The fastest way to find the missing endpoint is to determine the distance from the known endpoint to the midpoint and then performing the same transformation on the midpoint. In this case, the x-coordinate moves from 4 to 2, or down by 2, so the new x-coordinate must be 2-2 = 0. The y-coordinate moves from 4 to -5, or down by 9, so the new y-coordinate must be -5-9 = -14.

An alternate solution would be to substitute (4,4) for (x1,y1) and (2,-5) for (x,y) into the midpoint formula:

x=(x1+x2)/2

y=(y1+y2)/2

Solving each equation for (x2,y2) yields the solution (0,-14).

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2 years ago

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Evaluate the line integral c y3 ds, c.x = t3, y = t, 0 â‰¤ t â‰¤ 3

Temka [501]

$\displaystyle\int_{\mathcal C}y^3\,\mathrm dS=\int_{t=0}^{t=3}t^3\sqrt{1^2+(3t^2)^2}\,\mathrm dt=\int_0^3t^3\sqrt{1+9t^4}\,\mathrm dt$

Take $u=1+9t^4$ so that $\mathrm du=36t^3\,\mathrm dt$ . Then

$\displaystyle\int_{\mathcal C}y^3\,\mathrm dS=\frac1{36}\int_{u=1}^{u=730}\sqrt u\,\mathrm du=\frac{730^{3/2}-1}{54}$

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2 years ago