Of the infinitely many lines that are tangent to the curve of y = -sin x and pass through the origin, there is one that has the largest slope. Use Newton's method to find the slope of that line correct to six decimal places.


please help answer or solution i dont care

3/29/2006 4:41:37 PM

bmp

3/29/2006 5:18:33 PM

Quote :

"# Do not consecutively reply to a topic - If you are within the alloted timespan for editing a message that no one has replied to and you think of something to add, EDIT the message. DO NOT REPLY AGAIN. Constant abuse of this will not be tolerated. "

that wasn't a bump. that was padding.
Be patient

3/29/2006 5:24:03 PM

well thank you for writing something that has absolutely nothing to do with the topic, thanks for the bump though

3/29/2006 5:27:21 PM

i hope the abbreviation for your major does not end in E

3/29/2006 5:32:00 PM

Am I right in assuming obviously that the goal is to
1) find the slope of the line that is the largest that is still tangent to the curve y=-sin x

if so..I keep getting something like the line y=.222222x to still be tangent but its wrong

I don't know if theres a much easier way to think about this or not

[Edited on March 29, 2006 at 6:00 PM. Reason : v]

3/29/2006 5:32:20 PM

Hmmm.... the tangent to Y=-sin(x) at the point (a,-sin(a)) would have slope -cos(a). So generically we know it's a line,

y=mx+b=-cos(a)x+b

And that line goes through the origin, so b=0. Thus the line is

y=-cos(a)x

But what else do we know? We know y=-sin(a) when x=a so,

-sin(a)=-cos(a)a

Also known as,

tan(a)=a

Of course tangent is periodic with period Pi, so and there will be many solutions. We want the one that is in the interval (-Pi/2,Pi/2 ). That solution will clearly give the largest slope just think graphically for a moment or two.

To use Newton's method invent f(a)=tan(a)-a and make a guess of a in (-Pi/2,Pi/2 ) then interate like Newton's method says. Or ignore the stupid question and use a calculator to zoom in on the intersection of y=x and y=tan(x) like the rest of the civilized world ( I hate Stewart )

3/30/2006 6:54:18 PM

It looks like a=0 is correct. So the slope is -cos(0)= -1. Seems like I must have done something wrong...

3/30/2006 7:02:59 PM

mathman, everything you've done looks good to me (solve tan(a)=a for a), with the exception of wanting the solution in the interval (-pi/2,pi/2).

If you think graphically, you should see that you actually want the solution in the interval (pi,3pi/2).

3/30/2006 11:06:08 PM

Quote :

"largest slope"

as opposed to largest magnitude slope which is what I was thinking about implicitly.

Quote :

"If you think graphically, you should see that you actually want the solution in the interval (pi,3pi/2)."

I see it, my conceptual error was misreading the intial statement. Thanks.

[Edited on March 31, 2006 at 2:17 AM. Reason : .]

3/31/2006 2:17:21 AM

I see what you mean; I've seen the word "largest" used before when they really meant "largest in absolute value". In that case, it would be a slope of -1, like you said. (Of course, you wouldn't need Newton's method for that).

3/31/2006 2:38:56 AM