1Integration 1.1Definition of the Integral 1.1.8Exercises
1.1.8.1.
Hint.
Draw a rectangle that encompasses the entire shaded area, and one that is encompassed by the shaded area. The shaded area is no more than the area of the bigger rectangle, and no less than the area of the smaller rectangle.
We can improve on the method of Question 1 by using three rectangles that together encompass the shaded region, and three rectangles that together are encompassed by the shaded region.
Since the sum runs from 1 to 3, there are three intervals. Suppose \(2 = \Delta x = \frac{b-a}{n}\text{.}\) You may assume the sum given is a right Riemann sum (as opposed to left or midpoint).
Notice that the index starts at \(k=0\text{,}\) instead of \(k=1\text{.}\) Write out the given sum explicitly without using summation notation, and sketch where the rectangles would fall on a graph of \(y=f(x)\text{.}\)
You’ll want the limit as \(n\) goes to infinity of a sum with \(n\) terms. If you’re having a hard time coming up with the sum in terms of \(n\text{,}\) try writing a sum with a finite number of terms of your choosing. Then, think about how that sum would change if it had \(n\) terms.
Don’t be afraid to guess \(\De x\) and \(f(x\)) (review Definition 1.1.11). Then write out explicitly \(\sum\limits_{i=1}^{n} f(a+i\De x)\Delta x\) with your guess substituted in, and compare the result with the given sum. Adjust your guess if they don’t match.
The main step is to express the given sum as the right Riemann sum \(\sum\limits_{k=1}^{n} f(a+k\De x)\Delta x\text{.}\) Don’t be afraid to guess \(\De x\) and \(f(x\)) (review Definition 1.1.11). Then write out explicitly \(\sum\limits_{k=1}^{n} f(a+k\De x)\Delta x\) with your guess substituted in, and compare the result with the given sum. Adjust your guess if they don’t match.
The main step is to express the given sum in the form \(\sum_{i=1}^{n} f(x_i^*)\Delta x\text{.}\) Don’t be afraid to guess \(\De x\text{,}\)\(x_i^*\) (for either a left or a right or a midpoint sum — review Definition 1.1.11) and \(f(x\)). Then write out explicitly \(\sum_{i=1}^{n} f(x_i^*)\Delta x\) with your guess substituted in, and compare the result with the given sum. Adjust your guess if they don’t match.
The main step is to express the given sum in the form \(\sum\limits_{i=1}^{n} f(x_i^*)\Delta x\text{.}\) Don’t be afraid to guess \(\De x\text{,}\)\(x_i^*\) (probably, based on the symbol \(R_n\text{,}\) assuming we have a right Riemann sum — review Definition 1.1.11) and \(f(x\)). Then write out explicitly \(\sum\limits_{i=1}^{n} f(x_i^*)\Delta x\) with your guess substituted in, and compare the result with the given sum. Adjust your guess if they don’t match.
Draw a picture: the area we want is a trapezoid. If you don’t remember a formula for the area of a trapezoid, think of it as the difference of two triangles.
For part (b): don’t panic! Just take it one step at a time. The first step is to write down the Riemann sum. The second step is to evaluate the sum, using the given identity. The third step is to evaluate the limit \(n\rightarrow\infty\text{.}\)
The first step is to write down the Riemann sum. The second step is to evaluate the sum, using the given formulas. The third step is to evaluate the limit as \(n\rightarrow\infty\text{.}\)
The first step is to write down the Riemann sum. The second step is to evaluate the sum, using the given formulas. The third step is to evaluate the limit \(n\rightarrow\infty\text{.}\)
You’ve probably seen this hint before. It is worth repeating. Don’t panic! Just take it one step at a time. The first step is to write down the Riemann sum. The second step is to evaluate the sum, using the given formula. The third step is to evaluate the limit \(n\rightarrow\infty\text{.}\)
Using the definition of a right Riemann sum, we can come up with an expression for \(f(-5+10i)\text{.}\) In order to find \(f(x)\text{,}\) set \(x=-5+10i\text{.}\)
Recall that for a positive constant \(a\text{,}\)\(\diff{}{x}\left\{a^x\right\} = a^x \log a\text{,}\) where \(\log a\) is the natural logarithm (base \(e\)) of \(a\text{.}\)
The difference between the upper and lower bounds is the area that is outside of the smaller rectangles but inside the larger rectangles. Drawing both sets of rectangles on one picture might make things clearer. Look for an easy way to compute the area you want.
