MTH Calculus II Essex County College Division of Mathematics and Physics Lecture Notes # Sakai Web Project Material Introduction - - 0 - Figure : Graph of y sin ( x y ) = x cos (x + y) with red tangent line. With few exceptions, we have only dealt with relationships between two variables some functional, others not functional. In those rare cases where we had more than two variable relationships we invariably found some way to reduce it to two variables. By now you should have a good idea how to graph simple relationships between two variables, and also use what you ve learned in pre-calculus and calculus to refine these basic graphs. When confronted with difficult two variable problems, you should at least be able to use a computer to construct a graph. So if you re asked to graph y = sin x, I hope you don t say, give me a calculator! However, if you re asked to graph y sin ( x y ) = x cos (x + y), and find the line tangent to this curve where x and y are positive proper fractions, and x = y, you would use technology to do so. Yes, in this case I d certainly want to use a computer to help visualize the problem. I don t think anyone could do the graph by hand, but some of you may be able to find the tangent line requested without graphing the relationship. In any case This document was prepared by Ron Bannon (ron.bannon@mathography.org) using L A TEX ε. Last revised October 0, 00. Found this by implicit differentiation and I used the point where x = y = π/4. y π 4 = + π ( π x π ) 4
I would still like visual confirmation that I m right and that s where a computer can help. The graph of y sin ( x y ) = x cos (x + y) and the requested tangent line, for the impatient among us, is given in Figure (page ). Parameters Now we are going to intentionally introduce a third variable, often called a parameter into our two variable problems. Yes, as before we may decide to eliminate this third variable in an attempt to land on familiar ground of two variable relationships. However, no matter how frustrating this becomes, you should always know that technology can deal with our limited capacity to resolve abstractions that are not yet familiar to us. For example, there was a time when jumping from linear relationships to more complex quadratics relationships proved insurmountable in your academic careers, so you may have resorted to using a graphing utility. So please consider having a graphic calculator 3 nearby for those times your brain is too stretched to move on. As an example, suppose we have a simple linear equation, y = x + 3, where we now define x as a function of t as follows: x = f (t). Certainly, if x is a function of t, we can also state that y is a functions of t as follows: For this particular example we might say that y = g (t). x = f (t) = t, and y = g (t) = 4t + 3. Yes, you probably think this is pointless, and in fact taking a simple problem and making it impossible. For example, I think it s really easy to graph y = x + 3 (Figure ), but not so easy to graph (f (t), g (t)) (Figure 3, page 3). Let s do both by using a computer to see what happens. 3 This may mean your computer!
First, the traditional way. 3 5-4 -3 - - 0 3 4 5 - - -3 Now the unfamiliar way. Figure : Graph of y = x + 3, software image on right. 3 5-4 -3 - - 0 3 4 5 - - -3 Figure 3: Graph of (f (t), g (t)), software image on the right. Problem: Now let s try to eliminate the parameter in this particular example. Work: 3
3 Graphing a Parametric Equation You should carefully look into using software to graph parametric equations, however, you should still be able to reason through simple examples. Here s one x = f (t) = t + t, and y = g (t) = t, where 3 t 3. Here, I would like to suggest that you construct a simple table (Table, page 4) and then connect the dots in order. You ll see that if you follow t from 3 forward that the graph moves in a certain direction. For this example, the initial point is (f ( 3), g ( 3)) = (5, 6), and the termial point is (f (3), g (3)) = (, 6). Table : x = t + t and y = t point t x y No. 3 5 6 No. 6 4 No. 3 No. 4 0 0 0 No. 5 3 No. 6 0 4 No. 7 3 6 9 8 7 6 5 4 3-4 -3 - - 0 3 4 5 6 7 8 9 0 3 4 5 6 7 8 9 0 3 4 5 6 7 8 9 30 3 3 33 34 35 - - -3-4 -5-6 -7-8 -9-0 - - Figure 4: x = t + t and y = t in red, for 3 t 3. 4
Problem: Now s your chance to do some real work.. Verify Table (page 4) is correct.. Plot the points, in order, directly on the given graph (Figure 4, page 4). Notice the direction that these points follow. 3. Differentiate both f (t) and g (t) and try to make sense out of these derivatives and the way these points are moving. 4. Try to eliminate the parameter to get a relationship between x and y. 4 5. You should also see that the graph of the relationship between x and y may differ, as in this example. 4 This relationship between x and y describes a set of points. The parametric equations have a great advantage though if we consider t to be time, we can view these points as location in time and can easily follow these points to see how they re moving. 5
