# Curve + tangent

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• Jun 25th 2010, 07:58 AM
Ackbeet
By "where do they solve it", I mean "where are the solutions?" You've found them. There's really only one step left. What do you suppose that is?
• Jun 25th 2010, 08:03 AM
maggiec
Quote:

Originally Posted by Ackbeet
By "where do they solve it", I mean "where are the solutions?" You've found them. There's really only one step left. What do you suppose that is?

provide a general solution?
show that these values of x satisfy the original equation?
• Jun 25th 2010, 08:05 AM
Ackbeet
You've found the locations where the line y = x equals the original function. But how do you know that, at those points, y = x touches the original function?
• Jun 25th 2010, 08:58 AM
maggiec
Quote:

Originally Posted by Ackbeet
You've found the locations where the line y = x equals the original function. But how do you know that, at those points, y = x touches the original function?

at the points where y=x touches the original function the gradients are equal
• Jun 25th 2010, 08:59 AM
Ackbeet
True... but how do you know that?
• Jun 25th 2010, 09:01 AM
maggiec
Quote:

Originally Posted by Ackbeet
True... but how do you know that?

ermm.. because the line y=x is a tangent to the curve? Thus at the point where it is a tangent its gradient is equal to that of the curve?

I'm sorry Im doing really bad with this one!
• Jun 25th 2010, 09:02 AM
Ackbeet
Hint: evaluate the _____________ at the points of touching, and show that they're all equal to __ .
• Jun 25th 2010, 09:08 AM
maggiec
Quote:

Originally Posted by Ackbeet
Hint: evaluate the _____________ at the points of touching, and show that they're all equal to __ .

evaluate the gradient at the points of touching, and show that they're all equal to 1?

ie. when x = 5pi/2, dy/dx = 1
when x = 9pi/2, dy/dx = 1
...
• Jun 25th 2010, 09:09 AM
Ackbeet
You got it. I would say that once you've shown all that, you're done with this problem.
• Jun 25th 2010, 09:11 AM
maggiec
Thank you soooo much! You've been such a great help :)
• Jun 25th 2010, 09:12 AM
Ackbeet
You're very welcome. Have a good one!
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