Since \(f(x)\) is linear, there exist real numbers \(m\) and \(c\) such that \(f(x)=mx+c\text{.}\) It’s a little easier to first look at a single triangle from each sum, rather than the sums in their entirety.
For part (a), use the symmetry of the integrand. For part (b), the area \(\int \limits_{0}^1 \sqrt{1-x^2}\dee{x}\) is easy to find--how is this useful to you?
For two functions \(f(x)\) and \(g(x)\text{,}\) define \(h(x)=f(x)\cdot g(x)\text{.}\) If \(h(-x)=h(x)\text{,}\) then the product is even; if \(h(-x)=-h(x)\text{,}\) then the product is odd.
The integrand is similar to \(\dfrac{1}{\sqrt{1-x^2}}\text{,}\) so factoring out \(\sqrt{2}\) from the denominator will make it look like some flavour of arcsine.
Figure out where the two curves cross. To determine which curve is above the other, try evaluating \(f(x)\) and \(g(x)\) for some simple value of \(x\text{.}\) Alternatively, consider \(x\) very close to zero.
Draw sketches. The mechanically easiest way to answer part (b) uses the method of cylindrical shells, which is in the optional section 1.6. The method of washers also works, but requires you to have more patience and also to have a good idea what the specified region looks like. Look at your sketch very careful when identifying the ends of your horizontal strips.
(b) Hopefully, you sketched the ellipse in part (a). What was its smallest radius? Its largest? These correspond to the polar and equitorial radii, respectively.
(d) Remember that the absolute error is the absolute difference of your two results--that is, you subtract them and take the absolute value. The relative error is the absolute error divided by the actual value (which we’re taking, for our purposes, to be your answer from (c)). When you take the relative error, lots of terms will cancel, so it’s easiest to not use a calculator till the end.
The mechanically easiest way to answer part (b) uses the method of cylindrical shells, which we have not covered. The method of washers also works, but requires you have enough patience and also to have a good idea what \(\cR\) looks like. So it is crucial to first sketch \(\cR\text{.}\) Then be very careful in identifying the left end of your horizontal strips.
You can use ideas from this section to answer the question. If you take a very thin slice of the column, the density is almost constant, so you can find the mass. Then you can add up all your little slices. It’s the same idea as volume, only applied to mass.
Remember our rule: \(\int u \dee{v} = uv - \int v \dee{u}\text{.}\) So, we take \(u\) and use it to make \(\dee{u}\text{,}\) and we take \(\dee{v}\) and use it to make \(v\text{.}\)
You’ll probably want to use integration by parts. (It’s the title of the section, after all). You’ll break the integrand into two parts, integrate one, and differentiate the other. Would you rather integrate \(\log x\text{,}\) or differentiate it?
You know, or can easily look up, the derivative of arccosine. You can use a similar trick as the book did when antidifferentiating other inverse trigonometric functions in Example 1.7.9.
This example is similar to Example 1.7.10 in the text. The functions \(e^{x/2}\) and \(\cos(2x)\) both do not substantially alter when we differentiate or antidifferentiate them. If we use integration by parts twice, we’ll end up with an expression that includes our original integral. Then we can just solve for the original integral in the equation, without actually integrating.
You’ll want to do an integration by parts for (a)--check the end result to get a guess as to what your parts should be. A trig identity and some amount of algebraic manipulation will be necessary to get the final form.
Think, first, about how to get rid of the square root in the argument of \(f''\text{,}\) and, second, how to convert \(f''\) into \(f'\text{.}\) Note that you are told that \(f'(2) = 4\) and \(f(0) = 1\text{,}\)\(f(2) = 3\text{.}\)
The power of cosine is odd, so we can reserve one cosine for \(\dee{u}\text{,}\) and turn the rest into sines using the identity \(\sin^2 x + \cos^2 x =1\text{.}\)
Since the power of sine is odd (and positive), we can reserve one sine for \(\dee{u}\text{,}\) and turn the rest into cosines using the identity \(\sin^2 + \cos^2 x =1\text{.}\)
Since there are no secants in the problem, it’s difficult to use the substitution \(u=\sec x\) that we’ve enjoyed in the past. Example 1.8.12 in the text provides a template for antidifferentiating an odd power of tangent.
