I need some help trying to do the following:

Given $$L: (x,y)=(\sin t , 1+3\sin t)~, \quad 0<t< \pi $$

Find $a, b, c$ and $d$ so that

$$L_2: (x , y) = (-2,-5)+u(a;b)~, \quad c<u<d$$

represents the same points that $L$ does.

I don't know where I should begin. If somebody knows how to do it, please give me some hints.


I'm going to rewrite your question as $L(x,y)=(\sin(t),1+3\sin(t))$, for $0<t<π$.

And $L_2(x,y)=(-2,-5)+u(a,b)$ for $c<u<d$.

Well, if we analyze $L$ first, we see that its actually just a line because we have something of the form $(x, 1+3x)$ which just describes a line. This line segment runs from $(0,1)$ (at $t=0$) to $(1,4)$ (at $t=\frac{π}{2})$ right back to $(0,1)$ (at $t=0$).

So this graph is just the segment from $(0,1)$ to $(1,4)$.

Now, if we look at $(-2,-5)$, you may notice that it actually lies on the same line as $(0,1)$ and $(1,4)$.

Therefore, we can use $L_2(x,y)=(-2,-5)+u(1,3)$.

We find out that if we let $u=2$ and $u=3$, we get $(0,1)$ and $(1,4)$ respectively.

Therefore, we have $L_2(x,y)=(-2,-5)+u(1,3)$ for $2<c<3$.


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