What is the distance between the following polar coordinates?:

$(5,\frac{5\pi}{4}),(2,\frac{11\pi}{8})$

$(5,\frac{5\pi}{4}),(2,\frac{11\pi}{8})$

Audrina Jackson
2022-07-04
Answered

What is the distance between the following polar coordinates?:

$(5,\frac{5\pi}{4}),(2,\frac{11\pi}{8})$

$(5,\frac{5\pi}{4}),(2,\frac{11\pi}{8})$

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Differentiate the parametric function and find $\frac{\mathrm{d}y}{\mathrm{d}x}$ and $\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^{2}}$ in terms of "t" when:

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and $\frac{\mathrm{d}y}{\mathrm{d}t}=\mathrm{ln}|t+1|$

and used $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}$, which gives

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Differentiate the parametric function and find $\frac{\mathrm{d}y}{\mathrm{d}x}$ and $\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^{2}}$ in terms of "t" when:

$x=\frac{1}{t-1}$ and $y=\frac{1}{t+1}$

I have first started by finding $\frac{\mathrm{d}y}{\mathrm{d}x}$ by finding $\frac{\mathrm{d}x}{\mathrm{d}t}$ and $\frac{\mathrm{d}y}{\mathrm{d}t}$ which comes to

$\frac{\mathrm{d}x}{\mathrm{d}t}=\mathrm{ln}|t-1|$

and $\frac{\mathrm{d}y}{\mathrm{d}t}=\mathrm{ln}|t+1|$

and used $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}$, which gives

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{ln}|t+1|}{\mathrm{ln}|t-1|}$ now if I divide them by each other doesn't it equal to 0? What have I done wrong here?

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$(4,\frac{7\pi}{4}),(3,\frac{3\pi}{8})$

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Why is speed of parametric defined as

$speed=\sqrt{{\left(\frac{dx}{dt}\right)}^{2}+{\left(\frac{dy}{dt}\right)}^{2}}$

How is this derived? What is the principle here? Is this Pythag? Thinking of this in terms of vectors?

I know speed = |velocity|

Why is speed of parametric defined as

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How is this derived? What is the principle here? Is this Pythag? Thinking of this in terms of vectors?