Integrating even powers of tangent is surprisingly different from integrating odd powers of tangent. You’ll want to use the identity \(\tan^2x = \sec^2 x -1\text{,}\) then use the substitution \(u=\tan x\text{,}\)\(\dee{u}=\sec^2 x\dee{x}\) on (perhaps only a part of) the resulting integral. Example 1.8.16 show you how this can be accomplished.
Remember \(e\) is some constant. What are our strategies when the power of secant is even and positive? We’ve seen one such substitution in Example 1.8.15.
The beginning of this section has a template for choosing a substitution. Your goal is to use a trig identity to turn the argument of the square root into a perfect square, so you can cancel \(\sqrt{(\mbox{something})^2}=|\mbox{something}|\text{.}\)
Since \(\theta\) is acute, you can draw it as an angle of a right triangle. The given information will let you label two sides of the triangle, and the Pythagorean Theorem will lead you to the third.
You can draw a right triangle with angle \(\theta\text{,}\) and use the given information to label two of the sides. The Pythagorean Theorem gives you the third side.
As in Question 1, choose an appropriate substitution. Your answer should be in terms of your original variable, \(x\text{,}\) which can be achieved using the methods of Question 3.
As in Question 1, choose an appropriate substitution. Your answer will be a number, so as long as you change your limits of integration when you substitute, you don’t need to bother changing the antiderivative back into the original variable \(x\text{.}\) However, you might want to use the techniques of Question 4 to simplify your final answer.
Question 1 guides the way to finding the appropriate substitution. Since the integral is definite, your final answer will be a number. Your limits of integration should be common reference angles.
Question 1 guides the way to finding the appropriate substitution. Since you have in indefinite integral, make sure to get your answer back in terms of the original variable, \(x\text{.}\) Question 3 gives a reliable method for this.
In part (a) you are asked to integrate an even power of \(\cos x\text{.}\) For part (b) you can use a trigonometric substitution to reduce the integral of part (b) almost to the integral of part (a).
To antidifferentiate even powers of cosine, use the formula \(\cos^2\theta = \frac{1}{2}(1+\cos(2\theta))\text{.}\) Then, remember \(\sin(2\theta)=2\sin\theta\cos\theta\text{.}\)
Think of \(e^x\) as \(\left(e^{x/2}\right)^2\text{,}\) and use a trig substitution. Then, use the identity \(\sec^2 \theta = \tan^2 \theta +1\text{.}\)
If a quadratic function can be factored as \((ax+b)(cx+d)\) for some constants \(a,b,c,d\text{,}\) then it has roots \(-\frac{b}{a}\) and \(-\frac{d}{c}\text{.}\)
You can save yourself some work in developing your partial fraction decomposition by renaming \(x^2\) to \(y\) and comparing the result with Question 7.
Since the degree of the numerator is the same as the degree of the denominator, we can’t do our partial fraction decomposition before we simplify the integrand.
The mechanically easiest way to answer part (c) uses the method of cylindrical shells, which we have not covered. The method of washers also works, but requires you have enough patience and also to have a good idea what \(R\) looks like. So look at the sketch in part (a) very carefully when identifying the left endpoints of your horizontal strips.
The absolute error is the difference of the two values; the relative error is the absolute error divided by the exact value; the percent error is one hundred times the relative error.
You don’t have to find the actual, exact maximum the second derivative achieves--you only have to give a reasonable “ceiling” that it never breaks through.
As usual, the biggest part of this problem is finding \(L\text{.}\) Don’t be thrown off by the error bound being given slightly differently from Theorem 1.11.13 in the text: these expressions are equivalent, since \(\De x = \frac{b-a}{n}\text{.}\)
The final sentence in part (b) is just a re-statement of the error bounds we’re familiar with from Theorem 1.11.13 in the text. The information \(\big|s^{(k)}(x)\big|\le \dfrac{k}{1000}\) gives you values of \(M\) and \(L\) when you set \(k=2\) and \(k=4\text{,}\) respectively.
Since the cross-sections of the pool are semi-circular disks, a section that is \(d\) metres across will have area \(\frac{1}{2}\pi\left(\frac{d}{2}\right)^2\) square feet. Based on the drawing, you may assume the very ends of the pool have distance 0 feet across.
