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

`v_x=(dx)/(dt)`

and

`v_y=(dy)/(dt)`

But the question already gives us

`v_x=(dx)/(dt)=3`

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

`(dy)/(dt)=2x(dx)/(dt)+4(dx)/(dt)+0`

Since

`(dx)/(dt)=3`

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

`(dy)/(dt)=2(-1)(3)+4(3)=6`

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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