Exercise: Imagine you are teaching a course in calculus. Make a list of 10 questions that you would find hardest to answer with regard to the material in the first 9 chapters. I believe that making up questions is a more challenging endeavor than is answering them. AP Calculus AB: 7.3 Sketching Slope Fields - Duration: 46:51. Advanced Placement 60,896 views. Introduction to Slope Fields (Differential Equations 9) - Duration: 34:18.
Coordinate systems are tools that let us use algebraic methods tounderstand geometry. While the rectangular(also calledCartesian) coordinates that wehave been using are the most common, some problems are easier toanalyze in alternate coordinate systems.
A coordinate system is a scheme that allows us to identify any pointin the plane or in three-dimensional space by a set of numbers. Inrectangular coordinates these numbers are interpreted, roughlyspeaking, as the lengths of the sides of a rectangle. In polarcoordinates a point in the plane is identified by a pair of numbers $(r,theta)$.The number $theta$ measures the angle between the positive$x$-axis and a ray that goes through the point,as shown in figure 10.1.1; the number$r$ measures the distance from the origin to thepoint. Figure 10.1.1 shows the point withrectangular coordinates $ds (1,sqrt3)$ and polar coordinates $(2,pi/3)$, 2 units from the origin and $pi/3$ radians from thepositive $x$-axis.
Just as we describe curves in the plane using equations involving $x$and $y$, so can we describe curves using equations involving $r$ and$theta$. Most common are equations of the form $r=f(theta)$.
Example 10.1.1 Graph the curve given by $r=2$. All points with $r=2$ are atdistance 2 from the origin, so $r=2$ describes the circle of radius 2with center at the origin.
Example 10.1.2 Graph the curve given by $r=1+costheta$. We first consider$y=1+cos x$, as in figure 10.1.2. As $theta$ goesthrough the values in $[0,2pi]$, the value of $r$ tracks the value of$y$, forming the 'cardioid' shape of figure 10.1.2.For example, when $theta=pi/2$, $r=1+cos(pi/2)=1$, so we graph thepoint at distance 1 from the origin along the positive $y$-axis, whichis at an angle of $pi/2$ from the positive $x$-axis. When$theta=7pi/4$, $ds r=1+cos(7pi/4)=1+sqrt2/2approx 1.71$, and thecorresponding point appears in the fourth quadrant. This illustratesone of the potential benefits of using polar coordinates: the equationfor this curve in rectangular coordinates would be quite complicated.
Each point in the plane is associated with exactly one pair of numbersin the rectangular coordinate system; each point is associated with aninfinite number of pairs in polar coordinates. In the cardioidexample, we considered only the range $0le thetale2pi$, andalready there was a duplicate: $(2,0)$ and $(2,2pi)$ are the samepoint. Indeed, every value of $theta$ outside the interval $[0,2pi)$duplicates a point on the curve $r=1+costheta$ when$0letheta< 2pi$. We can even make sense of polar coordinates like$(-2,pi/4)$: go to the direction $pi/4$ and then move a distance 2in the opposite direction; see figure 10.1.3. As usual, a negative angle $theta$ means an anglemeasured clockwise from the positive $x$-axis. The point infigure 10.1.3 also has coordinates$(2,5pi/4)$ and $(2,-3pi/4)$.
The relationshipbetween rectangular and polar coordinates is quite easy tounderstand. The point with polar coordinates $(r,theta)$ hasrectangular coordinates $x=rcostheta$ and $y=rsintheta$; thisfollows immediately from the definition of the sine and cosinefunctions. Using figure 10.1.3 as anexample, the point shown has rectangular coordinates $ds x=(-2)cos(pi/4)=-sqrt2approx 1.4142$ and $ds y=(-2)sin(pi/4)=-sqrt2$. This makes it very easy to convertequations from rectangular to polar coordinates.
Example 10.1.3 Find the equation of the line $y=3x+2$ in polarcoordinates. We merely substitute: $rsintheta=3rcostheta+2$, or $ds r= {2over sintheta-3costheta}$.
Example 10.1.4 Find the equation of the circle $ds (x-1/2)^2+y^2=1/4$ in polarcoordinates. Again substituting:$ds (rcostheta-1/2)^2+r^2sin^2theta=1/4$. A bit of algebra turns thisinto $r=cos(t)$. You should try plotting a few $(r,theta)$ values toconvince yourself that this makes sense.
Example 10.1.5 Graph the polar equation $r=theta$. Here the distance fromthe origin exactly matches the angle, so a bit of thought makes itclear that when $thetage0$ we get the spiral of Archimedes in figure 10.1.4. When $theta< 0$, $r$ is alsonegative, and so the full graph is the right hand picture in thefigure.
Converting polar equations to rectangular equations can be somewhattrickier, and graphing polar equations directly is also not always easy.
10 1 Slope Field Sap Calculus Formulas
Example 10.1.6 Graph $r=2sintheta$. Because the sine is periodic, we knowthat we will get the entire curve for values of $theta$ in$[0,2pi)$. As $theta$ runs from 0 to $pi/2$, $r$ increases from 0to 2. Then as $theta$ continues to $pi$, $r$ decreases again to0. When $theta$ runs from $pi$ to $2pi$, $r$ is negative, and itis not hard to see that the first part of the curve is simply tracedout again, so in fact we get the whole curve for values of $theta$in $[0,pi)$. Thus, the curve looks something likefigure 10.1.5. Now, this suggeststhat the curve could possibly be a circle, and if it is, it wouldhave to be the circle $ds x^2+(y-1)^2=1$. Having made this guess, wecan easily check it. First we substitute for $x$ and $y$ to get$ds (rcostheta)^2+(rsintheta-1)^2=1$; expanding and simplifyingdoes indeed turn this into $r=2sintheta$.