Don’t get caught up in the interpretation of the integral. It’s nice to see how integrals can be used, but for this problem, you’re still just approximating the integral given, and bounding the error.
See Example 1.11.16. You’ll want to use a calculator for the approximation in (a), and for finding the appropriate number of intervals in (b). Remember that Simpson’s rule requires an even number of intervals.
You’ll have to differentiate \(f(x)\text{.}\) To that end, you may also want to review the fundamental theorem of calculus and, in particular, Example 1.3.5.
To have five decimal places of accuracy, your error must be less than 0.000005. This ensures that, if you round your approximation to five decimal places, they will all be correct.
In using Simpson’s rule to approximate \(\displaystyle\int_1^{x\vphantom{\frac{1}{2}}} \frac{1}{t} \dee{t}\) with \(n\) intervals, \(a=1\text{,}\)\(b=x\text{,}\) and \(\De x = \dfrac{x-1}{n}\text{.}\)
If an approximation \(A\) of the integral \(\int_1^2 \frac{1}{1+x^2} \dee{x}\) has error at most \(\varepsilon\text{,}\) then \(A-\varepsilon \leq \int_1^2 \frac{1}{1+x^2} \dee{x} \leq A+\varepsilon\text{.}\)
If you use Simpson’s rule to approximate \(\int_1^2 \frac{1}{1+x^2} \dee{x}\text{,}\) you won’t need very many intervals to get the requisite accuracy.
Second: when evaluating integrals, always check to see if you can use a simple substitution before trying a complicated procedure like partial fractions.
Recall \(\displaystyle\lim_{x \to \infty}\arctan x = \frac{\pi}{2}\text{.}\) For the other limits, use logarithm rules, and beware of indeterminate forms.
Break up the integral. The absolute values give you a nice even function, so you can replace \(|x-a|\) with \(x-a\) if you’re careful about the limits of integration.
Use integration by parts twice to find the antiderivative of \(e^{-x}\sin x\text{,}\) as in Example 1.7.10. Be careful with your signs--it’s easy to make a mistake with all those negatives.
There are two things that contribute to your error: using \(t\) as the upper bound instead of infinity, and using \(n\) intervals for the approximation.
First, find a \(t\) so that the error introduced by approximating \(\int_0^\infty \frac{e^{-x}}{1+x}\dee{x}\) by \(\int_0^t \frac{e^{-x}}{1+x}\dee{x}\) is at most \(\frac{1}{2} 10^{-4}\text{.}\) Then, find your \(n\text{.}\)
After your substitution, you should have a polynomial expression in \(u\)—but it might take some simplification to get it into a form you can easily integrate.
We notice that the integrand has a quadratic polynomial under the square root. If that polynomial were a perfect square, we could get rid of the square root: try a trig substitution, as in Section 1.9.
You should prepare your own personal internal list of integration techniques ordered from easiest to hardest. You should have associated to each technique your own personal list of signals that you use to decide when the technique is likely to be useful.
Part (a) can be done by inspection — use a little highschool geometry! Part (b) is reminiscent of the antiderivative of logarithm—how did we find that one out? Part (c) is an improper integral.
For part (b), first complete the square in the denominator. You can save some work by first comparing the derivative of the denominator with the numerator. For part (d) use a simple substitution.
For part (a), split the integral in two. One part may be evaluated by interpreting it geometrically, without doing any integration at all. For part (c), multiply both the numerator and denominator by \(e^x\) and then make a substitution.
2Applications of Integration 2.1Work 2.1.2Exercises
2.1.2.1.
Hint.
Watch your units: \(1 \,\mathrm{J} = 1\, \frac{\mathrm{kg}\cdot\mathrm{m}^2}{\mathrm{sec}^2}\text{,}\) but your mass is not given in kilograms, and your height is not given in metres.
Adding or subtracting two quantities of the same units doesn’t change the units. For example, if I have one metre of rope, and I tie on two more metres of rope, I have \(1+2=3\) metres of rope — not 3 centimetres of rope, or 3 kilograms of rope.
