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

Knight

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Mar 31, 2007
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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!
 
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 dydx\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 y(0)=50\displaystyle y(0)\,=\,50, how can i graph the rest of y\displaystyle y?
I need to know its exact position (inflection points and so on).
If the graph of dydx\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 dydx\displaystyle \frac{dy}{dx} . . .
Code:
      |       B                       D
    2 +       *                       *
      |     *   *                    *
      |   *       *                 *
      | *           *              *
    1 *A              *           *
      |                 *        *
      |                   *     *
      |                     * C*
    - + - - - + - - - + - - - * - - - + -
      |       1       2       3       4

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

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

Integrate: y=12x2+x+C1\displaystyle \:y \:=\:\frac{1}{2}x^2\,+\,x\,+\,C_1

Since y(0)=50\displaystyle y(0)\,=\,50, we have: 50=02+0+C1        C1=50\displaystyle \:50\:=\:0^2\,+\,0\,+\,C_1\;\;\Rightarrow\;\;C_1\,=\,50

. . Hence, on [0,1],  f(x)=12x2+x+50\displaystyle [0,\,1],\;f(x)\:=\:\frac{1}{2}x^2\,+\,x\,+\,50


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

Integrate: y=12x2+3x+C2\displaystyle \:y \:=\:-\frac{1}{2}x^2\,+\,3x\,+\,C_2

Since (1,2)\displaystyle (1,2) is on the line, we have: 2=1212+31+C2        C2=12\displaystyle \:2\:=\:-\frac{1}{2}\cdot1^2\,+\,3\cdot1\,+\,C_2\;\;\Rightarrow\;\;C_2\,=\,-\frac{1}{2}

. . Hence, on [1,3],f(x)=12x2+3x12\displaystyle [1,\,3],\:f(x)\:=\:-\frac{1}{2}x^2\,+\,3x\,-\,\frac{1}{2}


On the interval [3,4]\displaystyle [3,\,4], we have segment CD.\displaystyle CD.
. . and we have: dydx=2x6\displaystyle \:\frac{dy}{dx}\:=\:2x\,-\,6

Integrate: y=x26x+C3\displaystyle \:y \:=\:x^2\,-\,6x\,+\,C_3

Since (3,0)\displaystyle (3,0) is on the line, we have: 0=3263+C3        C3=9\displaystyle \:0 \:=\:3^2\,-\,6\cdot3\,+\,C_3\;\;\Rightarrow\;\;C_3\,=\,9

. . Hence, on the interval [3,4],  f(x)=x26x+9\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.

 
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