How can a draw the graph of y if i have the graph of dy/dx?

[Knight]

I have been given a graph of dy/dx which is just one line of all linear segments. Pretty basic. The grid also has a nubered axis. What i want to know is if they have given me that y(0)= 50, how can i graph the rest of y? I need to know its exact position (inflection points and so on).. Please help!

I hope you can understand this!

 

[soroban]

Re: How can a draw the graph of y if i have the graph of dy/

Hello, Knight!

I have been given a graph of \(\displaystyle \frac{dy}{dx}\) which is just one line of all linear segments.
The grid also has a numbered axis.
If they have given me that \(\displaystyle y(0)\,=\,50\), how can i graph the rest of \(\displaystyle y\)?
I need to know its exact position (inflection points and so on).

If the graph of \(\displaystyle \frac{dy}{dx}\) is all linear segments, it must be a piecewise function
. . with abrupt changes in direction and discontinuities.

I'll try to explain with an example.
Suppose this is the graph of \(\displaystyle \frac{dy}{dx}\) . . .

      |       B                       D
    2 +       *                       *
      |     *   *                    *
      |   *       *                 *
      | *           *              *
    1 *A              *           *
      |                 *        *
      |                   *     *
      |                     * C*
    - + - - - + - - - + - - - * - - - + -
      |       1       2       3       4

I remind you that we are dealing with slopes here . . .

On the interval \(\displaystyle [0,\,1]\), we have segment \(\displaystyle AB\).
. . and we have: \(\displaystyle \:\frac{dy}{dx}\:=\:x\,+\,1\)

Integrate: \(\displaystyle \:y \:=\:\frac{1}{2}x^2\,+\,x\,+\,C_1\)

Since \(\displaystyle y(0)\,=\,50\), we have: \(\displaystyle \:50\:=\:0^2\,+\,0\,+\,C_1\;\;\Rightarrow\;\;C_1\,=\,50\)

. . Hence, on \(\displaystyle [0,\,1],\;f(x)\:=\:\frac{1}{2}x^2\,+\,x\,+\,50\)


On the interval \(\displaystyle [1,\,3]\), we have segment \(\displaystyle BC\).
(I assume you can write the equation of that line segment.)
. . and we have: \(\displaystyle \:\frac{dy}{dx}\:=\:-x\,+\,3\)

Integrate: \(\displaystyle \:y \:=\:-\frac{1}{2}x^2\,+\,3x\,+\,C_2\)

Since \(\displaystyle (1,2)\) is on the line, we have: \(\displaystyle \:2\:=\:-\frac{1}{2}\cdot1^2\,+\,3\cdot1\,+\,C_2\;\;\Rightarrow\;\;C_2\,=\,-\frac{1}{2}\)

. . Hence, on \(\displaystyle [1,\,3],\:f(x)\:=\:-\frac{1}{2}x^2\,+\,3x\,-\,\frac{1}{2}\)


On the interval \(\displaystyle [3,\,4]\), we have segment \(\displaystyle CD.\)
. . and we have: \(\displaystyle \:\frac{dy}{dx}\:=\:2x\,-\,6\)

Integrate: \(\displaystyle \:y \:=\:x^2\,-\,6x\,+\,C_3\)

Since \(\displaystyle (3,0)\) is on the line, we have: \(\displaystyle \:0 \:=\:3^2\,-\,6\cdot3\,+\,C_3\;\;\Rightarrow\;\;C_3\,=\,9\)

. . Hence, on the interval \(\displaystyle [3,\,4],\;f(x)\:=\:x^2\,-\,6x\,+\,9\)


Therefore: \(\displaystyle \;f(x) \;=\;\begin{Bmatrix}\frac{1}{2}x^2\,+\,x\,+\,50 &\;\; & 0\,\leq\,x\,\leq\,1 \\
-\frac{1}{2}x^2\,+\,3x\,-\frac{1}{2} & & 1\,\leq\,x\,\leq\,3 \\ \\ \\
x^2\,-\,6x\,+\,9 & & 3\,\leq\,x\,\leq\,4
\end{Bmatrix}\)

The graph has three unconnected parabolic arcs.