Multiplying or dividing quantities of some units gives rise to a quantity with the product or quotient of those units. For example, if I buy ten pounds of salmon for \(\$\)50, the price of my salmon is \(\dfrac{50\text{ dollars} }{10\text{ pounds}} = \dfrac{50}{10} \frac{\text{dollars}}{\text{pound}} = 5 \frac{\text{dollars}}{\text{pound}}\text{.}\) (Not 5 pound-dollars, or 5 pounds.)
Hooke’s law says that the force required to stretch a spring \(x\) units past its natural length is proportional to \(x\text{;}\) that is, there is some constant \(k\) associated with the individual spring such that the force required to stretch it \(x\) m past its natural length is \(kx\text{.}\)
Definition 2.1.1 tells us the work done by the force from \(x=1\) to \(x=b\) is \(W(b) = \int_1^b F(x) \dee{x}\text{,}\) where \(F(x)\) is the force on the object at position \(x\text{.}\) To recover the equation for \(F(x)\text{,}\) use the Fundamental Theorem of Calculus.
For (a), \(\frac{c}{\ell-x}\) is meausured in Newtons, while \(\ell\) and \(x\) are in metres. For (b), notice the similarities and differences between the tube of air and a spring obeying Hooke’s law.
Since you’re given the area of the cross-section, it doesn’t matter what shape it has. However, the density of water is given in cubic centimetres, while the measurements of the tank are given in metres.
Consider the work done to lift a horizontal plate from 2 m below the ground to a height \(z\text{.}\) You’ll need to know the mass of the plate, which you can calculate from its volume, since its density is given to you.
When you pull the box, the force you’re exerting is exactly the same as the frictional force, but in the opposite direction. In (a), that force is constant. In (b), it changes. Check Definition 2.1.1 for how to turn force into work.
Remember that the work done on an object is equal to the change in its kinetic energy, which is \(\frac{1}{2}mv^2\text{,}\) where \(m\) is the mass of the object and \(v\) is its velocity. Hooke’s law will tell you how much work was done stretching the spring.
As in Question 18 in this section, the change in kinetic energy of the car is equal to the work done by the compressing struts. The only added step is to calculate the spring constant, given that a car with mass 2000 kg compresses the spring 2 cm in Earth’s gravity. You’re not calculating work to find the spring constant: you’re using the fact that when the car is sitting still, the force exerted upward by the struts is equal to the force exerted downward by the mass of the car under gravity.
To find the area of a horizontal layer of water, use some geometry. A horizontal cross-section of a sphere is a circle, and its radius will depend on the height of the layer in the tank.
The basic ideas you’ve used already with “cable problems” still work, you only need to take care that the density of the cable is no longer constant. The mass of a tiny piece of cable, say of length \(\dee{x}\text{,}\) is (density)\(\times\)(length) = \((10-x)\dee{x}\text{,}\) where \(x\) is the distance of our piece from the bottom of the cable.
You can formulate a guess by considering the work done on the ball versus the work done on the rope in Question 16, Solution 1. But be careful — the ball in that problem did not have the same mass as the rope.
There are two things that vary with height: the density of the liquid, and the area of the cross-section of the tank. Make a formula \(M(h)\) for the mass of a thin layer of liquid \(h\) metres below the top of the tank, using mass\(=\)volume\(\times\)density. The rest of the problem is similar to other tank-pumping problems in this section.
Theorem 1.11.13 gives error bounds for the standard types of numerical approximations. You won’t need very many intervals to achieve the desired accuracy.
Part (a) is asking the length of the pieces we’ve cut our interval into. Part (c) should be given in terms of \(f\text{.}\) Our final answer in (d) will resemble a Riemann sum, but without some extra manipulation it won’t be in exactly the form of a Riemann sum we’re used to.
You can antidifferentiate an odd power of cosine with a substitution; for an even power of cosine, use the identity \(\cos^2 x = \frac{1}{2}\big(1+\cos(2x)\big)\text{.}\)
If you’re not sure how to antidifferentiate, try the substitution \(u=kx\text{,}\)\(\dee{u}=k\dee{x}\text{,}\) keeping in mind that \(k\) is a constant. Interestingly, your final answer won’t depend on \(k\text{.}\)
Notice the term \(50\cos\left(\frac{t}{12}\pi\right)\) has a period of 24 hours, while the term \(200\cos\left(\frac{t}{4380}\pi\right)\) has a period of one year.
