parametric equations again

[Mel Mitch]

Hello again,
this time no trigs.....so if you can understand how to form the equation....please guide me.

Problem 1

A curve is defined parametrically by the equations

x= t^3 - 6t + 4 , y= t - 3 + 2/t
find the equation

Problem 2
(A)
A curve is given by the parametric equations:

x= (1-t)/(1+t), y=(1-t)(1+t)^2
find the equation

(b)
find dy/dx in terms of t

 

[galactus]

Problem 2
(A)
A curve is given by the parametric equations:

\(\displaystyle x= \frac{1-t}{1+t}, \;\ y=(1-t)(1+t)^{2}\)
find the equation

Solve x for t and sub into y.

\(\displaystyle x=\frac{1-t}{1+t} \;\ \Rightarrow t=\frac{1-x}{x+1}\)

\(\displaystyle y=\left(1-\frac{1-x}{x+1}\right)\left(1+\frac{1-x}{x+1}\right)^{2}=\frac{8x}{(x+1)^{3}}\)

(b)
find dy/dx in terms of t

\(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)

\(\displaystyle \frac{dy}{dt}= -(t+1)(3t-1)\)

\(\displaystyle \frac{dx}{dt}=\frac{-2}{(t+1)^{2}}\)

\(\displaystyle \frac{ -(t+1)(3t-1)}{\frac{-2}{(t+1)^{2}}}=\frac{(t+1)^{3}(3t-1)}{2}\)

Now, see what to do. Try the first one.

 

[soroban]

Hello, Mel Mitch!

I haven't solved #1, but I've made some observations . . .

\(\displaystyle \text{(1) A curve is defined parametrically by the equations: }\;\begin{array}{ccc}x &=& t^3 - 6t + 4 \\ y &=& t - 3 + \frac{2}{t} \end{array}\)
. . . \(\displaystyle \text{Find the equation.}\)

\(\displaystyle \text{I've noted that: }\;\begin{array}{ccc}x &=& (t-2)(t^2+2t-2) \\ \\[-3mm]y &=& \dfrac{(t-2)(t-1)}{t} \end{array}\)


\(\displaystyle \text{Is anyone inspired? } \hdots \text{ Anyone? }\;\text{Anyone?}\)

 

[Mel Mitch]

thank u galactus and soroban......going to do number 1........

 

[BigGlenntheHeavy]

\(\displaystyle x \ = \ t^{3}-6t+4 \ \ \ \ y \ = \ t-3+\frac{2}{t}\)

\(\displaystyle \frac{dx}{dt} \ = \ 3t^{2}-6 \ = \ 3(t^{2}-2) \ and \ \frac{dy}{dt} \ = \ 1-\frac{2}{t^{2}} \ = \ \frac{t^{2}-2}{t^{2}}\)

\(\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \ = \ \frac{1}{3t^{2}}, \ t \ doesn't \ = \ \pm\sqrt 2.\)

\(\displaystyle Now, \ from \ soroban \ inspiration, \ y \ = \ \frac{x(t-1)}{t(t^{2}+2t-2)}\)

\(\displaystyle Ergo, \ \frac{dy}{dx} \ = \ \frac{(t-1)}{t(t^{2}+2t-2)}\)

\(\displaystyle Hence, \ \frac{1}{3t^{2}} \ = \ \frac{t-1}{t(t^{2}+2t-2)}\)

\(\displaystyle This \ implies \ that \ 3t^{3}-3t^{2} \ = \ t^{3}+2t^{2}-2t. \ which \ implies \ t(2t-1)(t-2) \ = \ 0.\)

\(\displaystyle Now \ if \ t=0, \ y \ is \ undefined, \ and \ if \ t=2, \ y \ and \ x \ = \ 0. \ If \ t \ = \ \frac{1}{2}, \ then \ x=\frac{9}{8} \ and \ y \ = \ \frac{3}{2}.\)

\(\displaystyle I \ get \ y \ = \ \frac{4x}{3} \ when \ t \ = \ \frac{1}{2}.\)

 

[Mel Mitch]

thanks big.G

 

[Mel Mitch]

Now the question uses the same parametric equations.....
how could i find for the equations of the normals or the equations of the tangents, where the curve meets the x-axis.
using the below:

x= t^3 - 6t + 4 , y= t - 3 + 2/t

Now u dont have to do all for them the first equation either normal or tangent would be helpfulfor me to understand how complete the second equation and future problems.

 

[Deleted member 4993]

Now the question uses the same parametric equations.....
how could i find for the equations of the normals or the equations of the tangents, where the curve meets the x-axis.
using the below:

x= t^3 - 6t + 4 , y= t - 3 + 2/t

Now u dont have to do all for them the first equation either normal or tangent would be helpfulfor me to understand how complete the second equation and future problems.

Follow the method shown in (by arthur):

viewtopic.php?f=3&t=35434&p=137680#p137680