[just pretend the u's i's and j's in these equations have the squiggly line]
Let u = ti + t²j. Find the equation of the path. Consider vector v = [e^t]i + [e^(2t)]j. Show that the path of this vector is the same as u's path.
Assuming both vectors' equations start when t = 0, do these vectors ever coincide?
Please show working and thank you very much for any answers,
Let u = ti + t²j. Find the equation of the path. Consider vector v = [e^t]i + [e^(2t)]j. Show that the path of this vector is the same as u's path.
Assuming both vectors' equations start when t = 0, do these vectors ever coincide?
Please show working and thank you very much for any answers,
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Let u=(x,y) then x=t and y=t^2 so the cartesian equation of the path is y=x^2
Let v=(x,y) then x=e^t and y=e^(2t) giving y=x^2 which is the same as for u and to say that they
have the same path is not correct since for u, x can take negative values when t is negative but for
v x can only take positive values, since e^t>0.
The point on the first path when t=e^2 is the same as the point on the second path when t=2 but if the
the question means t is the same on both paths then you require t=e^t and you can discover whether
there are positive values of t by drawing graphs of y=e^t and y=t and looking for any intersection.
Let v=(x,y) then x=e^t and y=e^(2t) giving y=x^2 which is the same as for u and to say that they
have the same path is not correct since for u, x can take negative values when t is negative but for
v x can only take positive values, since e^t>0.
The point on the first path when t=e^2 is the same as the point on the second path when t=2 but if the
the question means t is the same on both paths then you require t=e^t and you can discover whether
there are positive values of t by drawing graphs of y=e^t and y=t and looking for any intersection.