Paremetric equations: arc length ∫ √[(dx/dt)^2 + (dy/dt)^2] dt

[pilot227]

Any help on this questions would be appreciated:

The arc length for a parametric curve is given by

∫ √[(dx/dt)^2 + (dy/dt)^2] dt

The limits of this definite integral are t1 to t2.

For a circle the parametric equations are

x = r·cos(t)

y = r·sin(t)

Demonstrate through the appropriately applied algebraic and Calculus techniques, for a unit circle that the length of quarter of the curve of a circle is ? 2 where t2 and t1 are 2? and 3? 2 respectively.

Thank you in advance. Also attached the questions if it adds clarity.

 

[Deleted member 4993]

Any help on this questions would be appreciated:
The arc length for a parametric curve is given by


∫ √(?? ??)^2+ (?? ??)^2 ??

The limits are t2 and t1


For a circle the parametric equations are

? = ?cos? ? = ?sin?

Demonstrate through the appropriately applied algebraic and Calculus techniques, for a unit circle that the length of quarter of the curve of a circle is ? 2 where t2 and t1 are 2? and 3? 2 respectively.

Thank you in advance. Also attached the questions if it adds clarity.

What are your thoughts?

Please share your work with us ...even if you know it is wrong.

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[pilot227]

Work so far

x= r cos t

I differentiated that to give -r sin t

y= r sin t

I differentiated that to give r cos t

Simplifying the equation it is effectively dx/dt+dy/dt

-r sin t + r cos t

Where I am stuck is do I now integrate -r sin t + r cos t?

The questions states a unit circle =1 so r = 1 but do I insert 1 in now as I am then just integrating 1? Or do I leave as r and integrate r?

 

[Deleted member 4993]

x= r cos t

I differentiated that to give -r sin t

y= r sin t

I differentiated that to give r cos t

Simplifying the equation it is effectively dx/dt+dy/dt ................................. Incorrect

-r sin t + r cos t

Where I am stuck is do I now integrate -r sin t + r cos t?

The questions states a unit circle =1 so r = 1 but do I insert 1 in now as I am then just integrating 1? Or do I leave as r and integrate r?

dy/dt = r*cos(t) → (dy/dt)² = [r*cos(t)]² → (dy/dt)² = r² * cos²(t)

dx/dt = -r*sin(t) → (dx/dt)² = [-r*sin(t)]² → (dx/dt)² = r² * sin²(t)

(dy/dt)² + (dx/dt)² = r² * [sin²(t) + cos²(t)] ....... continue

 

[mmm4444bot]

int[t1..t2] √[(dx/dt)^2 + (dy/dt)^2] dt

Please note the added grouping symbols (in red). Without them, your integrand was garbage.

Also attached the questions if it adds clarity.

The image certainly adds clarity because this is what you posted:

∫ √(?? ?&#119905:wink:^2+ (?? ?&#119905:wink:^2 ??

In the future, please use the Preview Post button to proofread your posts, before submitting them.

Also note that your attached image renders with small, blurry characters. This is because you did not first crop away all of the unnecessary blank white space, as well as the irrelevant, partially cut-off information at the bottom. This forum must reduce full-sized screen shots, in order to fit them within its post format. With a little extra effort, your posts will be readable. Thank you. :cool: