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Numerade Educator



Problem 67 Hard Difficulty

Suppose a curve is given by $\mathbf{r}(t)=\langle f(t), g(t)\rangle,$ where $f^{\prime}$ and $g^{\prime}$ are continuous, for $a \leq t \leq b .$ Assume the curve is traversed once, for $a \leq t \leq b$ and the length of the curve between $(f(a), g(a))$ and $(f(b), g(b))$ is $L$. Prove that for any nonzero constant $c$ the length of the curve defined by $\mathbf{r}(t)=\langle c f(t), c g(t)\rangle,$ for $a \leq t \leq b,$ is $|c| L$


You really get $|c| L$ is length of the given curve.


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Video Transcript

very prone. 67 I suppose with a curve are is a function of F and G and the derivative of effigies are all continuous. The improved that for a constant see the length of the curve is going by sea times l So length of the curve in a girl from A to B of f prime square pose G prime squared No, if our zip go to C f g f our cf c g Then using the length in grow, we get the integral from zero be a squarer cf prime squared plus C g prime squared d t could pull out the sea and we get in the girl Or maybe just our regular length formula, Which means you just see times l okay, now see is less than zero. Then let's say C equals negative. D So are of tea is negative. D f t common negativity. GFT Then again, we set up our length and grow to be negative. D f Prime Square plus negative G g Prime squared and the square just cancels. So again we could just Paul D. Integral from A to B and again it's just f prime square plus G prom square, which is l. So that's you go to D L. So since we proved that either away, you're gonna end up with ah de l or C L than the length. So you have CEO first, see greater than zero and D l for C less than zero where c is the goto negative d Then we can generalize this statement as the absolute value C L.

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