sheilaw said:
how to find the domain and limit of Vector Functions?
1.Let a vector function F be defined by
F(t)=8t^t i + (64-t)^(1/2) j+(t-8)(t-3)/(t-3)
The domain is the set of values of t for which the equation is defined. Eg, the domain would exclude t=3, because you can't divide by zero. Look at each term, and draw a number line showing the domain for each term - then combine them all, keeping just those t for which all three coordinates are defined.
Finding limits - you find the limit of each coordinate, and you have the limit of the vector function. Eg, the limit as u goes to 0 of (sin(u)/u, 2u+3, (u^3-u)/(u^2+u)) would be (1,3,-1).
sheilaw said:
2.r1(t)=(t,1-t,27+t^2) r2(s)=(9-s,s-8,s^2),find the angle between the two curves
i found t=3,s=6 and the intersection is (3,-2,36). i don't know what should i do next.
next, you need the direction of the two curves at that point. The direction of the curve is found by differentiating each coordinate. For example, the direction of (1/u, 2u+3, u^3-4u) when u=2 can be found by differentiating to get (-1/u^2, 2, 3u^2-4), then substituting u=2 to get (-1/4, 2, 8).
So, differentiate your first curve with respect to t, your second with respect to s, and substitute in your t=3 and s=6. Then you'll have two vectors, parallel to the two curves.
Do you know how to find the angle between two vectors using the dot product? Remember : \(\displaystyle a\cdot b = |a||b|\cos(\theta)\)