A cross section of \(S\) at location \(x\) is a circle with radius \(x^2\text{,}\) so area \(\pi x^4\text{.}\) Part (a) is asking for the average of this function on \([0,2]\text{.}\)
Remember force is the product of the spring constant with the distance it’s stretched past its natural length. The units given in the question are not exactly standard, but they are compatible with each other.
To find a definite integral of the absolute value of a function, break up the interval of integration into regions where the function is positive, and intervals where it’s negative.
Imagine cutting out the shape and setting it on top of a pencil, so that the pencil lines up with the vertical line \(x=a\text{.}\) Will the figure balance, or fall to one side? Which side?
In (a), the slices all have the same width, so the area of the slices is larger (and hence the density of \(R\) is higher) where \(T(x)-B(x)\) is larger.
You can use a trigonometric substitution to find the area, then a partial fraction decomposition to find the \(y\)-coordinate of the centroid. Remember \(\sin(1/2)=\pi/6\text{.}\)
Vertical slices will be easier than horizontal. An integration by parts might be helpful to find \(\bar x\text{,}\) while trigonometric identities are important to finding \(\bar y\text{.}\)
You can save quite a bit of work by, firstly, exploiting symmetry and, secondly, thinking about whether it is more efficient to use vertical strips or horizontal strips.
Draw a sketch. Rotating about a horizontal line is similar to rotating about the \(x\)-axis, but for the radius of a slice, you’ll need to know \(|y-(-1)|\text{:}\) the distance from the outer edge of the region (the boundary function’s \(y\)-value) to \(y=-1\text{.}\)
Go back to the derivation of Equation 2.3.5 (centroid for a region) to figure out what to do when your surface does not have uniform density. We will consider a rod \(R\) that reaches from \(x=0\) to \(x=4\text{,}\) and the mass of the section of the rod along \([a,b]\) is equal to the mass of the strip of our rectangle along \([a,b]\text{.}\)
Horizontal slices will help you, where symmetry doesn’t, to set up a rod \(R\) whose centre of mass is the same as one coordinate of the centre of mass of the circle. When you’re integrating, trigonometric substitutions are sometimes the easiest way, and sometimes not.
The model in the question gives you the setup to solve this problem. You know how to find the centre of mass of a rod — that’s Equation 2.3.4 — so all you need to find is \(\rho(y)\text{,}\) the density of the rod at position \(y\text{.}\) To find this, consider a thin slice of the cone at position \(y\) with thickness \(\dee{y}\text{.}\) Its volume \(V(y)\) is the same as the mass of the small section of the rod at position \(y\) with thickness \(\dee{y}\text{.}\) So, the density of the rod at position \(y\) is \(\rho(y)=\frac{V(y)}{\dee{y}}\text{.}\)
Use similar triangles to show that the shape of the lower (also upper) half of the hourglass is a truncated cone, where the untruncated cone would have had a height 10 cm.
To calculate the centre of mass of the upturned sand using the result of Question 29, you should find \(h=9.8\) (not\(h=10\) — think carefully about our model from Question 29) and \(k=8.8\text{.}\) For the centre of mass of the sand before turning, \(h=10\) and \(k=6\text{.}\)
The techniques of Section 2.1 get pretty complicated here, so it’s easiest to use the techniques we developed in Questions 6, 29, and 30 in this section. That is, (1) find the height of the centre of mass of the water in its starting and ending positions, and then (2) model the work done as the work moving a point mass with the weight of the water from the first centre of mass to the second.
The height change of the centre of mass is all that matters to calculate the work done against gravity, so you only have to worry about the height of the centres of mass.
The area of \(R\) is precisely one, so the error in your approximation is the error involved in approximating \(\int_0^{\sqrt{\pi/2}}2x^2\sin(x^2)\,\dee{x}\text{.}\)
You don’t need to solve the differential equation from scratch, only verify whether the given function \(y=f(x)\) makes it true. Find \(\diff{y}{x}\) and plug it into the differential equation.
Note \(\diff{}{x}\{f(x)\} = \diff{}{x}\{f(x)+C\}\text{.}\) Plug in \(y=f(x)+C\) to the equation \(\diff{y}{x}=xy\) to see whether it makes the equation is true.
