx in 8.36?

3. ### zulfi100 Thread Starter Active Member

Jun 7, 2012
386
1
Hi,
which is stated in 8.37.

Zulfi.

Nov 9, 2007
5,005
519

5. ### zulfi100 Thread Starter Active Member

Jun 7, 2012
386
1
Hi,
Thanks for your consistency in my problem. I didnt consider it significant. It says that dy/dx does not exist at origin. Actually i didnt consider that 8.36 is related to my prob. However i have answered you, now plz explain me why you think its related to answer of my question being asked in this post.

Zulfi.

6. ### studiot AAC Fanatic!

Nov 9, 2007
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519
The whole section of your text is about calculating the length of a curve between some point where x=a to some point where x=b

On page 417 of your text it defines the length, L, to be

$L = \int_a^b {\sqrt {\left( {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right)} } dx$

It says this requires a 'smooth function of x'

On page 418 there is a simple worked example using this definition (example2).

Page 418 then goes on to discuss what happens when the function is not smooth ("vertical tangents, corners and cusps")

and in example 3 shows what happens when dy/dx does not exist because the tangent is vertical so cannot be input to the formula.

You need to understand that a vertical tangent means that dy/dx is infinite so cannot be used in a normal formula.

It shows that one way round this problem (you should check that you really cannot input any values for dy/dx into the formulae for example 2) is to use the second formula they state. That is rearranging the equation for the curve so instead of y in terms of x they use x in terms of y, for the same curve.

So we rotate the curve through 90 degrees and a vertical tangent become a horizontal one dx/dy which does exist and equals 0.
This can be input to the second formula the gives you for arc length on page 417, namely

$L = \int_c^d {\sqrt {\left( {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} \right)} } dy$

Does this help?

Don't forget that the point x=a is now the point y=c and the point x=b is now the popint y=d. That is you need new limits of integration for the same length of curve.

Last edited: Dec 16, 2013
7. ### zulfi100 Thread Starter Active Member

Jun 7, 2012
386
1
Hi,
Thanks for your answer. You mean that we have a function y=x to the power 1/3. It represents a vertical tangent. If we find dy/dx then it would be
1/(3x to the power 2/3). At origin x= 0 so it would be infinity. In order to avoid this we do a 90 degree rotation so instead of y=x to the power 1/3 , we have
x= y3 (horizontal tangent) & now we will find dx/dy which would be 3y2 so it would be zero at origin.