Not sure exactly what math this is, but it's post-Algebra. Help please?
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Crimson Sunbird
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This is calculus. Find out the values of \(\displaystyle t\) at \(\displaystyle x=-2\) and \(\displaystyle x=2\). Differentiate \(\displaystyle x\) with respect to \(\displaystyle t\), then integrate \(\displaystyle y\frac{\mathrm dx}{\mathrm dt}\) with respect to \(\displaystyle t\) between the values of \(\displaystyle t\) at \(\displaystyle x=-2\) and \(\displaystyle x=2\).
Crimson Sunbird
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Right. The integrand will have to be integrated piecewise over intervals on which \(\displaystyle x\) is a one-to-one function of \(\displaystyle t\) as
\(\displaystyle \displaystyle\int_{x=-2}^{x=2}y\frac{\mathrm dx}{\mathrm dt}\,\mathrm dt\ =\ \int_{x=-2}^{x=-\frac2{\sqrt3}}y\frac{\mathrm dx}{\mathrm dt}\,\mathrm dt\,+\,\int_{x=-\frac2{\sqrt3}}^{x=\frac2{\sqrt3}}y\frac{\mathrm dx}{\mathrm dt}\,\mathrm dt\,+\,\int_{x=\frac2{\sqrt3}}^{x=2}y\frac{\mathrm dx}{\mathrm dt}\,\mathrm dt\)
\(\displaystyle \displaystyle\int_{x=-2}^{x=2}y\frac{\mathrm dx}{\mathrm dt}\,\mathrm dt\ =\ \int_{x=-2}^{x=-\frac2{\sqrt3}}y\frac{\mathrm dx}{\mathrm dt}\,\mathrm dt\,+\,\int_{x=-\frac2{\sqrt3}}^{x=\frac2{\sqrt3}}y\frac{\mathrm dx}{\mathrm dt}\,\mathrm dt\,+\,\int_{x=\frac2{\sqrt3}}^{x=2}y\frac{\mathrm dx}{\mathrm dt}\,\mathrm dt\)
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