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mariarad [96]
3 years ago
10

Reparametrize the curve with respect to arc length measured from the point where t = 0 in the direction of increasing t. (Enter

your answer in terms of s.) r(t) = e5t cos(5t) i + 5 j + e5t sin(5t) k
Mathematics
1 answer:
Alex Ar [27]3 years ago
7 0

Answer:

I would need more information to help you

Step-by-step explanation:

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2 years ago
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GaryK [48]
\bf \cfrac{x^3y}{xy^5}\cdot \cfrac{x^2y^9}{x^8}\implies \cfrac{x^3x^2y^1y^9}{x^1x^8y^5}&#10;\\ \quad \\&#10;\textit{now, using exponent rules, same base, different exponent}&#10;\\ \quad \\&#10;\cfrac{x^{3+2}y^{1+9}}{x^{1+8}y^5}\implies \cfrac{x^5y^{10}}{x^9y^5}\\&#10;----------------------------\\\bf \textit{now, keep in mind that}&#10;\\ \quad \\&#10;a^{-{ n}} \implies \cfrac{1}{a^{ n}}\qquad \qquad&#10;&#10;\cfrac{1}{a^{ n}}\implies a^{-{ n}}&#10;\\ \quad \\&#10;%  negative exponential denominator&#10;a^{{ n}} \implies \cfrac{1}{a^{- n}}&#10;\qquad \qquad &#10;&#10;\cfrac{1}{a^{- n}}\implies \cfrac{1}{\frac{1}{a^{ n}}}\implies a^{{ n}}&#10;\\ \quad \\&#10;thus\\&#10;----------------------------\\&#10;\bf \cfrac{x^5y^{10}}{x^9y^5}\implies \cfrac{x^5y^{10}}{1}\cdot \cfrac{1}{x^9}\cdot \cfrac{1}{y^5}\implies x^5y^{10}\cdot x^{\boxed{-9}}\cdot y^{\boxed{-5}}&#10;\\ \quad \\&#10;x^5x^{-9}y^{10}y^{-5}\implies x^{5-9}y^{10-5}\implies x^{-4}y^5&#10;\\ \quad \\&#10;\cfrac{1}{x^{\boxed{4}}}\cdot \cfrac{y^5}{1}\implies \cfrac{y^5}{x^4}
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3 years ago
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