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Graffer's question at Yahoo! Answers regarding finding the equation of a tangent line

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MarkFL

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Feb 24, 2012
13,775
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MarkFL

Administrator
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Feb 24, 2012
13,775
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: