Saturday, May 3, 2014

Calculus of a Single Variable, Chapter 6, 6.3, Section 6.3, Problem 30

Equation of a tangent line to the graph of function f at point (x_0,y_0) is given by y=f(x_0)+f'(x_0)(x-x_0).
Since every tangent passes through the origin (0,0) we have
0=f(x_0)+f'(x_0)(0-x_0)
x_0f'(x_0)=f(x_0)
Let us write the equation using usual notation for differential equations.
x (dy)/(dx)=y
Now we separate the variables.
(dy)/y=dx/x
Integrating the equation, we get
ln y=ln x+ln c
c is just some constant so ln c is also some constant. It is only more convenient to write it this way.
Taking antilogarithm gives us the final result.
y=cx
There fore, our functions f have form f(x)=cx where c in RR.
Graphically speaking these are all the lines that pass through the origin. Since the tangent to a line at any point is the line itself the required property is fulfilled.
Graph of several such functions f can be seen in the picture below.

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