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Aleksandr [31]
2 years ago
13

Find the slope of the tangent line of the curve r = cos (3theta) at theta = pi / 3

Mathematics
1 answer:
LiRa [457]2 years ago
4 0

The slope of the tangent line to the curve at a point <em>(x, y)</em> is d<em>y</em>/d<em>x</em>. By the chain rule, this is equivalent to

d<em>y</em>/d<em>θ</em> × d<em>θ</em>/d<em>x</em> = (d<em>y</em>/d<em>θ</em>) / (d<em>x</em>/d<em>θ</em>)

where <em>y</em> = <em>r(θ)</em> sin(<em>θ</em>) and <em>x</em> = <em>r(θ)</em> cos(<em>θ</em>). Then

d<em>y</em>/d<em>θ</em> = d<em>r</em>/d<em>θ</em> sin(<em>θ</em>) + <em>r(θ)</em> cos(<em>θ</em>)

d<em>x</em>/d<em>θ</em> = d<em>r</em>/d<em>θ</em> cos(<em>θ</em>) - <em>r(θ)</em> sin(<em>θ</em>)

Given <em>r(θ)</em> = cos(3<em>θ</em>), we have

d<em>r</em>/d<em>θ</em> = -3 sin(3<em>θ</em>)

and so

d<em>y</em>/d<em>x</em> = (-3 sin(3<em>θ</em>) sin(<em>θ</em>) + cos(3<em>θ</em>) cos(<em>θ</em>)) / (-3 sin(3<em>θ</em>) cos(<em>θ</em>) - cos(3<em>θ</em>) sin(<em>θ</em>))

When <em>θ</em> = <em>π</em>/3, we end up with a slope of

d<em>y</em>/d<em>x</em> = (-3 sin(<em>π</em>) sin(<em>π</em>/3) + cos(<em>π</em>) cos(<em>π</em>/3)) / (-3 sin(<em>π</em>) cos(<em>π</em>/3) - cos(<em>π</em>) sin(<em>π</em>/3))

d<em>y</em>/d<em>x</em> = -cos(<em>π</em>/3) / sin(<em>π</em>/3)

d<em>y</em>/d<em>x</em> = -cot(<em>π</em>/3) = -1/√3

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