Or as a sum of 3 t plus 5 t. So plus 3t plus 5t.

So these first two terms are the same thing as t times t plus 3.

So if I factor a t out, that becomes t times t squared divided by t is t.

So we get t squared plus a plus b t, I can write this is b plus a t as well. plus ab.

So I want this to be t is equal to a, and as t increases, I want this to be t equals b.

So we can view this as t to the 0.

So we could write that as T let me write it this way. We could write that as T of Rn, which is the same thing as the image of T.

So that's t.

So T, E, D is the same as F, E, D.

This might be a little confusing, image of T. So this is sometimes written as just im of T.

So T is F.

So t for time.

So, this is t plus 3 times t plus 5. or t plus 5 times t plus 3 either way

That's what T is equal to, so T is this transformation.

So 1 times t. And v1 is minus 2 times t.

So the convolution of sine of t with cosine of t is 1 2t sine of t.

So d y d t.

So b prime of t.

So they're thinking deep t

I don' t think so.

So when you've evaluate what's the limit of this as t approaches infinity?

So f of t is just going to be sine of t.

And so that is the acceleration vector as a function of t but they care at time t is equal to 3.

So s plus t, this transformation of x we defined as being equal to s of x, this vector plus t of x.

So that's T of 1, 0.

So T would look like that.

So I define some transformation T.

So I'll say b of t.

Don t be so impatient, Christian.

So we can rewrite this right here as a product of t plus 3.

So it's 3 s times the Laplace transform of t to the n minus 1, so t squared.

Well T minus 1 if you remember T is between zero and 1. So T is less than 1. So T minus one, this right here is going to be a negative number.

So if we view f of t as just sine of t or sine of 2t, then we can kind of backwards pattern match.

So now we know t to the nth power, t to the whatever power.

So our integral this was t equals 0 to t is equal to infinity.

So this is t equals 3 I'm sorry, this is t equals 4 years.

So let's take some values of t.

So t is equal to three years.

So Dirac delta of t minus c.

So that's where the T comes from.

So this is my f of t.

So when t 0, where are we?

So let's pick t is equal to 0. t is equal to pi over 2.

So it would be e to the t minus 2 cosine of t minus 2.

So the endpoints work, and if you think about it intuitively, as t increases, so when t is at a, this thing is going to be x of b, y of b.