The red marks show the slope \(y(x)\) would have at a point if it crosses that point. So, pick a value of \(y(0)\text{;}\) based on the red marks, you can see how fast \(y(x)\) is increasing or decreasing at that point, which leads you roughly to a value of \(y(1)\text{;}\) again, the red marks tell you how fast \(y(x)\) is increasing or decreasing, which leads you to a value of \(y(2)\text{,}\) etc (unless you’re already off the graph).
To draw the sketch similar to Question 8(d), don’t actually calculate every single slope; find a few (for instance, where the slope is zero, or where it’s negative), and use a pattern (for instance, the slope increases as \(y\) increases) to approximate most of the points.
Start by multiplying both sides of the equation by \(e^y\) and \(\dee{x}\text{,}\) pretending that \(\diff{y}{x}\) is a fraction, according to our mnemonic.
You need to solve for your function \(y(x)\) explicitly. Be careful with absolute values: if \(|y|=F\text{,}\) then \(y=F\) or \(y=-F\text{.}\) However, \(y=\pm F\)is not a function. You have to choose one: \(y=F\) or \(y=-F\text{.}\)
If your answer doesn’t quite look like the answer given, try manipulating it with logarithm rules: \(\log a + \log b = \log(ab)\text{,}\) and \(a\log b = \log(b^a)\text{.}\)
Be careful about signs. If \(y^2=F\text{,}\) then possibly \(y=\sqrt{F}\text{,}\) and possibly \(y=-\sqrt{F}\text{.}\) However, \(y=\pm\sqrt{F}\) is not a function.
Be careful about signs. If \(\log|y|=F\text{,}\) then \(|y|=e^F\text{.}\) Since you should give your answer as an explicit function \(y(x)\text{,}\) you need to decide whether \(y=e^F\) or \(y=-e^F\text{.}\)
The general solution to the differential equation will contain the constant \(k\) and one other constant. They are determined by the data given in the question.
The general solution to the differential equation will contain the constant \(k\) and one other constant. They are determined by the data given in the question.
To solve \(\frac{x-a}{x-b}=Y\) for \(x\text{,}\) move the terms containing \(x\) out of the denominator, then gather them on one side of the equals sign and factor out the \(x\text{.}\)
The general solution to the differential equation will contain a constant of proportionality and one other constant. They are determined by the data given in the question.
You do not need to know anything about investing or continuous compounding to do this problem. You are given the differential equation explicitly. The whole first sentence is just window dressing.
Suppose that in a very short time interval \(\dee{t}\text{,}\) the height of water in the tank changes by \(\dee{h}\) (which is negative). Express in two different ways the volume of water that has escaped during this time interval. Equating the two gives the needed differential equation.
Sketch the mercury in the tank at time \(t\text{,}\) when it has height \(h\text{,}\) and also at time \(t+\dee{t}\text{,}\) when it has height \(h+\dee{h}\) (with \(\dee{h} \lt 0\)). The difference between those two volumes is the volume of (essentially) a disk of thickness \(-\dee{h}\text{.}\) Figure out the radius and then the volume of that disk. This volume has to be the same as the volume of mercury that left through the hole in the bottom of the sphere, which runs out in the shape of a cylinder. Toricelli’s law tells you what the length of that cylinder is, and from there you can find its volume. Setting the two volumes equal to each other gives the differential equation that determines \(h(t)\text{.}\)
For any \(p \gt 0\text{,}\) determine first \(y(t)\) (in terms of \(p\) and \(c\)) and then the times (also depending on \(p\) and \(c\)) at which \(y=2\text{,}\)\(y=1\) and \(y=0\text{.}\) The condition that “the top half takes exactly the same amount of time to drain as the bottom half” then gives an equation that determines \(p\text{.}\)
For (b), imagine cutting up the triangle into its black and white parts, then sharing it equally among a certain number of friends. What is the easiest number of friends to share with, making sure each has the same area in their pile?
To adjust the starting index, either factor out the first term in the series, or subtract two series. For the subtraction option, consider Question 10.
To find the difference between \(\displaystyle\sum_{n=1}^\infty c_n\) and \(\displaystyle\sum_{n=1}^\infty c_{n+1}\text{,}\) try writing out the first few terms.
When you see \(\displaystyle\sum_k \Big(\cdots \textcolor{red}{k}\cdots \ \ -\ \ \cdots \textcolor{red}{k+1}\cdots\Big)\text{,}\) you should think “telescoping series.”
