• Mar 7th 2007, 12:46 PM
UMStudent
If http://lennes.math.umt.edu/webwork/1...rob7image1.png is a point on the graph of http://lennes.math.umt.edu/webwork/1...rob7image2.png and if <IMG alt="c > 0 \, ," src="http://lennes.math.umt.edu/webwork/152/tmp/png/5/854706/adams-prob7image3.png" align=middle> then c = ?

I have this problem I've tried a few things one solving for y but you end up taking the sqrt of a negative. I don't know what to do I'm pretty lost thanks for any help.
• Mar 7th 2007, 12:57 PM
Jhevon
Quote:

Originally Posted by UMStudent
If http://lennes.math.umt.edu/webwork/1...rob7image1.png is a point on the graph of http://lennes.math.umt.edu/webwork/1...rob7image2.png and if <IMG alt="c > 0 \, ," src="http://lennes.math.umt.edu/webwork/152/tmp/png/5/854706/adams-prob7image3.png" align=middle> then c = ?

I have this problem I've tried a few things one solving for y but you end up taking the sqrt of a negative. I don't know what to do I'm pretty lost thanks for any help.

Let's plug in the points (-4, c) and see what happens.

[(-4)^2]/64 + (c^2)/25 = 1

=> 16/64 + (c^2)/25 = 1
=> (c^2)/25 = 1 - 16/64
=> (c^2)/25 = 3/4
=> c^2 = 25(3/4) = 75/4
=> c = sqrt(75/4) = sqrt(75)/2 = sqrt(25*3)/2 = 5*sqrt(3)/2

how did you end up taking the squareroot of a negative?
• Mar 7th 2007, 01:09 PM
UMStudent
Thanks. I guess sometimes I just need to see something done. I completely understand now. I was stuck thinking one way to get the equation y = something so if you work that out their ends up being a negative. Thanks though .
• Mar 7th 2007, 01:15 PM
Jhevon
Quote:

Originally Posted by UMStudent
Thanks. I guess sometimes I just need to see something done. I completely understand now. I was stuck thinking one way to get the equation y = something so if you work that out their ends up being a negative. Thanks though .

i don't think so. because all i did was replace y with c, so solving for c would be the same as solving for y
• Mar 7th 2007, 01:29 PM
UMStudent
I see what I did now I never did 1 - 16/64. Because I left in x I didn't subsitute the -4 in for x.

But if you don't mind I have another question I'm trying to do
http://lennes.math.umt.edu/webwork/1...rob6image1.png I want to find f'(x) again and I was trying to follow steps in my book. I got sin(5 cos 7x)+(5 cos 7x)(cos x) but the programs says thats wrong. So I tried a differenet problem similiar to it
If http://lennes.math.umt.edu/webwork/1...rob5image1.png and got 6e^(x*cos(x))(-sin(x)) but it says thats wrong. I'm following the rules right out of the book and the trig functions but not sure. Are you able to see what i'm doing wrong here? Thanks
• Mar 7th 2007, 01:52 PM
Jhevon
Quote:

Originally Posted by UMStudent
I see what I did now I never did 1 - 16/64. Because I left in x I didn't subsitute the -4 in for x.

But if you don't mind I have another question I'm trying to do
http://lennes.math.umt.edu/webwork/1...rob6image1.png I want to find f'(x) again and I was trying to follow steps in my book. I got sin(5 cos 7x)+(5 cos 7x)(cos x) but the programs says thats wrong. So I tried a differenet problem similiar to it
If http://lennes.math.umt.edu/webwork/1...rob5image1.png and got 6e^(x*cos(x))(-sin(x)) but it says thats wrong. I'm following the rules right out of the book and the trig functions but not sure. Are you able to see what i'm doing wrong here? Thanks

ok, so we need the chain rule here. i'd have to see your working to tell you exactly where you went wrong. But anyway...

The Chain Rule:

d/dx[f(g(x))] = f'(g(x))*g'(x)

so to find the derivative of sin(5sin(7x)), we find the derivative of sine with the same function in brackets and then MULTIPLY by the derivative of the function in brackets. Here's how to do it.

g(x) = sin(5sin(7x))
=> g'(x) = cos(5sin(7x))*(7*5cos(7x)) .............7*5cos(7x) is the derivative of 5sin(7x)
= 35cos(7x)*cos(5sin(7x))

g(x) = 6e^xcosx .................the 6 is a constant, so we can forget about it for the time being and multiply by it when done. we need the product rule to find the derivative of xcosx (that's probably where you went wrong).
=> g'(x) = 6[(cosx - xsinx)e^xcosx]
= 6(cosx - xsinx)e^xcosx