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
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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