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  • 4 years ago

verify that the given point is on the curve nd find the lines that are a) tangent and normal to the curve at the given point. x^2cos^2y -siny=0 (0,pi) solve by implicit differentiation

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  1. ZeHanz
    • 4 years ago
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    This is the curve, I think:\[x^2\cos ^2 y-\sin y=0\] If you set x = 0, all you have left is: sin y = 0 This is true for y = k*pi (k is an integer), so it is true for y = pi, so (0, pi) is on the curve. Implicit differentiation is just a quick way to differentiate without first solving the equation for y. You do everything the normal way: use the product rule etc., just remember that wherever you see y, this is a function of x, or y(x). This means that the Chain Rule plays an important role. I.e.: (siny)' = cosy*dy/dx Implicit differentiation of your curve therefore gives:\[2x \cos ^2 y -2x \sin y \cos y \frac{ dy }{ dx }-\cos y \frac{ dy }{ dx }=0\]Solve for dy/dx and replace 2xsinycosy by sin2y:\[\frac{ dy }{ dx }=\frac{ 2x \cos ^2y }{ x \sin 2y-\cos y }\]Substitute x=0 and y = pi to see the slope of the tangent line in (0, pi). Once you have done that, you will also immediately see the line normal to the curve!

  2. anonymous
    • 4 years ago
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    okay,thanks. i will now solve it.

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