When you see \(\displaystyle\sum_n \Big(\cdots \textcolor{red}{n}\cdots \ \ -\ \ \cdots \textcolor{red}{n+1}\cdots\Big)\text{,}\) you should immediately think “telescoping series”. But be careful not to jump to conclusions — evaluate the \(n^{\rm th}\) partial sum explicitly.
The stone at position \(x\) has mass \(\dfrac{1}{4^x}\) kg, and we have to pull it a distance of \(2^x\) metres. From this, you can find the work involved in pulling up a single stone. Then, add up the work involved in pulling up all the stones.
The comparison test is Theorem 3.3.8. However, rather than trying to memorize which way the inequalities go in all cases, you can use the same reasoning as Question 3.
By the divergence test, for a series \(\sum a_n\) to converge, we need \(\lim\limits_{n \to \infty} a_n=0\text{.}\) That is, the magnitude (absolute value) of the terms needs to be getting smaller.
If \(f(x)\) is positive and decreasing, then the integral test tells you that the integral and the series either both increase or both decrease. So, in order to find an example with the properties required in the question, you need \(f(x)\) to not be both positive and decreasing.
Review Example 3.3.9 for developing intuition about comparisons, and Example 3.3.10 for an example where finding an appropriate comparison series calls for some creativity.
The truncation error arising from the approximation \(\displaystyle\sum_{n=1}^\infty \frac{e^{-\sqrt{n}}}{\sqrt n} \approx \sum_{n=1}^N \frac{e^{-\sqrt{n}}}{\sqrt n}\) is precisely \(E_N = \displaystyle\sum_{n=N+1}^\infty \frac{e^{-\sqrt{n}}}{\sqrt n}\text{.}\) You’ll want to find a bound on this sum using the integral test.
What does the fact that the series \(\displaystyle\sum_{n=1}^\infty\frac{na_n-2n+1}{n+1}\) converges guarantee about the behavior of \(a_n\) for large \(n\text{?}\)
What does the fact that the series \(\sum_{n=1}^\infty a_n\) converges guarantee about the behavior of \(a_n\) for large \(n\text{?}\) When is \(x^2\le x\text{?}\)
We are approximating a finite sum — not an infinite series. To get greater accuracy, use exact values for the first several terms in the sum, and use an integral to approximate the rest.
You know the geometric series expansion of \(\frac{1}{1-x}\text{.}\) What (calculus) operation(s) can you apply to that geometric series to convert it into the given series?
Equation 2.3.1 tells us the centre of mass of a rod with weights \(\{m_n\}\) at positions \(\{x_n\}\) is \(\displaystyle\bar x =\frac{\sum m_nx_n}{\sum m_n}\) .
If you don’t have these memorized, it’s good to be able to derive them. For instance, \(\log(1+x)\) is the antiderivative of \(\dfrac{1}{1+x}\text{,}\) whose Taylor series can be found by modifying the geometric series \(\sum x^n\text{.}\)
There is an important Taylor series, one of the series in Theorem 3.6.7, that looks a lot like the given series. Be careful about the limits of summation.
Split the series into a sum of two series. There is an important Taylor series, one of the series in Theorem 3.6.7, that looks a lot like each of the two series.
Use Theorem 3.6.3 to bound the error in a partial-sum approximation. This theorem requires you to consider values of \(c\) between \(x\) and \(x=0\text{;}\) since \(x\) could be anything from \(-2\) to \(1\text{,}\) you should think about values of \(c\) between \(-2\) and \(1\text{.}\)
You know the Maclaurin series for \(\log(1+y)\text{.}\) Use it! Remember that you are asked for a series expansion in powers of \(x-2\text{.}\) So you want \(y\) to be some constant times \(x-2\text{.}\)
For Newton’s method, recall we approximate a root of the function \(g(x)\) in iterations: given an approximation \(x_n\text{,}\) our next approximation is \(x_{n+1}=x_n-\dfrac{g(x_n)}{g'(x_n)}\text{.}\)
To gauge your error, note that from approximation to approximation, the first digits stabilize. Keep refining your approximation until the first two digits stop changing.
First, modify your known Maclaurin series for arctangent into a Maclaurin series for \(f(x)\text{.}\) This series is not hard to repeatedly differentiate, so use it to find a power series for \(f^{(10)}(x)\text{.}\)
