# 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#?

##### 1 Answer

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

#### Explanation:

We have parametric equations:

# { (x=5t^2), (y=t^3) :} #

defining the motion of a particle from

# 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 #