Mark's Calc III - Individual question feed

How to find the equation of a plane containing two lines?

Anonymous — Wed, 02 Jul 2014 10:32:36 -0500

On the Exam 1 Review sheet, question #13b asks to find an equation of the plane that contains the two lines p(t)=(3+2t,1=t,4+t) and q(t)=(-2+3t,2t,2+t). I found the intersection of these lines to be at (1,2,3). Would the next step be to find the cross product of the direction vectors and use the intersection point as my (a,b,c) in the general equation for a plane? *Comment*: I think you're on the right track!

Number 10 on the review sheet

SpaceManSpiff — Wed, 02 Jul 2014 14:01:57 -0500

Hey, guys. Here's where I am with this one. First I solved for a normal vector $$n=\langle p_{1}-p_{2} \rangle \times \langle p_{3}-p_{2}\rangle $$$$= (\langle 1,-2,3 \rangle - \langle -1,2,2 \rangle) \times (\langle 1,1,2 \rangle - \langle -1,2,2 \rangle)$$$$=\langle 2,-4,1 \rangle \times \langle 2,-1,0 \rangle$$$$=\langle 1,2,6 \rangle$$ Then I define a p nought $$p_{0}=p_{1}=\langle 1,-2,3 \rangle$$ So if I let the coefficients of the plane equation be the components of the normal vector I end up here. $$(x-1)+2(y+2)+6(z-3)=0$$$$x+2y+6z=21$$ So I guess the question is did everyone else get the same result? If not how did you do the problem? Is this the best method or is there a simpler way?

Finding the distance traveled over a length of time (of an object)

Kyouko — Wed, 02 Jul 2014 07:44:06 -0500

Questions 1c and 2 on the review test as us to find an integral that could be solved in order to find the distance traveled by an object over a parametric plot. I can visualize this process, however, I am not quite sure how to go about setting up an integral for this using both \vec{x} = t + cos(4t) and \vec{y} = -sin(4t). Here question 1c is asking us to find the integral described above for the following vector $$ \vec{p}(t) = \langle t + cos(4t), -sin(4t) \rangle $$ and $$ \vec{p'}(t) = \langle 1 - 4sin(4t), -4cos(4t) \rangle $$ I believe setting up the integral would set up using the derivative of the position function integrated over the time interval giving the integral $$ \int_0^{2\pi} \ <1 - 4sin(4t), -4cos(4t)> \mathrm{d}t $$ Any help would be appreciated

From Fundamental circular motion how do we get the tangent line when given t?

asmith14 — Mon, 30 Jun 2014 17:57:28 -0500

So I know fundamental circular motion is $\vec{p} = \langle{\cos(t)},{\sin(t)}\rangle$ and we are given the equation: $\vec{p}(t) = \langle{t} + \cos(4t), {− \sin(4t)}\rangle$ and $t=\pi/3$. I want to know how from fundamental motion we can write out a parametrization equation and with that find a line tangent to that point. *Comment*: Does this refer to a particular problem?

Finding a value so that it is perpendicular to a vector

asmith14 — Tue, 01 Jul 2014 13:09:26 -0500

I thought that if you are looking for a value that is perpendicular to a vector you could use the dot product and just set it equal to zero. Number 7 on the review for Exam 1 says to find a value of t so that $ \langle 1,t, t^2 \rangle$ is perpendicular to $\langle 1,1,1 \rangle $. When I did the dot product and set it equal to zero I used the quadratic formula and it will not work because there is a negative under the radical. $$ \langle 1,t, t^2 \rangle \cdot \langle 1, 1, 1 \rangle = 0 $$ $$ 1+t+t^2 = 0 $$ Then I put those numbers into the quadratic formula: $$ t = \frac{-1\pm\sqrt{1-4(1)(1)}}{(2)(1)} = \frac{-1\pm\sqrt{-3}}{2} $$ So am I doing it wrong, or am I doing it right but my numbers are wrong, or what is going on because this answer does not make sense to me?