An object moves along the coordinate line in accordance with the equation s(t) = 3t^4 - 4t^3 +2 , for t0. How close to the origin does the object come?

Question

An object moves along the coordinate line in accordance with the equation s(t) = 3t^4 - 4t^3 +2 , for t>0. How close to the origin does the object come?

amistre64 · Answer

i wonder if we could devise a manner in which to find the shortest vector from 0,0 to the curve

amistre64 · Answer

the shortest distance is perpendicular to the slope of the curve, which may be a fabrication in my own mind tho

amistre64 · Answer

if we graph it it appears to be closer and closer as it approaches 0

amistre64 · Answer

http://www.wolframalpha.com/input/?i=3t^4+-+4t^3+%2B2

amistre64 · Answer

but that graph may be misleading since it is not the same scale for x and y

amistre64 · Answer

if we make it a parametric curve tho:

$$x=t$$
$$y = 3t^4 - 4t^3 +2$$
$$r = \sqrt{x^2 + y^2}$$
$$r = \sqrt{t^2 + (3t^4 - 4t^3 +2)^2}$$
$$r = \sqrt{t^2 + 9t^8-24t^7+16t^6+12t^4-16t^3+4}$$
$$r = \sqrt{9t^8-24t^7+16t^6+12t^4-16t^3+t^2+ 4}$$
this should be the equation for distance from the origin

amistre64 · Answer

we could take the derivative of that and find its min and max to determine at what "t" has the shortest distance

amistre64 · Answer

but thats as far as i think i can take it for tonight :)  if im right that is

amistre64 · Answer

$$r = \sqrt{9t^8-24t^7+16t^6+12t^4-16t^3+t^2+ 4}$$
$$r' = \frac{72t^7-168t^6+96t^5+48t^3-48t^2+2t}{2\sqrt{9t^8-24t^7+16t^6+12t^4-16t^3+t^2+ 4}}= 0$$
at abt t = 0, t = .04, and t = .9