3. Easily Tricked? Technology will certainly prove helpful, but using your head should be your main goal. Learning this material will require you to not only use your head, but also to explore technology to see why your head may not always produce the correct answer. For example, suppose you re asked to graph f (t) = cos t g (t) = sin t, where t R. And you decide to eliminate the paramter t. Work: Okay, give it a shot. Now graph out the relationship between x and y. Now look at the actual graph (Figure 5) of the parametric equations. Notice anything strange? Now, looking back over the both f and g, you should note that x and 0 y. Tricky, I know. 4 Using Software Here s an interesting problem, graph x = cos (3t) and y = cos (t + cos (3t)). I really don t think you can do this 5 using a table and plotting points, so I want to recommend that you either use a calculator 6 or a computer 7. Since we have a site license for Mathematica, here s the Mathematica code, ParametricPlot[{Cos[3 t], Cos[t + Cos[3 t]]}, {t, 0, Pi}], that you should try-out today. Make sure your graph looks like the one in Figure 6. You should also play around with the range for t in this example. 5 At least I can t. I have literally no patients for this type of problem. 6 Yes, you may have to read your manual. 7 I m use an application called Grapher, which is free on Mac OS X. 6
- 0 Figure 5: Graph of (f (t), g (t)). - - 0 - Figure 6: x = cos (3t) and y = cos (t + cos (3t)) 7
5 Finding Derivatives Given that both x and y are functions of t, I think you ll agree that it is pretty easy to find f and g. 8 Actually we already did this in a prior example and tried to make sense out of them, especially as they related to the way the graph moves (has direction). Now, for another example, suppose x = f (t) = t 3 and y = g (t) = t t, you can differentiate both x and y with respect to t quite easily. Here goes... f (t) = dx dt = 3t, g (t) = dy dt = t. More difficult to find would be find dy/dx. Let s start by finding a relationship between x and y. That will put us on familiar ground. Work: Okay, that was easy and you should have found that y = x /3 x /3. So we have and if we like, we can even find dy dx = 3x /3 3x /3, d y dx = 4 9x 5/3 + 9x 4/3. Normally, we don t need derivatives beyond these. You should realize that this largely depends on your ability to find a relationship between x and y, and that, my friend, may be very difficult to do! So let s look at another way to do these 8 We re differentiating with respect to t, so f (t) = dx dt and g (t) = dy dt. 8
derivatives that does not depend on finding a relationship between x and y. The first derivative: x = f (t) () y = g (t) () F (x) = y relationship between x and y (3) F (x) = g (t) substitute (4) F (f (t)) = g (t) substitute (5) F (f (t)) f (t) = g (t) differentiate (6) F (x) f (t) = g (t) substitute (7) This (Equation 8) is often written as F (x) = g (t) /f (t) voilà (8) d y dx = dy dx = dy dt dx dt. (9) Taking this very same equation (Equation 9) we can find the second derivative. 9 ( ) d dy = d y dx dx dx second derivative (0) ( ) d dy dt dx dx dt So in our example above we have the first derivative dy dx = 3x /3 3x /3 = g (t) f (t) = t 3t = 3x /3 3x /3 And the second derivative ( ) d t d y dx = dt 3t 3t = t 4 9t 5 = 4 9x 5/3 + 9x 4/3 second derivative () just do the substitution just do the substitution Let s summarize. There s no need to find a relationship between x and y to find the find the derivative dy/dx. Just use this formula: dy dx = g (t) f (t). It should also be easier to analyze the derivative when it s a function of t. For this particular case we have dy dx = t 3t, 9 The trick here is to replace y in Equation 9 with dy/dx. 9
and we should note that we ll get a horizontal tangent at t =, that is, at the point (f () g ()) = (, ). You ll also get a vertical tangent at (f (0) g (0)) = (0, 0). Furthermore, our graph will be increasing for t <, and decreasing for t >. Here s the graph (Figure 7, page 0). - 0 3 Figure 7: Graph of x = f (t) = t 3 and y = g (t) = t t. The second derivative is a bit-more-tricky! d y dx = d dt ( ) dy dx dx dt However, I want to stress that the second derivative we found is pretty easy to analyze. We can see that we get a sign change at t = 0 and t =. Simple sign analysis will show that our curve will be concave down for 0 < t < and concave. That is, between the points (0, 0) and (8, 0) we should observe a concave down structure. 6 You Try!. Graph the parametric curve given by x = t 5 4t 3, and y = t. 0 0 Yes, you can use a calculator. 0
. Visually determine the point on the parametric curve graph in () where the tangent does not exist. 3. Determine the points on the parametric curve graph in () where the tangent is vertical. 4. Determine the single point on the parametric curve graph in () that has two different t values. What are these values. Also determine the the slopes of the line tangent at these two t values. 5. Graph the parametric curve given by x = t, and y = t 7 + t 5. 6. Using the second derivative, find where our parametric curve will be concave up or concave down. Your answer should be given in t. Yes, you can use a calculator.
