Question Find the unit tangent vector of r(t) = -5sin(5t)i +...

Question Find the unit tangent vector of \textbf{r}(t) = -5\sin(5t)\textbf{i} + 5\cos(t)\textbf{j} - 9t^4\textbf{k} at $t = \frac{\pi}{2}$. Give an exact answer. Sorry, that's incorrect. Try again? $\textbf{T}\left(\frac{\pi}{2}\right) = 0\textbf{i} + \frac{-5}{\sqrt{25 + \left(\frac{9\pi^2}{2}\right)^2}}\textbf{j} + \frac{\frac{9\pi^3}{2}}{\sqrt{25 + \left(\frac{9\pi^2}{2}\right)^2}}\textbf{k}$

Transcript

00:01 In this problem, we are given the vector r of t equals to t squared minus 25 i vector plus t times j vector.

00:15 And we are asked to calculate t of 5 as well as n of 5, that is the tangential and the normal vectors.

00:27 So now let us begin by t of 5.

00:30 We know that t of t equals to r dash of t over the magnitude of r dash of t.

00:39 So now let us differentiate r of t with respect to t.

00:43 We obtain 2 times t minus 0 times i vector plus 1 times j vector.

00:51 So this equals to 2 t i vector plus j vector.

00:56 Now let us find the magnitude of this vector.

01:00 So we have the magnitude of r dash of t to be equal to.

01:04 Square root of 2 t the whole squared plus 1 the whole squared and this equals to square root of 4 t squared plus 1.

01:16 So now since we need to calculate at t equals to 5 let us substitute this, we obtain r dash of 5 to be equal to 2 times 5 which is 10.

01:27 So we get 10 i vector plus j vector and the modulus of this, that is the magnitude of, that is the magnitude of, this equals to square root of 10 squared which is 100 plus 1 which is 1 again.

01:45 So we get square root of 101.

01:49 Substituting this in the formula of t of t we obtain t of t equals to 10 i plus j the whole divided by square root of 101.

02:03 So therefore this is our final answer.

02:13 Next, next we need to calculate n of t as well.

02:17 So first let us look at the formula of n of t...

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