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Use implicit differentiation to find an equation of the tangent line to the curve at...
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For y = f(x), the slope of the tangent at any point (x, y) is the value of `dy/dx` at that point.
The curve is defined in the problem by `y*sin(16x) = x*cos(2y)`
Using implicit differentiation
`y*cos(16x)*16 + sin(16x)(dy/dx) = x*(-sin 2y)*2*(dy/dx) + cos 2y`
=> `(dy/dx)(sin 16x + 2x*sin 2y) = (cos 2y - 16*y*cos 16x)`
=> `(dy/dx) = (cos 2y - 16*y*cos 16x)/(sin 16x + 2x*sin 2y)`
At the point (pi/2, pi/4)
`dy/dx = -4`
The equation of the tangent is `(y - pi/4)/(x - pi/2) = -4`
=> `y - pi/4 = -4x + 2*pi`
=> `4x + y + 7*pi/4 = 0`
The tangent to `y*sin(16x) = x*cos(2y)` at `(pi/2, pi/4)` is `4x + y + 7*pi/4 = 0`
Posted by justaguide on March 7, 2012 at 8:47 AM (Answer #1)
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