(We regard UP as being the POSITIVe direction.)

Now the relation between *x* and *y* is:

x^{2}+y^{2}= 20^{2}

Differentiating throughout with respect to time (since the value of *x* and *y* depends on *t*):

`d/(dt)x^2+d/(dt)y^2=0`

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

That is:

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

Now, we know

`(dy)/(dt)=-4`

and we need to know the horizontal velocity (`dx/(dt)`) when

`x = 16`.

The only other unknown is *y*, which we obtain using Pythagoras' Theorem:

`y=sqrt(20^2-16^2)=sqrt(144)=12`

So

`(16)(dx)/(dt)+(12)(-4)=0`

gives

`(dx)/(dt)=3\ "ms"^-1`.

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