Calculus - space parameterized

Calculus - space parameterized

Postby Gustavo » Sat Feb 09, 2019 2:57 pm

Consider the curve C in the space parameterized by

γ (t) = (sin πt, 2sen πt, cos πt),

with t ∈ R.

a) Determine the tangent lines of curve C at points t = 0 and t = 1/2;

b) Find the point of intersection between the lines determined in the previous item.
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Re: Calculus - space parameterized

Postby HallsofIvy » Wed Mar 06, 2019 11:13 am

I am puzzled at your posting this without showing any attempt yourself to do the problem. Did you just post here so someone could do your homework for you without even trying to do it yourself or do you really not understand anything at all about the problem? Neither speaks well of you!

In any case, this is pretty straight forward. Take the derivative of that vector valued function by taking the derivative of each component: [tex]\left(\pi cos(\pi t), 2\pi cos(\pi t), -\pi sin(\pi t)\right)[/tex]. At t= 0 the point on the curve is [tex]\left(sin(0), 2 sin(0), cos(0)\right)= \left(0, 0, 1\right)[/tex] and the derivative is [tex]\left(\pi cos(0), 2\pi cos(0), -\pi sin(0)\right)= \left(\pi, 2\pi, 0\right)[/tex]. So parametric equations for the tangent line at t= 0 are [tex]x(s)= \pi s[/tex], [tex]y(s)= 2\pi s[/tex] and [tex]z(s)= 1[/tex] with parameter s. That can also written as the vector valued function [tex]\left(\pi s, 2\pi s, 1\right)[/tex].

At t= 1/2, the point on the curve is [tex]\left(sin(\pi/2), 2 sin(\pi/2), cos(\pi/2)\right)= \left(1, 2, 0\right)[/tex] and the derivative is [tex]\left(\pi cos(\pi/2), 2\pi cos(\pi/2), -\pi sin(\pi/2)\right)= \left(\pi, 2\pi, -\pi \right)[/tex] so parametric equations or the tangent line at t= 1/2 are [tex]x(r)= 1+ \pi r[/tex], [tex]y(r)= 2+ 2\pi r[/tex], and [tex]z(r)= -\pi r[/tex] with parameter r. That can also be written as the vector valued fuction [tex]\left(1+ \pi r, 2+ 2pi r, -\pi r\right)[/tex].

Those two lines will intersect if [tex]x= \pi s= 1+ \pi r[/tex], [tex]y= 2\pi s= 2+ 2\pi r[/tex], and [tex]1= -\pi r[/tex]. Those are three equations in only two variables. Such equations do not normally have a solution because two lines in three dimensions do not normally intersect (they are typically skew). Here, from the last equation,[tex]r= \frac{-1}{\pi}[/tex]. Putting that into the first equation, [tex]\pi s= 1+ \pi(-1/\pi)= 1- 1= 0[/tex] so s= 0. Now the question is whether or not those two values, [tex]r= -1/\pi[/tex] and [tex]s= 0[/tex] satisfy the second equation: [tex]2\pi (0)= 0= 2+ 2\pi (-1/pi)= 2- 2[/tex]. Yes, [tex]r= -1/\pi[/tex] and [tex]s= 0[/tex] satisfy all three equations. The two tangent lines intersect at [tex]\left(0, 0, 1\right)[/tex].

Now, the question is, do you have any reason to believe this is correct? I have tried to be pretty clear in my reasoning. Did you understand it? Do you think you could now do a problem like this on your own? You are pretty sure to be given problems like this on a test where you won't be able to ask others to do it for you!

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