- #1
lovelylila
- 17
- 0
I have encountered a problem in my Calculus homework.
I have a position function, r(t)= (t^2)i + (4t)j and in my homework, I am asked to find the tangent to this curve at the point t=3. I did this by finding dy/dx, or 2t/4 @ t=3 is 6/4. However, I am also asked to relate this to the velocity vector for the position function @ t=3, but I don't understand the relationship. Would they share the same slope? Any help is very much appreciated! :-)
I have a position function, r(t)= (t^2)i + (4t)j and in my homework, I am asked to find the tangent to this curve at the point t=3. I did this by finding dy/dx, or 2t/4 @ t=3 is 6/4. However, I am also asked to relate this to the velocity vector for the position function @ t=3, but I don't understand the relationship. Would they share the same slope? Any help is very much appreciated! :-)