10.1 Slope Fieldsap Calculus

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.

Calculus 10th Edition answers to Chapter 1 - Limits and Their Properties - Section Project - Graphs and Limits of Trigonometric Functions - Page 90 d including work step by step written by community members like you. Textbook Authors: Larson, Ron; Edwards, Bruce H., ISBN-10: 1-28505-709-0, ISBN-13: 978-1-28505-709-5, Publisher: Brooks Cole. Figure 10.1 shows n = 1 and n =2 and n = 3 and n = a. 10.1 Two terms, then three terms, then full series: The infinite sum gives a jfnite answer, provided x is between -1 and 1. Then xn goes to zero: Now start with the function 1/(1 -x). How does it produce the series? Slope fields are visual representations of differential equations of the form dy / dx = f (x, y). At each sample point of a slope field, there is a segment having slope equal to the value of dy / dx. Any curve that follows the flow suggested by the directions of the segments is a solution to the differential equation.

Figure 10.1.1. Polar coordinates of the point $ds (1,sqrt3)$.

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.

Figure 10.1.2. A cardioid: $y=1+cos x$ on the left, $r=1+costheta$ on the right.

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 $(2,-3pi/4)$.

Figure 10.1.3. The point $(-2,pi/4)=(2,5pi/4)=(2,-3pi/4)$ in polar coordinates.

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.

Figure 10.1.4. The spiral of Archimedes and the full graph of $r=theta$.

Converting polar equations to rectangular equations can be somewhattrickier, and graphing polar equations directly is also not always easy.

10.1 Slope Fieldsap Calculus

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 $r=2sintheta$.

Figure 10.1.5. Graph of $r=2sintheta$. You can drag the red point in the graph onthe left, and the corresponding point on the right will follow.

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$ds r=1+theta^1/pi^2$

Ex 10.1.18$ds r=cotthetacsctheta$

Ex 10.1.19$ds r={1oversintheta+costheta}$

Ex 10.1.20$ds r^2=-2secthetacsctheta$

10.1 Slope Fieldsap Calculus

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)

Special Focus: Differential Equations by the College Board (pdf - 61 pages)

The Domain of Solutions to Differential Equations by the College Board (pdf - 8 pages)

10 1 Slope Field Sap Calculus Formula

Differential Equations (DE)

1) Introduction and Characteristics of Differential Equations

  • Explanation:
    • Notes Annotated
    • Khan Video: What is a differential equation? (don't worry about the classifications - after 8:00)
    • Rabbit versus Wolves interactive Simulator (Shodor - Java)
  • Practice Problems
    • Characteristics of Differential Equations WS Answers

Slope Fields

2) Create a slope field given a differential equation

  • Explanation:
    • Create a slope field in Grapher (on a Mac)
    • Websites:
      • Slope Field Calculator (java) by Marek Rychlik; (click on a point to show a particular solution [or type it in]& adjust many parameters)
      • Softwares: WINPLOT (Windows software to download)
  • Practice Problems:
    • Create a Slope Field given DE WS Answers

3) Reading and analyzing slope fields

  • Explanation:
    • Notes Annotated
  • Practice Problems:
    • Reading Slope Fields WS Answers

Particular Solutions to Differential Equations (DE)

4) Graphing a particular solution given a slope field and a point

  • Explanation:
    • Notes Annotated
  • Practice Problems:
  • Sketch a solution curve on a Slope Field WS Answers
  • Match Slope Field with Possible Solution WS Answers

5) Verifying particular solutions to DE through substitution:

  • Explanation:
    • Notes Annotated
  • Practice:
    • Match the DE with a possible solution WS Answers
    • 2007 Q4 - part b

6) Finding a particular solution through Separable Differential Equations:

  • Explanation:
    • Notes Annotated
    • Kahn Videos:
  • Practice Problems
    • Separable Differential Equations WS #1 Answers
    • Separable Differential Equations WS #2 Answers

7) Domain of solutions to DE

10 1 Slope Field Sap Calculus Calculator

  • Explanation:
    • Notes Annotated

8) DE Word problems

  • Explanation:
    • Notes Annotated
    • Websites:
      • Models Exponential Growth (includes a java app)
      • Models Logistic Growth (includes a java app)
  • Practice Problems:
    • Worksheet Answers
10 1 slope field sap calculus formula

9) Local Linearization

  • Explanation:
    • Notes Annotated
    • Interactive website: This Java website shows how a function looks like a line when you zoom in on the graph. It also visually shows how the value of the function is similar to the value of the tangent line (drag the red point).
  • Practice Problems:
  • Linear Approximation WS Answers

10 1 Slope Field Sap Calculus Formulas

10) Review WS Answers

10 1 Slope Field Sap Calculus Equation

11) Differential Equations Connection Project

AP Questions: Slope fields have been a topic on the AP Calculus BC Exam since 1998 and on the AP Calculus AB Exam since 2004.

  • 2012 Q5 - second derivative of a DE and separable equation
  • 2011 Q5 - separable equations & linear approximation
  • 2010 (Form B) Q5 - slope field and separable
  • 2010 Q6 - separable equation & linear approximation
  • 2008 Q5 - slope field and separable equation
  • 2007 (Form B) Q5 - slope field, second derivative, max/min
  • 2007 Q4 - solve for a constant by substitution of a DE
  • 2006 (Form B) Q5 - slope field and separable equation
  • 2006 Q5 - slope field and separable equation
  • 2005 (Form B) Q6 - - slope field and separable equation
  • 2005 Q6 - slope field, linear approximation and separable equation
  • 2004 (Form B) Q5 - slope field and separable equation
  • 2004 Q6 - slope field and separable equation
  • 2003 (Form B) Q6 - separable equation
  • 2003 Q5 - part b and c - separable equation
  • 2002 (Form B) Q5 - max/min and separable equation
  • 2001 Q6 - separable equation
  • 2000 Q6 - separable equation
  • 1998 Q4 - linear approximation and separable equation