Calculus 3 and vectors

Question

Calculus III? I'm confused about vectors stuff. I'm taking cal 3. It uses vectors a lot. My teacher told me something that in cal 3, we don't this idea about slope or tangent in 3d instead we use vectors. Something like that he said. May you answer this question for me, is a vector a line? That's what I understood. I have read a lot articles.I understood that we can't use tangent lines in 3D , because they're planes in 3D which they will cut in different zones of surface. That's why we use vectors, because they look like a line on surface, not a plane. Am I right? What's parametric? What is a vector? I only know that vector has magnitud and direction. I don't find any relationship between vectors and calculus 3; surface thing.

Answer 1

It seems to me that you must accustom yourself to work with parametrical equations and make the correspondence with cartesian equations.

Let me take an example :

Consider the curve with parametric equation $M(t)=(x(t)=t,y(t)=t^2)$, where $M$ is a point (this equation could be written as the cartesian equation $y=x^2$ of a parabola). Differentiate $x$ and $y$ with respect to $t$, you get the vector $V_t=(1,2t)$ ; imagine you want the tangent line at point $M_1=(x_1=1,y_1=1)$ ; knowing that $\vec{V_1}=(1,2)$, you know that $\vec{V_1}$ is a directing vector of this line ; you will get the equation of this tangent line by writing that the current point $M=(x,y)$ of this line is such that $\vec{M_0M}=\binom{x-x_0}{y-y_0}$ is proportional (= have same direction) to vector $\vec{V_1}=(1,2)$

• either by writing that their determinant is $0$ (the most correct and generalizable way):

$$\det(\vec{M_0M},\vec{V_1})=\begin{vmatrix}(x-1) & 1\\(y-1) & 2\end{vmatrix}=0 \ \iff$$

$$(x-1)2=(y-1)1 \ \ \iff \ \ y=2x-1$$

• or by writing the proportionality of their coordinates in the following way :

$$\frac{y-1}{x-1}=\frac{2}{1}=2 ;$$

here you find back what you are accustomed to, i.e., thinking in terms of slopes (the slope of the tangent is its derivative, which is easy by differentiating cartesian equation $y=x^2$ with respect to $x$ to obtain $y'=2$ when $x=1$).

Now (exercise !) reparameterize the same curve by

$$(2t;4t^2)$$

and check that you get the same tangent line but with a tangent vector which will have the same direction as before but will have a different length because with this second parameterization you go faster on this curve.

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Edit 1 :

For example, with the new parametrization, it takes time $t=1/2$ (half a minute) to be at point $(1,1)$ whereas before, with parameterization $(t,t^2)$, you were arriving at this point in time $t=1$ (one minute). Think to the tangent vector as the force (see Edit 2) acting on your car while driving it along this trajectory. Naturally, the force has higher intensity when you drive faster (here its intensity is multiplied by $2$).

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Edit 2 : You have to take sometimes physicist glasses at first and in a second step try to see it defined in mathematical terms :

At first, a vector $\vec{V}$ can be understood intuitively as a force acting,

1) either at the origin.

2) as attached to a point A (in this case a mathematician will say it is $(A,\vec{V})$

3) As acting along a straight line "condemned to slide along this line". In this case, from the mathematician's point of view, a sliding vector is an "equivalence class" defined by:

$$(A,\vec{V}) \ \underbrace{\equiv}_{equivalent} \ (B,\vec{V}) \ \ \iff \ \ \vec{AB} \ \text{is proportional to} \ \vec{V}.$$