F24221Midterm2study
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University of Wisconsin, Madison**We aren't endorsed by this school
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MATH 221
Subject
Mathematics
Date
Jan 5, 2025
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4
Uploaded by SargentMoose4627
Math 221 - Midterm 2 TopicsSections Covered: 2.6, 2.8, 3.1, 3.2, 3.3, 3.4, 3.5, 3.7, 3.9, 4.1, 4.2, 4.3Implicit Differentiation:To take an implicit derivative, we pretend one variable is a function of another, then takederivatives using Chain Rule.Example:Suppose a curve is given by the equationxy2+y−x3= 2, and we want to finddydxat the point (−2,2).Solving forywould be a pain here, but we can pretend that we already knowyis a function ofx.So, we writex(y(x))2+y(x)−x3= 2, and take the derivative of both sides:ddx(x(y(x))2+y(x)−x3) =ddx2ddx(x(y(x))2) +ddxy(x)−ddxx3= 0y(x)2+ 2xy(x)y′(x) +y′(x)−3x2= 0.Now, we can rearrange to solve fory′(x):y′=3x2−y22xy+ 1.Now, to find the derivative at (−2,2), we just plug in to findy′=−8/7.Related Rates:Here are steps to approach a related rates problem.Let’s use a running example of just a cubeexpanding, so that the area increases at a constant rate of 6 cm3/sec, and we want to find how quickly each side lengthis increasing when the volume is 8cm3.1.Draw a picture.Never skip this step! This is the most useful step!2.Write down any rate(s) that you know because they are given by the problem.In our example, we know the rateat which the volume is increasing, so we write downdVdt= 6.3.Identify the rate we want to know.In our example, we want to know how quickly the side lengths are increasing,so we want to knowdℓdt.4.Use your picture to write down a relation between the quantities.In our example, we want to relate the volume tothe side lengths, and these are related byV=ℓ3.5.Take the implicit derivative with respect to timet.In our example, we getdVdt= 3ℓ2dℓdt.6.Solve for the rate we want.In our example, whenV= 8,ℓ= 2, sodℓdt=63·22=12.Derivatives and the Shape of a Graph:•Whenf′(x)>0, the function isincreasing, whenf′(x)<0, the function isdecreasing.•Whenf′′(x)>0, the function isconcave up, so the graph is curling upwards. Whenf′′(x)<0, the function isconcave down, so the graph is curling downward.•A point wheref′′(x) goes from positive to negative is called aninflection point.Finding Maximum and Minimum Values:We can use what we know about derivatives and the shape of a graphto find maximum and minimum values of functions. There are a couple different kinds of problems that can show up:•Finding theglobal(orabsolute) maximum and minimum values of a functionf(x) on a closed interval [a, b].•Finding thelocal(orrelative) maximum and minimum values of a functionf(x), on any interval.The methods between these are similar, but have key differences.1. For both problems, we always start by finding thecritical pointsof the functionf(x). A critical point is a valueofxwhere either•f′(x) = 0, or•f′(x) is undefined.1
This is because a local max/min can only happen at a point wheref′(x) changes sign, which can only happen at acritical point.For example, iff(x) =x4/3−2x2/3, thenf′(x) =43√x3−433√x=43(3√x−1/3√x). When we setf′(x) = 0, we get3√x= 1/3√x, sox2/3= 1. Cubing both sides we getx2= 1, sox=±1. When we look for wheref′(x) is undefined,we see that it is undefined whenx= 0. So, the critical points arex= 0,±1.2. Now, it depends on which problem you’re solving:(a) If you’re trying to find global maxes/mins on a closed interval [a, b], you should find the values off(x) at•Allcritical points, and•the two endpointsaandb.The largest value corresponds to the absolute maximum, and the smallest is the absolute minimum.(b) If you’re trying to find local maxes/mins, you need to check each critical point to decide if it is a local max,local min, or neither. You have two options:•The First Derivative Test:Make a sign chart for thef′(x). Anywhere the derivative goes from positiveto negative,f(x) goes from increasing to decreasing, so that is a local max. Anywhere the derivative goesfrom negative to positive,f(x) goes from decreasing to increasing, so that is a local min. Anything else isneither.•The Second Derivative Test:Take the second derivativef′′(x), and check the sign at each criticalpoint.Iff′′(x)>0, the graph off(x) is concave up (like a cup), so the point is a localminimum.Iff′′(x)<0, the graph off(x) is concave down (like a frown), so the point is a localmaximum.Optimization Problems:Here’s steps to approach an optimization problem. Let’s take a running example where weare trying to build a fishtank for a pet mola mola (sunfish). We need the the tank to be a box with glass sides and aplastic top and bottom, and we want the box to be as long as it is wide. In order for our sunfish to thrive, the volumeof the tank must be 2 m3. However, the glass sides of the tank are more expensive than the plastic top and bottom: theglass is $480/m2, while the top and bottom are $120/m2. We want to find the minimum cost to build our aquarium.1.Draw a picture.Just like with related rates, this is the most useful step!2.Write down any quantities you know because