## Required: y-coordinate of point(s) of tangency

Limits, differentiation, related rates, integration, trig integrals, etc.
jaybird0827
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### Required: y-coordinate of point(s) of tangency

Here is the problem:

Find the y-coordinate of the points on the graph of
$y\,=\,ln\,\|x|^3$
from which the tangents to the curve pass through the origin.

Attempted solution:

Drew the graph of
$y\,=\,ln\,\|x|^3$
and sketched two lines, each containing the origin and intersecting the graph in one point.

$y\,=\,ln\,\|x|^3$ ... given

$y\,=\,ln\,\|x^3|$ ... equivalent

$y\,=\,ln\,\|u|$ ... u-substitution

$dy\,=\,\frac{du}{u}$ ... differentiated

$u\,=\,x^3$ ... u-function

$du\,=\,3x^2\,dx$ ... differentiated

$\frac{dy}{dx}\,=\,\frac{3x^2}{x^3}\,=\,\frac{3}{x}$ ... chain rule

How do I find the point that I'm looking for on the graph,

$y\,=\,ln\,\|x|^3?$

I think I'm looking for the equation of each line that contains (0,0) and a point of tangency.
According to my sketch, there seems to be two such points, one in Quadrant I and one in Quadrant IV.

The answer in the back of the textbook is $3.$
Last edited by jaybird0827 on Fri Jun 19, 2009 9:08 pm, edited 1 time in total.

stapel_eliz
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If you plug their solution in, you can see that it doesn't work. When x = 3, then the slope between the point on the curve and the origin is not equal to the derivative of the curve at that point. So the book's answer would appear to be in error.

You have the derivative, 3/x (or 3/|x|, really, because the absolute-value bars in the original function mean that you have two curves, symmetric about the y-axis). At some point (x, ln|x3) on the curve, the slope of the line through that point and the origin is equal to the slope of the tangent; that is:

. . . . .$\frac{\ln(x^3)}{x}\, =\, \frac{3}{x}$

I think this is where the book's answer arose: from not completing the solution. The above says that ln|x3| = 3, so x3 = e3, and then x = e, and the slope 3/x would equal 3/e. Checking:

. . . . .$\frac{\ln|e^3|}{e}\, =\, \frac{3}{e}$

So I think this was what your book intended.

jaybird0827
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### Re:

stapel_eliz wrote:If you plug their solution in, you can see that it doesn't work. When x = 3, then the slope between the point on the curve and the origin is not equal to the derivative of the curve at that point. So the book's answer would appear to be in error.

Too often we have found that to be the case. Growl.

stapel_eliz wrote:You have the derivative, 3/x (or 3/|x|, really, because the absolute-value bars in the original function mean that you have two curves, symmetric about the y-axis). At some point (x, ln|x3) on the curve, the slope of the line through that point and the origin is equal to the slope of the tangent; that is:

. . . . .$\frac{\ln(x^3)}{x}\, =\, \frac{3}{x}$

D'oh! The fact that the graph has two curves occurred to me later, would you believe, during one of my usual waking periods in the wee hours. That makes sense.

So, we agree that I should have come up with

. . . . .$\frac{d}{dx}\,(ln\,|x^3|)\,=\,\frac{3}{|x|}$

stapel_eliz wrote:I think this is where the book's answer arose: from not completing the solution. The above says that ln|x3| = 3, so x3 = e3, and then x = e, and the slope 3/x would equal 3/e. Checking:

. . . . .$\frac{\ln|e^3|}{e}\, =\, \frac{3}{e}$

So I think this was what your book intended.

That helped. Your explanation is very clear. Thank you so much!
Oh, and thanks also for the free lesson on indenting LaTex formatting from the left margin. Elegant! I will add it to my repertoire.