MHB Find Equation of Tangent Line at (4π, f(4π)) - Graffer's Question

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The equation of the tangent line to the curve y=f(x)=x(10cos x-2sin x) at the point (4π, f(4π)) is derived using the point-slope formula. The function value at this point is calculated as f(4π)=40π. The derivative f'(x) is determined, and specifically f'(4π) is found to be 2(5-4π). Consequently, the equation of the tangent line is expressed as y=2(5-4π)(x-4π)+40π, which simplifies to slope-intercept form as y=2(5-4π)x+32π². A plot of the curve alongside its tangent line at the specified point is also provided.
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Re: graffer's question at Yahoo! Answers regarding finding the euqation of a tangent line

Hello graffer,

Using the point-slope formula, with the point $(4\pi,f(4\pi))$ and the slope $f'(4\pi)$, the equation of the tangent line is:

$y=f'(4\pi)(x-4\pi)+f(4\pi)$

Using the given function definition, we find:

$f(4\pi)=4\pi(10\cos(4\pi)-2\sin(4\pi))=4\pi(10)=40\pi$

$f'(x)=x(-10\sin(x)-2\cos(x))+(1)(10\cos(x)-2\sin(x))=2((5-x)\cos(x)-(5x+1)\sin(x))$

$f'(4\pi)=2((5-4\pi)\cos(4\pi)-(5\cdot4\pi+1)\sin(4\pi))=2(5-4\pi)$

Putting it all together, we find the equation of the tangent line is:

$y=2(5-4\pi)(x-4\pi)+40\pi$

In slope-intercept form, this is:

$y=2(5-4\pi)x+32\pi^2$

Here is a plot of the curve and its tangent line at the given point:

23ll5xy.jpg
 
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