Parametric Equations for Curve y = 2 - 7x^3 at Point (0,2,25)

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Homework Statement


Find parametric equations for the curve y = 2 - 7x3 through the point (0, 2, 25), parallel to the xy-plane.


Homework Equations





The Attempt at a Solution



The answer is:
x(t) = t
y(t) = 2-7t^3
z(t) = 25

but I don't know how to get there. Any help please?
 
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What's the problem? It looks like they want you to take the curve y=2-7x^3 in the x-y (z=0) plane and translate up to the z=25 plane and express in parametric form. This is far from the hardest problem you've posted.
 
well.. I don't know how to get the parametric equations from this..
 
Here's a 'simpler' example. The curve is y^2=x in the plane. Pick t=y. Then y(t)=t and x(t)=y(t)^2=t^2 is the parametric version.
 
I know how to do y^2
 
-EquinoX- said:
I know how to do y^2

How is your problem different?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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