<<

Slope Fields Review: THINGS YOU NEED TO KNOW!!!

Basics:  Fields are a graphical representation of the of a relation (not necessarily a . dy  You generate a slope field by plugging in x and y to the formula and generating dx for those individual points, then plotting them on the graph. o Positive slopes go up from left to right, negative ones go down o Larger numbers have steeper slopes o Zero slope is a horizontal segment o A non-existent slope value is not graphed.  Slope fields generally show what the of the formula looks like, because the lines trace out the shape of the y .  Example: dy o xy dx y

 dy  1 1 0 dx 1,1 

x dy           3   2   1 dx  3, 2









Sketching Solution :  The solution curve matches the slope where they tell you to begin.  It then conforms roughly to the slopes – they “guide” where the curve goes.  NEVER go outside the boundary of the field, put arrows at the end, or stop short of the edge of the field. The AP always counts these mistakes as the whole sketch being wrong. o Example: Sketch a solution curve to through the point (–2, 0). The sketch is above (the heavy blue on the slope field).

Separation of Variables:  This is basically an integration process to solve a differential (an equation dy solved for . dx  Step one: algebraically separate variables (including the dy and the dx) so that all of the x’s are on one side, and all the y’s are on the other. o Note: the dy and the dx can never, ever be in the denominator.  Integrate both sides of the equation. o Do not forget the “+ C” after integrating. If you do you will lose half the points for the problem on the AP test.  If you are given an initial value, plug in x and y to solve for C.  Finally, solve for y.  Note: Remember how to integrate negative exponents, as well as positives. This distinction comes up frequently. o Integrating a function to the negative one power gives you a natural log o Integrating any other negative power, just use the anti-.

EXAMPLE: dy x  Sketch the slope field for  dx y  Sketch the solution curve through (0, –2)  Solve for y for the solution curve through (0, –2)

Solution: y

 ydy xdx  ydy xdx  yx22 x C1          22 22  y x C2 2  y  x  C2 , since y = –2, the solution curve is negative  y  x2  C , plug in (0, –2)  2

 yx 2  4

Note that this was fairly easy integration – it generally will not be too difficult an integration problem, but be prepared for negative exponents and/or natural logs to result.