# find the slope of the curve

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• Jun 21st 2009, 03:18 PM
vtong
find the slope of the curve
The equation sin xy = y defines y implicitly as a function of x.
Find the slope y'(pi/3,1/2) at the point x=pi/3, y=1/2.

i know that y' = (ycos(xy))/(1-xcos(xy))

but i just dont know how to solve the problem when i plug in x and and y to find the slope (Blush)

please help

ty:)
• Jun 21st 2009, 03:32 PM
artvandalay11
Quote:

Originally Posted by vtong
The equation sin xy = y defines y implicitly as a function of x.
Find the slope y'(pi/3,1/2) at the point x=pi/3, y=1/2.

i know that y' = (ycos(xy))/(1-xcos(xy))

but i just dont know how to solve the problem when i plug in x and and y to find the slope (Blush)

please help

ty:)

y=sin(xy) so $y'=\frac{y\cos{xy}}{1-x\cos{xy}}$ and plug in to get $y'=\frac{\frac{1}{2}\cos{\frac{\pi}{6}}}{1-\frac{\pi}{3}\cos{\frac{\pi}{6}}}$ which is the slope of your function
• Jun 21st 2009, 04:02 PM
VonNemo19
Quote:

Originally Posted by vtong
The equation sin xy = y defines y implicitly as a function of x.
Find the slope y'(pi/3,1/2) at the point x=pi/3, y=1/2.

i know that y' = (ycos(xy))/(1-xcos(xy))

but i just dont know how to solve the problem when i plug in x and and y to find the slope (Blush)

please help

ty:)

The derivative of a function evaluated at a given point is the slope of the line tangent to the graph of the function at that point.

So, You've got a derivative, and you've got a point. Evaluate the derivative at that point and that will give the slope.

i.e. Plug in $\frac{\pi}{3}$ wherever you see an x, and $\frac{1}{2}$ whereever you see a y.

Note* $\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$.