can someone poke a hole in this?

• March 26th 2009, 01:20 PM
helothar
can someone poke a hole in this?
I take AP Calculus AB at my high school and I think I have found a flaw in my textbook. In a section about the length of curves, there is this problem asking the reader to find a curve based on a given integral and the part b of the problem asks how many curves are possible. The back of the book says that there is only one curve possible, but I think I have found another.
The problem is
Find a curve through the point (1,1) whose length integral is
$L= \int_{1}^{4} \sqrt{1 + \frac {1}{4x}}dx$

so following the formula for the length of a curve, (dy/dx)^2=(1/4x) so (dy/dx) = + or - $\frac {1}{2\sqrt{x}}$. based on those dy/dx's, y is equal to either the square root of x plus a constant or the negative square root of x plus a constant, depending on whether you use the positive dy/dx or the negative one. After solving for constants I find two solutions that work:
$y= \sqrt{x}
$
and $y= -\sqrt{x} + 2$

but my text book only thinks the first one is an acceptable answer apparently. can someone help me out?
• March 26th 2009, 06:00 PM
skeeter
Quote:

Originally Posted by helothar
I take AP Calculus AB at my high school and I think I have found a flaw in my textbook. In a section about the length of curves, there is this problem asking the reader to find a curve based on a given integral and the part b of the problem asks how many curves are possible. The back of the book says that there is only one curve possible, but I think I have found another.
The problem is
Find a curve through the point (1,1) whose length integral is
$L= \int_{1}^{4} \sqrt{1 + \frac {1}{4x}}dx$

so following the formula for the length of a curve, (dy/dx)^2=(1/4x) so (dy/dx) = + or - $\frac {1}{2\sqrt{x}}$. based on those dy/dx's, y is equal to either the square root of x plus a constant or the negative square root of x plus a constant, depending on whether you use the positive dy/dx or the negative one. After solving for constants I find two solutions that work:
$y= \sqrt{x}
$
and $y= -\sqrt{x} + 2$

but my text book only thinks the first one is an acceptable answer apparently. can someone help me out?

textbook answers have been wrong on many occasions.
• March 26th 2009, 07:19 PM
helothar
yeah i realize that i guess i was just wondering if i was missing something