Question

The curve y=f(x) called Tractrix has the following property" the derivative at any point x satisfies the formula:
f'(x) = -f(x)
(1-f(x)^2)^(1/2)
Consider the region under the tractrix, within the limits 0<=x<=a. Find the volume of the solid obtained by revolving this region around the x=axis in terms of the constant c=f(a). [Hint: use u-sunstitution]

Answer 1

can you rewrite the equation, is that meant to be a fraction?

Answer 2

Yes with -f(x) as the numerator

Answer 3

k

Answer 4

i solved for f(x) as if it were a regular D.E.. unfortunately its quite late here i have to sleep

Answer 5

Thanks anyways!

Answer 6

\[\frac{dy}{dx} = \frac{y}{\sqrt{1-y^2}}\]\[-\int\limits{\frac{\sqrt{1-y^2}}{y} dy} = x\]let\[u = \sqrt{1-y^2}\]\[\frac{du}{dy} = \frac{-y}{\sqrt{1-y^2}}\]\[dy = \frac{-du \sqrt{1-y^2}}{y}\]\[= \frac{-udu}{y}\] we now have
\[x = \int\limits{\frac{u^2}{1-u^2}}du \] um not sure if that sub was helpful.. im tired :/