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Marina CMI [18]
3 years ago
12

Find an equation of the tangent line to the curve y=arccos(x/2) at the point (1,4pi)

Mathematics
1 answer:
marishachu [46]3 years ago
5 0
The derivative of inverse trigonometric identities follow<span> a set of rules. </span>
<span>For </span>arc cos<span> x, the </span>dy/dx is equal to -1/ square root of (1 - x2). 
<span>In this case, we still have 1/2 x inside the angle quantity so we multiply the </span>dy/dx above with 1/2. Hence, y' = is -1<span>/ 2 square root of (1 - x2); x = 1, y' is infinity. this means the line is horizontal. The answer hence is y = 1.</span>
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Find all unit vectors that are orthogonal to the vector u = 1, 0, −4 .
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Step-by-step explanation:

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u = 1, 0, -4

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u = i + 0j - 4k

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If v = v₁ i + v₂ j + v₃ k is one of those vectors that are orthogonal to u, then

u. v = 0                    [<em>substitute for the values of u and v</em>]

=> (i + 0j - 4k) . (v₁ i + v₂ j + v₃ k)  = 0               [<em>simplify</em>]

=> v₁ + 0 - 4v₃ = 0

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Plug in the value of v₁ = 4v₃ into vector v as follows

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Equation (i) is the generalized form of all vectors that will be orthogonal to vector u

Now,

Get the generalized unit vector by dividing the equation (i) by the magnitude of the generalized vector form. i.e

\frac{v}{|v|}

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|v| = \sqrt{(4v_3)^2 + (v_2)^2 + (v_3)^2}

|v| = \sqrt{17(v_3)^2 + (v_2)^2}

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This is the general form of all unit vectors that are orthogonal to vector u

where v₂ and v₃ are non-zero arbitrary real numbers.

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3 years ago
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