Question

A path is parameterized by x(t)=102t and y(t)=136t, for 0<=t<=3. Which integral gives its arc length?

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Accepted Answer

$$\int_{0}^{3} \sqrt{(102)^{2}+(136)^{2}}\,dt$$

Answer

$$\int_{0}^{3} \sqrt{(102)^{2}-(136)^{2}}\,dt$$

Answer

$$\int_{0}^{3} (102+136)\,dt$$

Answer

$$\int_{0}^{3} \sqrt{1+(136)^{2}}\,dt$$

Practice: Integration - Parametric Arc Length Setup