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The length of the curve determined by the equations $x=t-1$ and $y=\sqrt{t}$ from $t=0$ to $t=4$ is.

I think that is is a parametric cruve and Paul's Online Notes the equation for the lenght of a parametric curve is:

$$\int_a^b \sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2}\; \mathrm{d}t$$

Which is $$\int^4_0 \sqrt{\left(1\right)^2 + \left(\frac{1}{2\sqrt{t}}\right)^2}\, \mathrm{d}t$$

The answer sheet states this as the answer:

$$\int_0^4 \sqrt{\left(t-1\right)^2 + \left(\sqrt{t}\right)^2}\,\mathrm{d}t $$

I am correct?

Thomas
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yiyi
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1 Answers1

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Yes, you are correct: $$ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = 1^2 + \left(\frac{1}{2\sqrt{t}}\right)^2 $$ which is what you need in the formula for the length.

Thomas
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