Answer to Question #106440 in Calculus for Abhijeet Thakur

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 : $\frac { \sqrt { 10 } }{ 3 } +3\sqrt { 2 } +3\ln { \frac { 1+\sqrt { 10 } }{ 3(\sqrt { 2 } -1) } }$

EXPLANATION.

The length L of the curve $G= \left\{ (x(t),y(t)):a\le x\le b \right\}$ is calculated by the formula

$L=\int _{ a }^{ b }{ \sqrt { { \left[ x'(t) \right] }^{ 2 }+{ \left[ y'(t) \right] }^{ 2 } } } dt$ . To determine the values of $a$ and $b$ we find the value of the parameter $t$ corresponding to the coordinates of the points of intersection of the parabola and the line. To do this , we solve the equation $3(3t^2) +6t-3=0$ or $(3t+1)^2-4=0$ . Hence $a=-1, b=\frac { 1 }{ 3 }$ .$x'(t)=6t,\quad y'(t)=6,\ \sqrt { { \left[ x'(t) \right] }^{ 2 }+{ \left[ y'(t) \right] }^{ 2 } } =\sqrt { 36{ t }^{ 2 }+36 }$

$L=6\int _{ -1 }^{ 1/3 }{ \sqrt { 1+t^{ 2 } } } dt$ $=F(1/3)-F(-1)$ , where $F(t)= 6 \left [ {\frac{t}{2}} \sqrt{1+t^2} +\frac{1}{2} \ln \left | t+\sqrt{1+t^2} \right | \right ]$ , $F(1/3)=\left[ \frac { \sqrt { 10 } }{ 3 } +3\ln { \frac { 1+\sqrt { 10 } }{ 3 } } \right]$ ,$F(-1)=\left[ -3\sqrt { 2 } +3\ln { (\sqrt { 2 } -1) } \right]$