Change of axis in case of Horizontal tangent

studiot

Joined Nov 9, 2007
4,998
Can somebody plz guide me why x & y axis are inter-changed in case of Horizontal tangent in the following link in Fig 8.37?
It tells you the answer in 8.36, just before 8.37.

What does it say about dy/dx in 8.36?
 

Thread Starter

zulfi100

Joined Jun 7, 2012
656
Hi,
Thanks for your attention. Plz explain me how you say that answer is in 8.36. In my view answer is in the statement:
plotted with x as a function of y
which is stated in 8.37.
Plz guide me about your understanding.

Zulfi.
 

Thread Starter

zulfi100

Joined Jun 7, 2012
656
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.
 

studiot

Joined Nov 9, 2007
4,998
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:

Thread Starter

zulfi100

Joined Jun 7, 2012
656
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.

Thanks again for your help.

Zulfi.
 
Top