7 Answers to You Try!. Graph the parametric curve given by x = t 5 4t 3, and y = t. 5 4 3-8 -7-6 -5-4 -3 - - 0 3 4 5 6 7 8 Figure 8: Graph of x = t 5 4t 3, and y = t.. Visually determine the point on the parametric curve graph in () where the tangent does not exist. At the sharp corner, (0, 0). If you take a look at the first derivative dy dx = t 5t 4 t, you ll see that this occurs when t = 0. The derivative is also undefined at t = ± /5, but that s not where the sharp corner is. 3. Determine the points on the parametric curve graph in () where the tangent is vertical. From the work above, we can see that it occurs when t = ± /5, this gives: ( ( ) 5/ ( ) ) ( 3/ 4, ( ) 5/ ( ) ) 3/ + 4, 5 5 5 5 5 5 4. Determine the single point on the parametric curve graph in () that has two different t values. What are these values. Also determine the the slopes of the line tangent at these two t values. By inspection this appears to happen when y = 4, and that s the same as saying t = ±. The slope of the tangent line at this point is ±/8.
5. Graph the parametric curve given by x = t, and y = t 7 + t 5. 5 4 3-0 - - -3-4 -5 Figure 9: Graph of x = t, and y = t 7 + t 5. 6. Using the second derivative, find where our parametric curve will be concave up or concave down. Your answer should be given in t. d y dx = 35t3 + 5t 4 Only one real solution to the second derivative being zero, and that occurs when t = 0. Simple sign analysis gives us concave up for t > 0, and concave down for t < 0. Now, look at the graph. Not easy, I know... 8 Area As always, I want to stress that you graph even if it s just a rough sketch the area you re trying to determine before you set-up an integral. It may not be easy, but technology is available that helps, and it s incumbent upon you to learn your tools. For example if you re asked to find the area between the x axis and the curve defined by y = x ; or the area 3 defined between y = e x and the positive x-axis... you should always try to visualize it first... the answer may be staring at you. Let s move on to parametric equations. Typically, in the past, we had some integral that will look like this b a F (x) dx. () Now, we are going to have parametric equations of the form x = f (t) and g (t) = y, where Answer: It s just half of a unit circle, so the area is π/. No, I did not do the integration. 3 You won t be able to do this one, but it s well known (related to the normal function) to mathematical statisticians and the answer is π/. Again, no need to do the integration, but I did need to know a very famous integral related to this. 3
α t β. Let s make a simple substitution into the Integral () above and we ll get b a F (x) dx = b a y dx = β α g (t) f (t) dt. (3) Example: Determine the area under (between the graph and the x-axis) the parametric curve given by the following parametric equations. x = (t sin t) y = ( cos t) 4π t 4π. To do this problem you need to graph (Figure 0, page 4) it first. Notice that you have four 4 3 30-5 -0-5 -0-5 0 5 0 5 0 5 30 - Figure 0: Graph of x = (t sin t), y = ( cos t), 4π t 4π, not properly scaled! equal sections, so I ll just look at one section for this example I ll use 0 t π and multiply its area by four. 4 b a 4π F (x) dx = 4 = 4 9 Arc Length 0 π 0 y dx a = (0 sin 0) and b = (π sin π) ( cos t) ( cos t) dt dx = ( cos t) dt = 48π Yes, you too can do this integration. Well, you should know by now that I am also going to approach arc length from a visual perspective too. For example, suppose I were to ask you to determine the length of arc for the parametric equations g (t) = y = sin t and f (t) = x = cos t for 0 t π? Yes, graph this one and you ll see that it s fairly easy. 4 Now let s do the calculus! If you eliminate the parameter you ll get y = x where x. You can now use the arc length formula that you learn prior. That is, if the derivatives are continuous on the interval 4 Half the unit cirlce, so its length is π. 4
of interest, then the length of the curve is Or if you prefer, L = L = b a d c f (a) f (b) + + ( ) dy dx. dx ( ) dx dy. dy Let s now instead try to develop an analogous model for the parametric form. Same as we did in the past, but this time we ll use parameter t. b ( ) dy L = + dx a dx f (b) ( g = + ) (t) f f (t) dt (t) = f (a) (f (t)) + (g (t)) dt Giving it a try on our example. L = = = = = π 0 π 0 π 0 π + ( ) dy dx dx (f (t)) + (g (t)) dt ( sin t) + (cos t) dt dt You re going to get confused, so please draw a picture before proceeding forward on arc length problems. There s also some (many) details I am leaving off here and you may want to read the book for a more complete picture. 0 Surface Area You may recall that we also did surface area and found that b [ ] dy S = π r (x) + dx. Here y is a function of x. dx a This is easily transformed to the parametric form as follows S = π f (b) f (a) g (t) (f (t)) + (g (t)) dt. 5
