Find the curve given the tangent

[madachi]

Homework Statement

Given that the tangent to the curve [itex] c(t) [/itex] at any point on the curve is [itex] T(t) = (-sin(t), cos(t) )[/itex], find [itex] c(t) [/itex] if the curve passes through the point [itex] (0,0) [/itex].

The Attempt at a Solution

I try to let
[itex] c(t) = ( x(t), y(t) ) [/itex]
Then
[itex] c'(t) = ( x'(t), y'(t) ) [/itex]
[itex]| c'(t) | = \sqrt{[x'(t)]^2 + [y'(t)]^2 } [/itex]
And
[itex] T(t) = \frac{c'(t)}{|c'(t)|} [/itex]

However this is complicated and consequently I am not sure how to solve it. I am also not sure how to "use" the point given since (0, 0) correspond the the values x and y respectively rather than t.

Thanks.

 

[rock.freak667]

If you had c'(t) =<2t,1> then c(t) = <t²+A,t+B> where A,B=constant.


But you have

c'(t)/|c'(t)| = <-sint,cost>

so what is c(t) ? (note that |c'(t)| is just a constant)

At (0,0), t=0. So what is c(t) now?

 

[madachi]

If you had c'(t) =<2t,1> then c(t) = <t²+A,t+B> where A,B=constant.


But you have

c'(t)/|c'(t)| = <-sint,cost>

so what is c(t) ? (note that |c'(t)| is just a constant)

At (0,0), t=0. So what is c(t) now?

c(t) = (cos(t) + C, sin(t) + K)
t= 0, point is (0,0)

So x(t) = 1 + C = 0, C = -1
And y(t) = 0 + K, K=0

So c(t) = (cos(t) -1 , sin(t) ) ?

 

[rock.freak667]

c(t) = (cos(t) + C, sin(t) + K)
t= 0, point is (0,0)

So x(t) = 1 + C = 0, C = -1
And y(t) = 0 + K, K=0

So c(t) = (cos(t) -1 , sin(t) ) ?

Yes but you forgot out |c'(t)| which is the distance from the center to any tangent. In this case it would just be the same as |T(t)|