r(t)=(2t-1)i+(4-t)j ; t=3 a. sketch the curve in the xy-plane determined by r(t) and indicate the orientation b, find r't and sketch r(t) and r'(t) for the indicated value t.

\[x(t)=2t-1\]\[y(t)=4-t\]use the formula for slope\[m={y(a)-y(b)\over x(a)-x(b)}\]to get the formula for the line in point-slope form\[y=y_1=m(x-x_1)\]

that should help you draw the curve

okay, that makes sense.. when i integrate, the t goes away.. how am i supposed to draw a line tangent to curve at point t .. if there is no t?

why are we integrating?

sorry! i mean when i derive! to get r'(t)

well you have a vector from that derivative, but the direction of it just doesn't depend on t (which should make sense since t is a straight line)

*since r is a straight lint

*line

the only thing you want to do is put the tangent vector in the right place, so find the point r(3) and draw the tangent there (it should coincide with the line of course)

also, i'm kind of confused on how to use the slope formula for looking at an equation? i think i've only done it when i have two points on x and two points on y so (y2-y1)/(x2-x1) but here i have two equations, and i'm kind of confused... :(

but you have expressions for x and y all the same what is since \[\vec r(t)=(2t-1)\hat i+(4-t)\hat j\]we have\[y(t)=2t-1\]\[x(t)=4-t\]so what is y(0) for example ?

y(0) = -1

x and y are switched above, my bad...

so, do i just pick random numbers to plug into my equation, to get the points? according to the book, i'm supposed to get rid of my parameters?

\[\vec r(t)=(2t-1)\hat i+(4-t)\hat j\]\[x(t)=2t-1\]\[y(t)=4-t\]so now you can get \(x(0),y(0),x(1),y(1)\) whatever the parameters will go away when we are done, just continue along... (you can see that by plugging in the numbers we are already eliminating t)

so x(o)= -1, x(1) = 1 y(o)=4, y(1) =3 does .. this mean that for r(o) its (-1,4) and (1, 3) and those are the two points i use in my slope form to get a slope of -1/2 ??`

and r(1) = (1,3) i mean

yes the slope is -1/2, r(0) is where x(0) and y(0) come from now you need to use point-slope form (at least that is how the book seems to want you to do it)

\[y-y(a)=m(x-x(a))\]where a is any value of t that should give you an equation that you can get into the form \(y=mx+b\) which is easy to graph

is it... \[y=(-1/2)x +1/2 ??\] but then, why do i care about t=3?, esp if i'm graphing in terms of x?

i guess i'm having a hard time relating the equation of the line i just drew to the vector..

you will see it in a second! the point t=3 is where they ask you to draw the tangent from, so that is only about part b, which we are about to do... first, what is r'(t) ?

r'(t)=2i+j

careful... \(\vec r'(t)=2\hat i-\hat j\)

|dw:1331588740537:dw| is that the line? o.O

thats a 1/2 haha

oops! you're right haha thats what i have on my notes, just copied it down wrong

the equation you have has the wrong intercept:\[y-y(0)=-\frac12(x-x(0))=y-4=-\frac12_+1)\to y=-\frac12x+\frac72\]but yeah, something like your graph (note that it moves to the right as t increases, so they draw arrows usually...)

|dw:1331589007546:dw|

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