1. ## Parametric Equations

I need some help on some of these problems...we very briefly went over the theory in class, with no real explanation or any examples!

1) Consider a parametric curve given by x(t)=t^2+28t+13 and y(t)=t^2+28t+4. How many units of distance are covered by the point P(t) = (x(t),y(t)) between t=0, and t=6 ?

I came up with 168, using the integral from 0 to 6 of sqrt(x'(t)^2+y'(t)^2)dt, but it's wrong

2) Find the length of the arc formed by x^2=5y^3 from point A to B, where A=(0,0) and B=(25,5).

I've never seen a problem like this, nor can I seem to find an example of one. Can anyone give me a push in the right direction?

3)Find the length of the parametrized curve formed by x(t)=0t^3+6t^2+6t and y(t)=-4t^3-6t^2+0t, where t goes from 0 to 1

I used the same formula as in #1, and ended up with a messy integral I couldnt really evaluate

4) Suppose a curve is traced by the parametric equations x= 5sin(t) and y=24-10cos(t)^2-20sin(t). At what point (x,y) is the tangent line horizontal?

I know that the tangent line is horizontal when the derivative is zero, but I don't know what to derive and such.

Thank you for any help!

2. Originally Posted by mistykz
I need some help on some of these problems...we very briefly went over the theory in class, with no real explanation or any examples!

1) Consider a parametric curve given by x(t)=t^2+28t+13 and y(t)=t^2+28t+4. How many units of distance are covered by the point P(t) = (x(t),y(t)) between t=0, and t=6 ?

I came up with 168, using the integral from 0 to 6 of sqrt(x'(t)^2+y'(t)^2)dt, but it's wrong
The above integral should have worked but I think the is easier.

note that
$\displaystyle x=t^2+28t+13$
$\displaystyle y=t^2+28t+4$ subrace them from each other and we get

$\displaystyle x-y=9 \iff y=x-9$

x(0)=13 and x(6)=217

now the arc length integral is

$\displaystyle \int_{13}^{217}\sqrt{1+1}dx=\sqrt{2}(217-13)=204\sqrt{2}$

Good luck.

3. Hello, mistykz!

1) Consider a parametric curve given by: .$\displaystyle \begin{array}{cccc} x(t)&=&t^2+28t+13 & {\color{blue}[1]}\\ y(t) &=& t^2+28t+4 & {\color{blue}[2]}\end{array}$

How many units of distance are covered by the point $\displaystyle P(t) \,= \,(x,\:y)$
between $\displaystyle t=0 \text{ and }t=6$ ?

$\displaystyle \begin{array}{cccc}\text{From {\color{blue}[1]}:} & t^2+28t & =& x - 13 \\ \text{From {\color{blue}[2]}:} & t^2+28t &=&y-4 \end{array}\quad\Rightarrow\quad y - 4 \:=\:x-13\quad\Rightarrow\quad y \:=\:x-9$

The graph is a straight line.

$\displaystyle \begin{array}{ccccc}\text{When }t=0\!: & x = 13 & y = 4 & \Rightarrow & (13,4) \\ \text{When }t=6\!: & x = 217 & y = 208 & \Rightarrow & (217,208) \end{array}$

The distance is: .$\displaystyle d \;=\;\sqrt{(217-13)^2 + (208-4)^2} \;=\;\sqrt{204^2 + 204^2} \;=\;\boxed{204\sqrt{2}}$