Question

Find the range of parametrics

Answer 1

How would I find the range of x(t) and y(t)?

Answer 2

@jim_thompson5910 @Hero @Spacelimbus

Answer 3

@TuringTest

Answer 4

I might be mislead by my english vocabulary, but isn't the range of a function just anything along the y-axis in a given interval?

Answer 5

That is what I think it is but I am not sure.

Answer 6

I think they just want you to find the max/min values of x(t) and y(t), which you could do by taking the derivative and setting it to 0 like in calc I, or you could probably eyeball it

Answer 7

....and checking the endpoints

Answer 8

Could you further explain or guide me through the process?

Answer 9

well, what is the max value of x(t) on \(0\le t\le\pi\) ?\[x(t)=\sin(t^2)\]\[x'(t)=2t\cos(t)=0\implies t=\{0,\frac\pi2\}\]

Answer 10

sorry those could be either max's or mins
you have to check the intervals or do the second derivative test to find out which

Answer 11

So the second derivative would be -2(2 sin(t) + t cos(t))

Answer 12

oh I messed up...

Answer 13

Oh yeah forgot the (t^2) inside

Answer 14

\[x(t)=\sin(t^2)\]\[x'(t)=2t\cos(t^2)\implies t=\{0,\sqrt\frac\pi2,\sqrt\frac{3\pi}2\}\]

Answer 15

yeah, and that changes the possible values of t

Answer 16

I forgot how to find maxes and mins. Could you remind me? It is the first derivative you take and then you make a number line, right?

Answer 17

\[x(t)=\sin(t^2)\]\[x'(t)=2t\cos(t^2)\implies t=\{0,\sqrt\frac\pi2,\sqrt\frac{3\pi}2,\sqrt\frac{5\pi}2\}\]for a critical point t=a such that x'(t)=0,
if x''(a)>0 then the graph is concave up, and t=a is a min of x(t)
if x''(a)<0 then the graph is concave down, and t=a is a max of x(t)

Answer 18

I'm not going to describe the other way (checking the ointervals) because it takes a bit too long to explain properly in my opinion

Answer 19

Could I do this by graphing it? I am allowed to use a calculator

Answer 20

intervals*

Answer 21

yeah, I guess
I'm not sure how good your calculator is (i.e. do you need exact values and will your calculator give them? most calculators wont say \(t=\sqrt\frac\pi2\) is a critical point, but rather that 1.253314137 is (which does not really illustrate the concept very well I'd say)

Answer 22

So would I say the range in from 0 to sqrt(5pi/2) ? I don't really understand how we are going to be getting the range from here

Answer 23

plug in each critical point that we found
look at the value of x(t) for each , and just check the highest and lowest values
that is the range
(I guess you don't need the second derivative test after all)

Answer 24

And this will find only the x range? Then I will have to do it for the y separately?

Answer 25

yes

Answer 26

for y you should be able to see easily that the endpoints are the max and min since it is a linear function

Answer 27

But it is a line wouldn't that mean there are no endpoints?

Answer 28

you are given that t is the interval \(0\le t\le\pi\) so the line is not infinite

Answer 29

Oh true. So would I plug 0 and pi in? T would be =-.5 if you solved it

Answer 30

T=-0.5 ??
what is T ? you mean t
we want to know the max and min for y(t), not t

Answer 31

so plug in 0 and pi for\[y(t)=2t+1\]

Answer 32

Oh yeah So the derivative would give us 2

Answer 33

Dude, no need to differentiate. On the interval [0,2pi] the range of sin(u) is [-1,1]. u=t^2, so our interval is [0,pi^2]. pi^2>2pi, so the range is still [-1,1].

Answer 34

@Herp_Derp oh yes, that is a better way to see it
@daniellerner yeah. but that wont tell us much since that suggests that there is no max or min since \(2\neq0\)
hence we skip that and notice that because it is a line we just plug the endpoints into the function

Answer 35

(of course that only works for a finite interval, otherwise there *would* be no max or min)
yes that is the range of y
herp_derp has given a smarter way to see the range of x, so I suggest you make sense of his answer for that

Answer 36

well, it's not a proof, and since you need more info anyway (like where the first min is) I suggested differentiating for a rigorous method

I said at the beginning that you could eyeball these though...

Answer 37

the range is what y can vary from
since y(t) is continuous, and the y values vary (continuously) from -1 to 1 on \(0\le t\le\pi\) (as does any cos(u) function as herp derp pointed out) wesay the range is [-1,1]

Answer 38

we say*

Answer 39

I meant x(t) above, my bad....

Answer 40

Alright thanks for your help. That made a lot of sense

Answer 41

welcome :)