How do you find the distance travelled from t=0 to t=3 by a particle whose motion is given by the parametric equations #x=5t^2, y=t^3#?

Question

Charles Arocha · Accepted Answer

# s = (181sqrt(181)-1000)/27 ~~ 53.15 #

We have parametric equations: # { (x=5t^2), (y=t^3) :} # defining the motion of a particle from #t=0# to #t=3#, so the total distance travelled is the arclength, which we calculate for parametric equations using: # s = int_alpha^beta \ sqrt( (dx/dt)^2+(dy/dt)^2 ) \ dt # # \ \ = int_0^3 \ sqrt( (10t)^2+(3t^2)^2 ) \ dt # # \ \ = int_0^3 \ sqrt( t^2(100+9t^2 )) \ dt # # \ \ = int_0^3 \ tsqrt( 100+9t^2 ) \ dt # In order to evaluate this integral, we can perform a substitution, Let # u = 9t^2 +100 => (du)/dt = 18t # And we change the limits of integration: # t={ (0), (3) :} => u={ (100), (181) :}# So then: # s = int_100^181 \ (1/18) \ sqrt( u ) \ dt # # \ \ = 1/27 \ (181^(3/2) - 100^(3/2)) # # \ \ = (181sqrt(181)-1000)/27 # # \ \ ~~ 53.15 #

Charles Arocha · Answer

undefined To find the distance traveled by the particle from $t = 0$ to $t = 3$ when its motion is given by the parametric equations $x = 5t^2$ and $y = t^3$, you can use the formula for arc length in parametric equations. The arc length formula for a parametric curve $(x(t), y(t))$ from $t = a$ to $t = b$ is given by:

$$ s = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt $$

Using the given parametric equations, you calculate $\frac{dx}{dt}$ and $\frac{dy}{dt}$, then substitute them into the formula and integrate from $t = 0$ to $t = 3$. This yields the distance traveled by the particle.

$$ \frac{dx}{dt} = 10t $$
$$ \frac{dy}{dt} = 3t^2 $$

Substituting these derivatives into the formula and integrating from $t = 0$ to $t = 3$ gives:

$$ s = \int_{0}^{3} \sqrt{(10t)^2 + (3t^2)^2} \, dt $$

After calculating the integral, you'll obtain the distance traveled by the particle from $t = 0$ to $t = 3$.