We differentiate the expression with respect to t:

`x^2/72.5+y^2/71.5=1`

`(2x(dx)/(dt))/72.5+(2y(dy)/(dt))/71.5=0`

We need to find y. We do this by substituting `x = 3.2` into the original expression:

`3.2^2/72.5+y^2/71.5=1`

Solving this gives:

`(71.5)(3.2)^2+72.5y^2=(71.5)(72.5)`

`732.16+72.5y^2=5183.75`

`y^2=(5183.75-732.16)/(72.5)`

`y^2=61.401`

`y=+-7.836`

The question tells us to take the positive value only. Substituting our known values gives:

`(2(3.2)(12.9))/72.5+(2(7.836)(dy)/dt)/71.5=0`

`(71.5)(2)(3.2)(12.9)+` `(72.5)(2)(7.836)(dy)/(dt)=0`

`(dy)/(dt)=-5.195`

This means the velocity in the y-direction is −5195 km/h.