# A bend in the road, find a tangent and normal components of acceleration of the car.

So, I am given a beginning vector $$r(t) =[t,t^2]$$ for $$-2 \leq t \leq 2$$ through a sharp bend in the road.

What I have done so far: I found velocity with the derivative of r(t) and got vector: $$[1,2t]$$ I found acceleration with the derivative of v(t) and got vector: $$[0,2]$$

What I cannot figure out: The changing direction, or $$N = \frac{\frac{dT}{dt}}{\left|\frac{dT}{dt}\right|}$$

conceptually, I understand how this works. In execution, however. . . .I am at a loss.

The book yields $$N = \frac{[-2t,1]}{\sqrt{1+4t^2}}$$.

I am not sure HOW the book came to this conclusion, but here is what I have tried so far in this:

1. Basic intuitive approach. $$\frac{[-2t,1]}{\sqrt{1+4t^2}} = \frac{[0,1]}{\sqrt{1+4t^2}}$$ Why I did this: $$T = \frac{v}{|v|}$$, right? So that gives us $$\frac{[1,2t]}{\sqrt{1+4t^2}}.$$

So, deriving $$1$$ and $$2t$$ I would get $$\frac{[0, 2]}{\sqrt(1+4t^2)}.$$

which would be $$T$$, and so I need to find $$\frac{[-2t,1]}{\sqrt{1+4t^2}}$$, which is derivative of $$T$$ divided by the magnitude of $$T$$.

That is basically $$\frac{\sqrt{0+4}}{{1+4t^2}}.$$ Which gives me $$[0,1]$$ for $$N$$ once we cancel out the $$\sqrt{1+4t^2}$$.

Clearly, this was wrong as the book came to another conclusion.

I'm now attempting a longer approach, but I still fear it is incorrect.

any assistance or insight would be appreciated.

• The title and the question don't match. The title says that you need to "find a tangent and normal components of acceleration". But in the body of the question your goal seems to be the unit normal vector $\mathbf{N}(t)$. Please clarify, and edit either the title or the question. – zipirovich Sep 27 '18 at 17:42

So, deriving $$1$$ and $$2t$$ I would get …

And that's where it went wrong. You completely ignored the denominator of the expression for $$\mathbf{T}(t)$$, which of course then resulted in a wrong answer.

Look at it this way.You have found $$\mathbf{T}(t)$$ to be $$\mathbf{T}(t)=\frac{[1,2t]}{\sqrt{1+4t^2}}=\left[\frac{1}{\sqrt{1+4t^2}},\frac{2t}{\sqrt{1+4t^2}}\right].$$ So the components of $$\mathbf{T}(t)$$ that you must differentiate are $$\frac{1}{\sqrt{1+4t^2}}$$ and $$\frac{2t}{\sqrt{1+4t^2}}$$, NOT "$$1$$" and "$$2t$$". It's just like with usual functions in Calculus I: the derivative of a fraction $$f(x)=\frac{p(x)}{q(x)}$$ is typically found by the Quotient Rule; but you can simply ignore the denominator and say that "$$f'(x)=\frac{p'(x)}{q(x)}$$".

Anyways, in your solution that's the main mistake. Although what happens next isn't quite clear either. Make sure that you clearly label all vectors and quantities in your work. For example, then you say

… I would get $$\frac{[0,2]}{\sqrt(1+4t^2)}$$ which would be $$T$$

but that's not true: that wouldn't be $$\mathbf{T}$$. You've already found $$\mathbf{T}(t)$$, see above. By deriving it, you will find its derivative $$\frac{d\mathbf{T}(t)}{dt}$$, which is the numerator for the formula for the unit normal vector $$\mathbf{N}(t)=\frac{\frac{d\mathbf{T}}{dt}}{\left|\frac{d\mathbf{T}}{dt}\right|}$$.

• I understand, alright. Thank you. – The_Senate Sep 27 '18 at 18:12
• @The_Senate: You're very welcome! If this helps, you could vote on it, and maybe even accept this answer... – zipirovich Sep 27 '18 at 18:20
• Can you just do one thing for me? Say : "accept this answer, in the name of the Senate." ? – The_Senate Sep 28 '18 at 19:34