Question

find the length of the portion of the parabola x=3t^2, y=6t cut off by the line 3x+y-3=0

Answer 1

Answer:

Step-by-step explanation:

Answer 2

The length of portion of parabola be 8.92

Step-by-step explanation:

The parabola, $x=3t^2,y=6t$

The parabola cut by the line $3x+y-3=0$.

Now find the point of intersection.

$3(3t^2)+6t-3=0$

$9t^2+6t-3=0$

$3t^2+2t-1=0$

$t=-1,\dfrac{1}{3}$

So, $-1\leq t\leq \frac{1}{3}$

Length of the portion of parabola cut by line.

$L=\int ds$

$L=\int_{-1}^{1/3}\sqrt{x'^2+y'^2}dt$

$L=\int_{-1}^{1/3}\sqrt{(6t)^2+6^2}dt$

$L=6\int_{-1}^{1/3}\sqrt{t^2+1}dt$

$L=6\left[\dfrac{1}{2}x\sqrt{x^{2}+1}+\dfrac{1}{2}\log\left|\sqrt{x^{2}+1}+x\right|\right]_{-1}^{\frac{1}{3}}$

$L\approx 8.92$

Hence, the length of portion of parabola be 8.92

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