How am I supposed to do this problem? Calc III?

Question

You don't have to give me the answer, i just would like to know where to start:

Show that the curves parameterized by (i, j, k are vectors)
r1(t)=(.5t^2+t)i+(t+1)j-tk
r2(t)=sin(t)i+e^tj+tan(t)k
intersect the points (0,1,0) and that the vectors tangent to the two curves at (0,1,0) are parallel.

Answer 1

For the first part:

Simply check for t = 0;

r1(0) = (.5*(0)^2+0)i + (0+1)j-0k

so, r1(0) = 0i+1j+0k or, (0,1,0) in vector form

r2(0) = sin(0)i+e^0j+tan(0)k

so, r2(0) = 0i + 1j + 0k or, as above (0,1,0) in vector form

so you know at t = 0 that the 2 curves intersect said point. (I found that they intersect (0,1,0) at t= 0 by setting each constituent parameter to its matched value)

finding the tangent is simple. The equation for a line tangent to a curve should be in your calc book. However, since you know the point you are looking at, you can find the line tangent to the curve by deriving the parametrized equations wrt t.

r1' = (t+1)i + 1j - 1k
r2' = cos(t)+ e^tj + sec^2(t)

go from there.

Answer 2

You need to figure out what t gives you those coordinates in both the functions. since your point is (0,1,0), you want to find a t such that:
.5t^2+t = 0
t+1 = 1
-t = 0
This one is easy to see that t = 0 gives the point you are looking for. Do a similar thing with r2. Keep in mind that the t for each function can be different.

To see if they are parallel, you need to find the tangent vector to the curves at the point. If the tangent vectors point in the same direction they are parallel.

Answer 3

that's simple really you just add a fomula x72 v.3 + 1 litre of carbonised acide ( the v. 2 of the acide)