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Readme [11.4K]
3 years ago
5

When rolling a number cube,

Mathematics
1 answer:
Gnoma [55]3 years ago
6 0

Answer:

yyyyyyyyyyyyyyydilligaf

Step-by-step explanation:

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What will be the result of substituting 2 for x in both expressions below?
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Find the arc-length parametrization of the curve that is the intersection of the elliptic cylinder x 2 + y 2/2 = 1 and the plane
storchak [24]

Answer:

f(\theta) = (cos(\frac{\theta}{\sqrt2}), \sqrt2 sin(\frac{\theta}{\sqrt2}), cos(\frac{\theta}{\sqrt2})-2)

0 ≤ Ф ≤ 4π.

Step-by-step explanation:

since x²+y²/2 = 1, then x²+s² = 1, with s = (y/√2)². Hence, (x,s) = (cos(Ф),sin(Ф)) and (x,y,z) = (cos(Ф),√2 sin(Ф), cos(Ф)-2). This expression evaluated in zero gives as result (1,0,-1). The derivate of this function is (-sin(Ф),√2 cos(Ф), -sen(Ф))

the norm of the derivate is √(sin²(Ф) + 2cos²(Ф)+sin²(Ф)) = √2. In order to make the norm equal to 1, i will divide Ф by √2, so that a √2 is dividing each term after derivating.

We take

f(\theta) = (cos(\frac{\theta}{\sqrt2}), \sqrt2 sin(\frac{\theta}{\sqrt2}), cos(\frac{\theta}{\sqrt2})-2)

Note that

  • f(0) = (1,0,-1)
  • f'(\theta) = (\frac{sin(\frac{\theta}{\sqrt2})}{\sqrt2}, cos(\frac{\theta}{\sqrt2})}, \frac{sin(\frac{\theta}{\sqrt2})}{\sqrt2})

Whose square norm is 1/2cos²(Ф/2)+sen²(Ф/2)+1/2cos²(Ф/2) = 1. This is te parametrization that we wanted.

The values from Ф range between 0 an 4π, because the argument of the sin and cos is Ф/2, not Ф, Ф/2 should range between 0 and 2π.

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3 years ago
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