they are given by the problem.In our example, we know that thevolume of the tank is 2 m3. If the side lengths of the aquarium areℓ,w, andh– for the length, width, and height– then we have thatℓ·w·h= 2. We also know that the box is as long as it is tall, soℓ=w, so we can rewrite theabove ashw2= 2.3.Write down the quantity you want to maximize/minimize.In our example, we want to minimize our total cost. Weneed to pay for two plastic pieces that areℓ×wrectangles, and we need to pay for two glass pieces that areℓ×hrectangles, and two that arew×hrectangles. So, our total cost is 2·120ℓ·w+ 2·480ℓ·h+ 2·480w·h. Using thatℓ=w, our total cost becomes 2·120w2+ 2·480wh+ 960wh= 2·120w2+ 4·480wh.4.Use the known information to solve for one of the variables.In our example, we can solve forh= 2/w2.5.Substitute into the function we’re trying to optimize.Our total cost above becomes 2·120w2+ 8·480/w.6.Find the maximum/minimum values of the function.Use the methods from the previous section: take the derivative,find the critical points, and determine which ones give the actual answer.In our example, we take the derivative to get 4·120w−8·480/w2. Setting this equal to 0, we getw3=8·4804·120= 8,sow= 2. We also get a critical point atw= 0, because there the derivative is not well defined, butw= 0 doesn’tmake sense for a length.7.ANSWER THE QUESTION.In our case, the question isn’t what the width should be, the question is what theminimum cost is. Plugging back in, we find that the total cost is 8·120 + 4·480 = 6·480 = $2880.Asymptotes:A vertical asymptote is a valueawhere limx→af(x) =∞.(Whether it is positive or negative doesn’tmatter, and it doesn’t matter if it’s the same sign on both sides.) To find vertical asymptotes:1.First identify the possible values wheref(x) is not defined.Usually,f(x) will be a fraction of two functions, andthis will be the places where the denominator is 0.2.Check each possible value to see what the limit is.Youmustcheck if limx→af(x) =∞, and be careful because somepoints might be removable discontinuities!Horizontal and slant asymptotes both have to do withlimx→±∞f(x). To find horizontal and slant asymptotes:1.Write down the limitslimx→∞f(x),limx→−∞f(x) .2
2.Pull out the largest powers ofxin the numerator and denominator.For example, iff(x) =3x+ 2x2−x, we rewrite asx(3 + 2/x)x2(1−1/x)=3 + 2/xx(1−1/x).3. If the limit is a number, it is a horizontal asymptote. If the power ofxin the numerator is one more than in thedenominator, there is a slant asymptote, and you can find it by doing long division.Antiderivatives:We know how to take the following antiderivatives.1. The antiderivative ofxnisxn+1/(n+ 1).2. The antiderivative of sin(x) is−cos(x), and the antiderivative of cos(x) is sin(x).3. The antiderivative of sec2(x) is tan(x), and the antiderivative sec(x) tan(x) is sec(x).4. Ifcis a constant, andF(x) is an antiderivative off(x), thencF(x) is an antiderivative ofcf(x).5. IfF(x) is an antiderivative off(x), andG(x) is an antiderivative ofg(x), thenF(x) +G(x) is an antiderivative off(x) +g(x).6.REMEMBER +C!!Definite Integrals and Areas:The definite integralZbaf(x)dxis defined to be the area under the graph off(x)betweenx=aandx=b. . . sort of. In calculus we adopt two funny conventions:•Area above thex-axis counts as “positive area”, and area below thex-axis counts as “negative area”.•Also, if you integrate by going the other way, you get “negative area”, soZabf(x)dx=−Zbaf(x)dx.Sometimes, we can find definite integrals using geometry.•For a constant functiony=c, the area under the curve will be the area of a rectangle.•For the graph ofy=mx+c, the area under the curve will be the area of a triangle or trapezoid.•For the graph ofy=√r2−x2, the area under the curve will be the area of (some portion of) a half-circle.When we can’t find definite integrals using geometry, we approximate them using rectangles.Example:The table gives values for a function for various values ofx, correct to two decimal places.xf(x)010.20.980.40.920.60.830.80.7110.57We can estimate the definite integralZ10f(x)dxusing 5 rectangles with left endpoints.In thiscase, we would getarea≈0.2·1 + 0.2·0.98 + 0.2·0.92 + 0.2·0.83 + 0.2·0.71 = 0.888.Now, since the functionf(x) is decreasing on this interval, and since we used left endpoints, thiswill be an overestimate of the area.On the other hand, using right endpoints, we getarea≈0.2(0.98 + 0.92 + 0.83 + 0.71 + 0.57) = 0.802.This will be an underestimate.Below is a picture of this function and the rectangles approximating it.0.20.40.60.810.20.40.60.813
Fundamental Theorem of Calculus:The Fundamental Theorem of Calculus gives us a powerful tool for findingdefinite integrals.It says that ifF(x) is an antiderivative off(x), thenZbaf(x)dx=F(b)−F(a).In other words,integrating is “the same” as anti-differentiating.Example:Let’s find the total area under the graph of sin(x) fromx= 0 tox=π. We know that−cos(x) is anantiderivative for sin(x), so by FTCZπ0sin(x)dx=−cos(x)π0=−cos(π)−(−cos(0))=−(−1)−(−1)= 2.So, the total area under the graph of sin(x) fromx= 0 tox=πis 2. There is no way we could have found that usingjust geometry! This is what makes calculus so powerful!4