Finding unit tangent and principal unit normal vectors

[ksdhart]

Hi all. I'm having more difficulties with vectors. Exercise #29 from Section 11.3 says:

For each of the vector-valued functions in Exercises #29-34, find the unit tangent vector and the principal unit normal vector at the specified value of t.

29) $\displaystyle \vec{r}(t)=<t,t^2>$ @ t = 1

My book gives two theorems for finding the unit tangent vector and the principal unit normal vector, and I tried to follow them, but encountered problems with the principal unit normal vector. The unit tangent vector is:

$\displaystyle \vec{T}(t) = \frac{\vec{r}\,'(t)}{||\vec{r}\,'(t)||}$

So I found the derivative of the given vector:

$\displaystyle \vec{r}\,'(t)=<1,2t>$

$\displaystyle ||\vec{r}\,'(t)||=\sqrt{1^2+\left(2t\right)^2}= \sqrt{1+4t}$

$\displaystyle \vec{T}(t)=\left(1+4t\right)^{-\frac{1}{2}}\cdot <1,2t>$

$\displaystyle \vec{T}(1)=\left(1+4\right)^{-\frac{1}{2}}\cdot <1,2>=<\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}>=<\frac{\sqrt{5}}{5},\frac{2\sqrt{5}}{5}>$

And that matches the answer in the back of the book for the unit tangent vector at t=1. Then the next theorem from my book says the principal unit normal vector is:

$\displaystyle \vec{N}(t)=\frac{\vec{T}\,'(t)}{||\vec{T}\,'(t)||}$

So I took the derivative of the unit tangent vector:

$\displaystyle \vec{T}(t)=<\frac{1}{\sqrt{1+4t}},\frac{2t}{\sqrt{1+4t}}>$

$\displaystyle \vec{T}\,'(t)=<-\frac{2}{\left(1+4t\right)^{\frac{3}{2}}},\frac{4t+2}{\left(1+4t\right)^{\frac{3}{2}}}>$

$\displaystyle ||\vec{T}\,'(t)||=\sqrt{\frac{4}{\left(1+4t\right)^3}+\frac{\left(4t+2\right)^2}{\left(1+4t\right)^3}}$

$\displaystyle \vec{T}\,'(1)=<-\frac{2}{\left(1+4\right)^{\frac{3}{2}}},\frac{4+2}{\left(1+4\right)^{\frac{3}{2}}}>=<-\frac{2\sqrt{5}}{25},\frac{6\sqrt{5}}{25}>$

$\displaystyle ||\vec{T}\,'(1)||=\sqrt{\frac{4}{\left(1+4\right)^3}+\frac{\left(4+2\right)^2}{\left(1+4\right)^3}}=\sqrt{\frac{4}{125}+\frac{36}{125}}=\frac{2\sqrt{2}}{5}$

$\displaystyle \vec{N}(1)=\frac{5}{2\sqrt{2}}\cdot <-\frac{2\sqrt{5}}{25},\frac{6\sqrt{5}}{25}>=<-\frac{\sqrt{10}}{10},\frac{3\sqrt{10}}{10}>$

Every step along the way looks fine to me... but I don't get the same answer as the back of the book. Did I miss up somewhere or is the book wrong? Their answer is:

$\displaystyle \vec{N}(1)=<-\frac{2\sqrt{5}}{5},-\frac{\sqrt{5}}{5}>$

 

[pka]

My book gives two theorems for finding the unit tangent vector and the principal unit normal vector, and I tried to follow them, but encountered problems with the principal unit normal vector. The unit tangent vector is:
$\displaystyle \vec{T}(t) = \dfrac{\vec{r}\,'(t)}{||\vec{r}\,'(t)||}$
So I found the derivative of the given vector:

You have wasted a lot of effort by ignoring the @t=1.
You can simply stop where I stopped the quote.
$\displaystyle \vec{T}(1) = \dfrac{\vec{r}\,'(1)}{||\vec{r}\,'(1)||}$

 

[Ishuda]

Hi all. I'm having more difficulties with vectors. Exercise #29 from Section 11.3 says:



My book gives two theorems for finding the unit tangent vector and the principal unit normal vector, and I tried to follow them, but encountered problems with the principal unit normal vector. The unit tangent vector is:

$\displaystyle \vec{T}(t) = \frac{\vec{r}\,'(t)}{||\vec{r}\,'(t)||}$

So I found the derivative of the given vector:

