Question

What is the instantaneous velocity of an object moving in accordance to `f(t)= (tsin(2t-pi/2),cost)` at `t=(5pi)/4`?

Answer 1

An object whose position is given by the parametric equation `f(t)=(x(t),y(t))` has a velocity of `v(t)=f'(t)=(x'(t),y'(t))`.

Here `x(t)=tsin(2t-pi/2)` so:

`x'(t)=(d/dtt)sin(2t-pi/2)+t(d/dtsin(2t-pi/2))`

`color(white)(x'(t))=sin(2t-pi/2)+tcos(2t-pi/2)(d/dt(2t-pi/2))`

`color(white)(x'(t))=sin(2t-pi/2)+2tcos(2t-pi/2)`

And `y(t)=cost` so

`y'(t)=-sint`

We find that `v((5pi)/4)=(x'((5pi)/4),y'((5pi)/4))`:

`x'((5pi)/4)=sin((5pi)/2-pi/2)+(5pi)/2cos((5pi)/2-pi/2)`

`color(white)(x'((5pi)/4))=sin(2pi)+(5pi)/2cos(2pi)`

`color(white)(x'((5pi)/4))=(5pi)/2`

And

`y'((5pi)/4)=-sin((5pi)/4)`

`color(white)(y'((5pi)/4))=1/sqrt2`

So the velocity vector at `t=(5pi)/4` is `v'((5pi)/4)=((5pi)/2,1/sqrt2)`.

The magnitude of the velocity (the speed) is given by `sqrt((x'((5pi)/4))^2+(y'((5pi)/4))^2)`, or:

`=sqrt((25pi^2)/4+1/2)=1/2sqrt(25pi^2+2)`