A simple example using parametric equations follows. Example: Show that the surface area of a unit sphere is 4π. Let s use f (t) = sin t and g (t) = cos t on π/ t π/. 5 Here s the integral. π/ S = π = π = 4π π/ π/ π/ cos t (cos t) + ( sin) dt cos t dt Now why don t you try doing this with y = x for x. 6 You try... You should be able to do the following problems. Yes, it s a good idea to use a graphing aid. 7. Graph f (t) = x = e t cos 5t, g (t) = y = e t sin 5t, 0 t π.. Find an equation of the tangent(s) to the curve x = cos t + cos t, y = sin t + sin t at the point (, ). Then graph the curve and the tangent(s). 3. Find the length of the loop of the curve x = 3t t 3, y = 3t. 4. Find the exact area of the surface obtained by rotating the curve x = 3t t 3, y = 3t about the x-axis for 0 t. 5. Find the area enclosed by the x-axis and the curve x = + e t, y = t t. 5 Yes, you need to graph this! 6 It is really just as easy. 7 Mathematica, or whatever you prefer. I prefer Grapher, but you need to make your own decisions. 6
Answers to You try.... Graph f (t) = x = e t cos 5t, g (t) = y = e t sin 5t, 0 t π. 5 0 5 0 5 0 5 0 5 Figure : Graph of (f (t), g (t)) for 0 t π. This would be pure hell to do by hand.. Find an equation of the tangent(s) to the curve x = cos t + cos t, y = sin t + sin t at the point (, ). Then graph the curve and the tangent(s). You ll need to figure out where this point occurs on the graph. Yes, I suggest you graph it before doing anything else, including the point and the tangent. By inspection, this point occurs at t = π/. 8 To find the slope at this point you will need the derivative evaluated at this point. Here s the derivative, dy cos t + cos t = dx sin t sin t. And here s the value of the derivative at this point. dy =. dx So the line tangent is t= π y = (x + ) y = x + 3 8 You ll need to use your head here. Just try to reason on this one, mainly because it is easier than thuggishly doing the math. 7
0 Figure : Graph of (x = cos t + cos t, y = sin t + sin t) for 0 t π, and the line tangent to this curve at the point (, ). 3. Find the length of the loop of the curve x = 3t t 3, y = 3t. You should, of course, figure out what the loop is and the best way to do this is to visualize it. The way to do this is by graphing various ranges for t, and also trying to figure out the range by looking at the equations. Just factoring the expression for x will give you a good idea... the loop occurs for 3 t 3. Again, use a graphing device. Here s the graph. 5 0 5-0 -5 0 5 0 5 Figure 3: Graph of ( x = 3t t 3, y = 3t ) for 3 t 3, where the loop occurs between ( 3, 3 ). 8
Now let s compute the length of this loop. = = 3 3 3 3 = 3 (36t ) + (9 8t + 9t 4 ) dt 3t + 3 dt 4. Find the exact area of the surface obtained by rotating the curve x = 3t t 3, y = 3t about the x-axis for 0 t. We just did this graph, but now t is being restricted to 0 t. Here s the graph. 3 - - 0 3 4 5 And here s the integration. Figure 4: Graph of ( x = 3t t 3, y = 3t ) for 0 t. = π = 48π 5 0 3t (36t ) + (9 8t + 9t 4 ) dt 5. Find the area enclosed by the x-axis and the curve x = + e t, y = t t. I think you re getting the point about graph. And the area between this curve and the x-axis is found by integration. A = 0 ( t t ) e t dt = 3 e Okay, I m tired and I used Mathematica! 9
3 Figure 5: Graph of ( x = + e t, y = t t ) for 0 t. Assignment:. You should read 0. and 0., and then be able to do the following problems. 0. Curves Defined by Parametric Equations:, 5, 9, 4, 6,, 4, 5, 8, 9, 33, 37, 40, 4, 45. 0. Calculus with Parametric Curves:, 4, 6, 7, 0,, 4, 5, 8,, 6, 3, 36, 37, 39, 4, 45, 5, 65, 74. WebAssign problems, similar to the ones above, will be posted and you need to get started right away!. The second unit exam will be on section sections 7.6, 7.7, 7.8, 8., 8., 0., 0.. The date will be discussed in class. Calculators, including cell phones and computers, are not allowed. One sheet of notes is allowed, but it can not contain any worked problems! 9 9 I will collect your note sheet, along with the exam. If I see any worked problems you will be assigned a zero on the exam. 0