Question

Find the decomposition of a(t) into tangenital and normal components at the point indicated

Answer 1

\[r(t)=<e^t,1-t>\]\[t=0\]

Answer 2

Okay, let's compute our derivatives:$$\mathbf{r}(t)=(e^t,1-t)\\\mathbf{r}'(t)=(e^t,-1)\\\mathbf{r}''(t)=(e^t,0)$$Can you determine our unit tangent vector from this?

Answer 3

The second one!

Answer 4

\[a_T = \frac{ r'(t)*r''(t) }{ ||r'(t)|| }\]
\[a_N = \frac{ ||r'(t) \times r''(t)|| }{ ||r'(t)|| }\]

* (dot product)

Answer 5

@FutureMathProfessor that's not our unit tangent vector, that's just *a* tangent vector. To make it unit length, find its magnitude:$$\|\mathbf{r}'(t)\|=\sqrt{e^{2t}+1}$$and now we scale \(\mathbf{r}'\) by its reciprocal:$$\mathbf{T}(t)=\frac1{\sqrt{e^{2t}+1}}\mathbf{r}'(t)=\left(\frac{e^t}{\sqrt{e^{2t}+1}},-\frac1{\sqrt{e^{2t}+1}}\right)$$

Answer 6

@Euler271 that is incorrect by the way :-)

Answer 7

So then if I plugged in t=0, the vector would be
(1/sqrt(2),-1/sqrt(2))

Answer 8

You're finding the unit vector. I'm finding the actual magnitude (times the unit vector)

Not sure what the question is asking

Answer 9

@Euler271 that formula is incorrect for \(a_N\)... you need \(\|r''(t)\|\) in that denominator hehe.

Answer 10

@FutureMathProfessor correct! Now, compute our acceleration vector at \(t=0\):$$\mathbf{r}''(0)=(1,0)$$Okay, now recall that for a twice-differentiable vector valued function \(\mathbf{r}'\) we know that our acceleration vector lies in the plane spanned by the unit tangent and unit normal vectors:$$$$