Question

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.

Answer 1

\[x(t)=2t-1\]\[y(t)=4-t\]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\[y=y_1=m(x-x_1)\]

Answer 2

that should help you draw the curve

Answer 3

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?

Answer 4

why are we integrating?

Answer 5

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

Answer 6

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)

Answer 7

*since r is a straight lint

Answer 8

*line

Answer 9

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)

Answer 10

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... :(

Answer 11

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 ?

Answer 12

y(0) = -1

Answer 13

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

Answer 14

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?

Answer 15

\[\vec r(t)=(2t-1)\hat i+(4-t)\hat j\]\[x(t)=2t-1\]\[y(t)=4-t\]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)

Answer 16

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 ??`

Answer 17

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

Answer 18

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)

Answer 19

\[y-y(a)=m(x-x(a))\]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

Answer 20

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?

Answer 21

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

Answer 22

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) ?

Answer 23

r'(t)=2i+j

Answer 24

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

Answer 25

is that the line? o.O

Answer 26

thats a 1/2 haha

Answer 27

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

Answer 28

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...)