This is a different situation to the other examples. This time we have y in terms of x, and there are no expressions given in terms of "t" at all.

To be able to find magnitude and direction of velocity, we will need to know




But the question already gives us


so all we need to find is `(dy)/(dt)`.

To find this, we differentiate the given function with respect to t throughout using the techniques we learned back in implicit differentiation:

y = x2 + 4x + 2




and we want to know the velocity at `x = -1`, we substitute these two values and get:


So now we have vy = 6 cm s-1.

So the magnitude of the velocity is given by:

`v=sqrt((v_x)^2+(v_y)^2)=sqrt(3^2+6^2)` `=6.7082`

The direction of the velocity is given by:

`theta_v=arctan((v_y)/(v_x))=arctan(6/3)` `=63.432^"o"`

So the velocity is 6.7 cm s-1, in the direction `63.4^@`.

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