Let c be the curve given parametrically in terms of arc length by: x=3s/5+1, y=4s/5-2 (0

Question

Let c be the curve given parametrically in terms of arc length by:  x=3s/5+1, y=4s/5-2 (0<=s<=10).  Find T=T(s).  Sketch the curve and the vector T(5).

blockcolder · Answer

Is T the unit tangent vector?

anonymous · Answer

Sorry, yes.

blockcolder · Answer

A little clarification might be needed.
$$x=\frac{3s}{s+1}, y=\frac{4s}{s-2}$$
Are these the parametric equations?

anonymous · Answer

$$x=\frac35s+1, y=\frac45s-2$$

anonymous · Answer

I'm having a hard time because they're already in terms of s (arc length) and a seem to be a straight line...

alexwee123 · Answer

parametircs arent that hard
just plug in the same value of s into the x and y equations to find the x and y values then just plug them in

alexwee123 · Answer

the grpah

anonymous · Answer

So I've done that, and I get a line... but the vector at T(5)... does it go from (0,0) to (4,2)?

alexwee123 · Answer

vectors can be anywhere on the graph so it can go from anywhere as long as it retains its direction and length

anonymous · Answer

I thought it was supposed to be a tangent vector... but I can't figure out how to graph it...

alexwee123 · Answer

well if you want to graph then you have to know what you're graphing
so first you would differentiate the parametric equations

alexwee123 · Answer

and then put them together into the vector

anonymous · Answer

When I differentiate, I get $$x=\frac35,y=\frac45$$

anonymous · Answer

which gives me the vector $$\left[ \frac35, \frac45 \right]$$

anonymous · Answer

So let's say I graph that... now how do I plug in T(5)?

alexwee123 · Answer

to be honest w/ you I only started learning about this too yesterday and 
my math teacher never really gave me problems like this that doesn't have a constant. so idont really know what to do w/ s=5 but i can tell you if it did  have a constant from the differentiation then you would plug that in but since i don't know that much i can't really give you a definite answer 
sorry :/

anonymous · Answer

Ok, no problem.  Thanks for trying.  I went through like 5 other really hard problems that took multiple pages... and then there's this problem.  It almost seems like a trick question.

alexwee123 · Answer

well im not sure about this but good luck w/ ur other problems :)