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Bogdan [553]
3 years ago
12

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = t2 +

35, y = ln(t2 + 35), z = t; (6, ln(36), 1)
Mathematics
1 answer:
Fed [463]3 years ago
8 0

Answer:

Attached below is the detailed solution

answer : x(t), y(t), z(t) = ( 6 + t/6,  In (36) + t/18, 1 + t )

Step-by-step explanation:

x = \sqrt{t^2 + 35 }  , y = In(t^2 + 35 ),  Z = t;(6, ln(36), 1)

Determine the parametric equations for the tangent line to the curve with the above given parametric equations

attached below is the detailed solution

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Which could be the first step in simplifying this expression? Check all that apply. (x cubed x Superscript negative 6 Baseline)
PtichkaEL [24]

Answer:

(x^{-3} )^{2}

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Step-by-step explanation:

(x^{3} x^{-6} )^{2} is the expression given to be solved.

First of all let us have a look at <u>3 formulas</u>:

1.\ p^a \times p^b = p^{(a+b)}\\2.\ (p^a \times q^b)^c = (p^{a})^c \times (q^{b})^c\\3.\ (p^a)^b = p^{a\times b}

Both the formula can be applied to the expression((x^{3} x^{-6} )^{2}) during the first step while solving it.

<u>Applying formula (1):</u>

(x^{3} x^{-6} )^{2}

Comparing the terms of (x^{3} x^{-6} ) with p^a \times p^b

p=x, a =3, b=-6

\Rightarrow x^{3+(-6)}\\\Rightarrow x^{3-6}\\\Rightarrow x^{-3}

So, (x^{3} x^{-6} )^{2} is reduced to (x^{-3} )^{2}

<u>Applying formula (2):</u>

Comparing the terms of (x^{3} x^{-6} )^{2} with (p^a \times q^b)^c

p=q=x, a =3, b=-6, c=2

\Rightarrow (x^{3})^2\times (x^{-6})^2\\\text{Applying Formula (3)}\\x^6 x^{-12}

So, (x^{3} x^{-6} )^{2} is reduced to x^6 x^{-12}.

So, the answers can be:

(x^{-3} )^{2}

x^6 x^{-12}

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There are several important information's already given in the question. Based on those information's the answer to the question can be easily deduced.
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Then
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So
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             2a

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simplify:

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