Exercises 10.1
Ex 10.1.1Plot these polar coordinate points on one graph:$(2,pi/3)$, $(-3,pi/2)$, $(-2,-pi/4)$, $(1/2,pi)$, $(1,4pi/3)$, $(0,3pi/2)$.
Find an equation in polar coordinates that has the samegraph as the given equation in rectangular coordinates.
Ex 10.1.2$ds y=3x$(answer)
Ex 10.1.3$ds y=-4$(answer)
Ex 10.1.4$ds xy^2=1$(answer)
Ex 10.1.5$ds x^2+y^2=5$(answer)
Ex 10.1.6$ds y=x^3$(answer)
Ex 10.1.7$ds y=sin x$(answer)
Ex 10.1.8$ds y=5x+2$(answer)
Ex 10.1.9$ds x=2$(answer)
Ex 10.1.10$ds y=x^2+1$(answer)
Ex 10.1.11$ds y=3x^2-2x$(answer)
Ex 10.1.12$ds y=x^2+y^2$(answer)
Converting polar equations to rectangular equations can be somewhattrickier, and graphing polar equations directly is also not always easy.
10 1 Slope Field Sap Calculus Formulas
Example 10.1.6 Graph $r=2sintheta$. Because the sine is periodic, we knowthat we will get the entire curve for values of $theta$ in$[0,2pi)$. As $theta$ runs from 0 to $pi/2$, $r$ increases from 0to 2. Then as $theta$ continues to $pi$, $r$ decreases again to0. When $theta$ runs from $pi$ to $2pi$, $r$ is negative, and itis not hard to see that the first part of the curve is simply tracedout again, so in fact we get the whole curve for values of $theta$in $[0,pi)$. Thus, the curve looks something likefigure 10.1.5. Now, this suggeststhat the curve could possibly be a circle, and if it is, it wouldhave to be the circle $ds x^2+(y-1)^2=1$. Having made this guess, wecan easily check it. First we substitute for $x$ and $y$ to get$ds (rcostheta)^2+(rsintheta-1)^2=1$; expanding and simplifyingdoes indeed turn this into $r=2sintheta$.
Exercises 10.1
Ex 10.1.1Plot these polar coordinate points on one graph:$(2,pi/3)$, $(-3,pi/2)$, $(-2,-pi/4)$, $(1/2,pi)$, $(1,4pi/3)$, $(0,3pi/2)$.
Find an equation in polar coordinates that has the samegraph as the given equation in rectangular coordinates.
Ex 10.1.2$ds y=3x$(answer)
Ex 10.1.3$ds y=-4$(answer)
Ex 10.1.4$ds xy^2=1$(answer)
Ex 10.1.5$ds x^2+y^2=5$(answer)
Ex 10.1.6$ds y=x^3$(answer)
Ex 10.1.7$ds y=sin x$(answer)
Ex 10.1.8$ds y=5x+2$(answer)
Ex 10.1.9$ds x=2$(answer)
Ex 10.1.10$ds y=x^2+1$(answer)
Ex 10.1.11$ds y=3x^2-2x$(answer)
Ex 10.1.12$ds y=x^2+y^2$(answer)
Sketch the curve.
Ex 10.1.13$ds r=costheta$
Ex 10.1.14$ds r=sin(theta+pi/4)$
Ex 10.1.15$ds r=-sectheta$
Ex 10.1.16$ds r=theta/2$, $thetage0$
Ex 10.1.17$ds r=1+theta^1/pi^2$
10 1 Slope Field Sap Calculus Equation
Ex 10.1.18$ds r=cotthetacsctheta$
10 1 Slope Field Sap Calculus Formula
Ex 10.1.19$ds r={1oversintheta+costheta}$
Ex 10.1.20$ds r^2=-2secthetacsctheta$
In the exercises below, findan equation in rectangular coordinates that has the samegraph as the given equation in polar coordinates.
Ex 10.1.21$ds r=sin(3theta)$(answer)
Ex 10.1.22$ds r=sin^2theta$(answer)
Ex 10.1.23$ds r=secthetacsctheta$(answer)
Ex 10.1.24$ds r=tantheta$(answer)
Slope Fields
Key Questions
Example:
How do you draw the slope field for #dy/dx = x - y#?
The slope field is a cartesian grid where you draw lines in various directions to represent the slopes of the tangents to the solution. Therefore by drawing a curve through consecutive slope lines, you can find a solution to the differential equation.
Take the example of #dy/dx# at #(3, 4)#. Here we see that
#dy/dx = 3 - 4 =-1#
So you would draw a line of slope #-1# at #(3,4)#. Repeat this for maybe 4 by 4 points to get the following slope field
Hopefully this helps!
If you are allowed to use a software, then a software called GeoGebra gives us the slope field below.
I hope that this was helpful.
Slope field can be used to visualize the flow of the solution curves of its corresponding differential equation. At each point of the solution curve, the curve will have the slope indicated in the slope field.
Questions
