Derivative as Tangent

The derivative of a vector function of one variable f(t) at a point is tangent to the graph at that point. One must be careful here. The vector f'(t) starts at the origin (either (0,0) or (0,0,0) ). To see that f'(t) is tangent to the graph at the point, the vector must be translated to begin at the point, that is, consider the vector f(t)+f'(t).

In[17]:=

  Clear[x,y,f]
  x[t_]:=t
  y[t_]:=t^2
  f[t_]:={x[t],y[t]}

In[18]:=

  p1=ParametricPlot[f[t],{t,-5,5}];

In[19]:=

  f'[2]

Out[19]=

  {1, 4}

In[20]:=

  p2=vectorPlot[{{0,0}},{f'[2]}];

In[21]:=

  Show[p1,p2];

As you can see, we do not have a tangent vector. But we will when we translate it to begin at the point on the curve.

In[22]:=

  p3=vectorPlot[{{0,0}+f[2]},{f'[2]+f[2]}];

In[23]:=

  Show[p1,p3,AspectRatio->Automatic];

Let's look at some more.

In[24]:=

  p5=vectorPlot[Table[f[i],{i,5}],Table[f[i]+f'[i],{i,5}]];

In[25]:=

  Show[p1,p5,AspectRatio->Automatic,PlotRange->{0,40}];

Looking at this should also give you a good sense that tangent vectors approximate the graph. We make use of this when we define arclength of a curve as the "sum" of the lengths of the tangent vectors

PLEASE NOTE WHEN GRAPHING VECTORS:
Use the option AspectRatio->Automatic so that the lengths of the vectors appear properly. When you do not, all the lengths look the same:

In[26]:=

  Show[p1,p5,PlotRange->{0,40}];

Up to Vector Differential Calculus