$\displaystyle \vec{r}\,'(t)=<1,2t>$

$\displaystyle ||\vec{r}\,'(t)||=\sqrt{1^2+\left(2t\right)^2}= \sqrt{1+4t}$

$\displaystyle \vec{T}(t)=\left(1+4t\right)^{-\frac{1}{2}}\cdot <1,2t>$

$\displaystyle \vec{T}(1)=\left(1+4\right)^{-\frac{1}{2}}\cdot <1,2>=<\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}> = <\frac{\sqrt{5}}{5},\frac{2\sqrt{5}}{5}>$

And that matches the answer in the back of the book for the unit tangent vector at t=1. Then the next theorem from my book says the principal unit normal vector is:

$\displaystyle \vec{N}(t)=\frac{\vec{T}\,'(t)}{||\vec{T}\,'(t)||}$

So I took the derivative of the unit tangent vector:

$\displaystyle \vec{T}(t)=<\frac{1}{\sqrt{1+4t}},\frac{2t}{\sqrt{1+4t}}>$

$\displaystyle \vec{T}\,'(t)=<-\frac{2}{\left(1+4t\right)^{\frac{3}{2}}},\frac{4t+2}{\left(1+4t\right)^{\frac{3}{2}}}>$

$\displaystyle ||\vec{T}\,'(t)||=\sqrt{\frac{4}{\left(1+4t\right)^3}+\frac{\left(4t+2\right)^2}{\left(1+4t\right)^3}}$

$\displaystyle \vec{T}\,'(1)=<-\frac{2}{\left(1+4\right)^{\frac{3}{2}}},\frac{4+2}{\left(1+4\right)^{\frac{3}{2}}}>=<-\frac{2\sqrt{5}}{25},\frac{6\sqrt{5}}{25}>$

$\displaystyle ||\vec{T}\,'(1)||=\sqrt{\frac{4}{\left(1+4\right)^3}+\frac{\left(4+2\right)^2}{\left(1+4\right)^3}}=\sqrt{\frac{4}{125}+\frac{36}{125}}=\frac{2\sqrt{2}}{5}$

$\displaystyle \vec{N}(1)=\frac{5}{2\sqrt{2}}\cdot <-\frac{2\sqrt{5}}{25},\frac{6\sqrt{5}}{25}>=<-\frac{\sqrt{10}}{10},\frac{3\sqrt{10}}{10}>$

Every step along the way looks fine to me... but I don't get the same answer as the back of the book. Did I miss up somewhere or is the book wrong? Their answer is:

$\displaystyle \vec{N}(1)=<-\frac{2\sqrt{5}}{5},-\frac{\sqrt{5}}{5}>$

Check the calculation of the norm of
||$\displaystyle \vec{r}$'(t)|| = [1 + 4 t]^1/2?

Note that if t=1 then t=t².

 

[ksdhart]

You have wasted a lot of effort by ignoring the @t=1.
You can simply stop where I stopped the quote.
$\displaystyle \vec{T}(1) = \dfrac{\vec{r}\,'(1)}{||\vec{r}\,'(1)||}$

I see what you mean, and I do appreciate the hint. The only problem is that that shortcut doesn't help me with the second part of the exercise, finding the principal unit normal vector. I can't use T(1) in my calculations to find N(1). I needed to find T(t) anyway, so I didn't really waste any effort at all. I just did the steps in a different order than you would have.

Check the calculation of the norm of
||$\displaystyle \vec{r}$'(t)|| = [1 + 4 t]^1/2?

Note that if t=1 then t=t².

I'm sorry, but I don't understand what you mean by this. I may just be being obtuse and not seeing a nice hint, but I don't know how that helps me at all. My answer for the unit tangent vector agrees with the back of the book, but things go awry after when I try to find principal unit normal vector.

 

[Ishuda]

...
I'm sorry, but I don't understand what you mean by this. I may just be being obtuse and not seeing a nice hint, but I don't know how that helps me at all. My answer for the unit tangent vector agrees with the back of the book, but things go awry after when I try to find principal unit normal vector.

You have (and use)
$\displaystyle ||\vec{r}\,'(t)||=\sqrt{1^2+\left(2t\right)^2}= \sqrt{1+4t}$
but isn't the correct value
$\displaystyle ||\vec{r}\,'(t)||=\sqrt{1^2+\left(2t\right)^2}= \sqrt{1+4t^2}$

Note that
$\displaystyle ||\vec{r}\,'(1)|| = \sqrt{1+4*(1)} = \sqrt{1+4*(1)^2} = \sqrt{5}$
However, the derivative will certainly depend on whether the t or t² is correct.

 

[ksdhart]

Oh, wow! I can't believe I didn't notice that before. Thanks for pointing out my error. Everything fell into place once I fixed that oversight.