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trapecia [35]
3 years ago
12

Find parametric equations for the tangent line at the point (cos(3pi/6),sin(3pi/6), (3pi/6))

Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
5 0
r(t)=(\cos{t},\sin{t},t)
\\r'(t)=(-\sin{t},\cos{t},1)
\\t= \frac{3\pi}{6} = \frac{\pi}{2}
\\ r'(t)=(-\sin{\frac{\pi}{2}},\cos{\frac{\pi}{2}},1)
\\ r'(t)=(-1,0,1)
\\
\\x=0+t(-1)=-t
\\y=1+t\times 0=1
\\z= \frac{\pi}{2} +t
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Answer:

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Step-by-step explanation:

OK lets try again.

The slope of the secant = slope of the tangent at a certain point ( The Mean Value Theorem).

Slope of the secant = f(5) - f(2) / (5 - 2)

= [(25-3) / (5-1)  - (4-3) / (2-1)] / 3

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The derivative at c = the slope of the tangent at c.

Finding the derivative:

f'(x) =  [2x(x - 1) - (x^2 - 3) ]/ (x - 1)^2    (where x = c).

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So equating the slopes:

(x^2 - 2x + 3) / (x - 1)^2 = 3/2

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x^2 - 2x - 3 = 0

(x - 3)(x + 1) = 90

x = 3 , -1

x can't be -1 because we are working between the values 2 and 5 so

x = c = 3.

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