Calculus: Preparation Problem Solutions 1. If f(t) = 4e 0.75t, for what t does f(t) = 0.5? Leave your answer in terms of a logarithm and fractions. Solution The equation implied by this question is and the goal is to solve for t. 0.5 = 4e 0.75t Firstly, any time you have decimals which are easily convertable to fractions, you should rewrite them in fraction form. Decimals are fine for number crunching, but when we are undertaking algebraic manipulation and analysis, we will see things much more clearly with fractions. Thus, our rewritten equation is 1 = 4e 3 4 t Recalling algebra class, this is an exponential equation. We need to clear away as much of the noise around the exponential term, dividing by 4. (Then we can undo the exponentiation by using the inverse operation: taking the logarithm.) 1 8 = e 3 4 t Now we bring the exponential term down by taking a logarithm. The base of the logarithm is the base of the exponential term. In this case, that base is e. This is the base of the natural logarithm, so we use ln rather than log e (the latter is not incorrect, but it is not good manners). Like any similar application of an inverse operation, we must do this to both sides of the equation: ln ( ) 1 = 3 8 4 t I always remember the saying the answer to a logarithm is an exponent. Does the previous equation fit that mold? Yes it does. I can feel good about this. Now, we have a simple final step to move the linear coefficient of 3 4 away from the t: t = 4 3 ln ( 1 8 ) Bonus thinking We ve done what was asked. What if you were on a desert island, with no calculators or log tables, and needed a decimal estimate for t? Here s the thought-processs of my back-of-the-envelope estimate. Note that this kind of mental math is very valuable during prototype design and quick sanity checks of designs or calculations. ( 1 t = ln 8 4 3 ) ( ) = ln 8 4 3 = ln ( 4) = ln 16 I know e.7, and so.7 t 16..7 < 9 and.7 3 < 7, so I might roughly estimate t.5. For the record, t.7758873. My guess was ok, but 3 would be slightly better as a rough estimate.
. At time t = 0 seconds, a projectile is at a position 5m away from some fixed point. The projectile is moving away from the fixed point at a speed of 0 m/s. Write a function d(t) whose value is the distance between the projectile and fixed point at time t. Solution You need to recognize quickly that this is a linear function. Because we want an explicit function, we need our linear equation in slope-intercept form. We have a constant rate of change, 0 m/s; that s the slope. We have a condition that when t = 0, d = 5; that s a point, and it is specifically the y-intercept in this case. We can move directly to y = mx + b. The argument, t, to the function d(t) means that t is the independent variable (i.e., x) and d is the dependent value (i.e., y). d(t) = 0t + 5 You could argue, given the English description and no mention of constraints for the domain of the function, that d(t) = 0t + 5. You need to be very proficient with this sort of question. Most complicated functions ultimately boil down to a lot of little linear pieces connected with ratios, trig function, powers, exponents, and logarithms. You can t focus on the complicated parts if every linear part takes a lot of mental power to create. If you need to practice this, make sure to do drill, drill, drill. 3. A projectile is moving at a constant speed of 15 m/s and on a path directly away from some fixed reference point. At time t = 6 seconds, the projectile is observed to be 15m away from the reference point. Write a function d(t) whose value is the distance, at time t, between the projectile and the reference point. How far from the reference point was the projectile at time t = 0. Solution This function also involves a constant rate of change, and so it is also a linear function. The difference is that the condition given is for a non-zero t. We don t automatically have the y-intercept, we have some other point (t, d) = (6, 15). This calls for point-slope form, y y 1 = m(x x 1 ): d 15 = 15(t 6) Solving for d yields: Thus the function definition is: d = 15t 90 + 15 = 15t + 35 d(t) = 15t + 35 Because d(0) = 35, the projectile was 35m away from the reference point when t = 0. Again, you could argue that we should write d(t) = 15t + 35. We should quickly justify to ourselves that this is reasonable; this is our sanity check. At t = 6, the projectile is at 15m and moving farther away. We d expect the projectile to be closer at t = 0, which it is. Page
AP Calculus Preparation 4. Graph the following function over its domain, being t + 4 h(t) = t + 4 1 016-017 careful to label axes appropriately: if if if if <t<4 t=4 4<t 5 7 < t 10 Solution Pay careful attention to several features of the graph below: Axes are labeled. Open and closed circles are chosen correctly and included in the correct places. No curve is drawn in areas outside the domain. The valid domain of t is (, 5] (7, 10]. (Do you understand the interval notation I just used? A parenthisis is exclusive, a bracket is inclusive.) Figure 1: Graph of piecewise function h(t) Page 3
5. Consider the equation: [ 4x + 10 4 = x 10 ]. (a) Solve for real values of x, if any. Leave radicals unsimplified and unapproximated. Solution This is a problem which provides ample opportunity for you to mess up along the way. A little mistake can ruin all that comes after. Sometimes, that s just how problems are, including some which may be on your AP exam, although the algebra typically doesn t get as hairy as this problem. If you had a general idea what to do, but just couldn t seem to get your answers in part A to match part B, don t feel too bad. I had to work the problem twice to get my answers to align with my calculator check. This is why calculators are a useful invention... but again, you must be able to handle tricky multistep algebra, because in the real world you are going to be dealing with more letters and less numbers. With a radical equation, our goal is to remove radicals by squaring both sides of the equation (for square roots) or raising to some other appropriate power (e.g., third power for cube roots). Before doing so, we usually want to combine radical terms when possible and isolate one radical term on one side of the equation. At first, we can t combine any radical terms, and one is already isolated. Thus, we are ready to square both sides: ( 4x + 10 4) = ( x 10) Squaring the right hand side is easy, but you must remember that you are squaring the entire left side. You must treat it as a binomial, and so you must use the distributive property on ( 4x + 10 4) ( 4x + 10 4). The first step below really ought to be done mentally, and normally you would move straight to the second equation below: [ ] 4x + 10 4x + 10 4 4x + 10 4 4x + 10 + [ 4 4] = x 10 4x + 10 8 4x + 10 + 16 = x 10 Now we must isolate the remaining radical and square both sides again: x + 36 = 8 4x + 10 Note that I mentally grouped like terms without a lot of writing. Now, to make life easier let s reduce coefficients. You might divide by to leave all coefficients as integers, but I prefer to create some relatively-simple fractions here by dividing all terms by 8. x 4 + 9 = 4x + 10 Now to square both sides. You really want to be able to do the fraction arithmetic in your head, but using paper is OK. x 16 + 9 4 x + 81 = 4x + 10 4 Now we have a quadratic equation. We need to get it into standard form. Again, you want to be able to group terms mentally here. You might want to first multiply all terms by 4 to leave only one fraction: x + 9x + 81 = 16x + 40 4...but I went straight to here: On to the quadratic formula: x 4 7x + 41 = 0 x = 7 ± 49 41 1/ Page 4
We could approximate this by hand, but I told you not to do so. That said, it s not too bad to do in this case: x = 14 ± 8 = 14 ± 4 Because they show up so much in trig (and thus in real-world problems), you should know that 1.41 and 3 1.73. We can use the former fact to approximate x, knowing that 4 1.41 = 5.64: x = 14 ± 5.64 So our two solutions for x are 19.64 and 8.36. Thus we definitely have two candidate solutions. How many, if any, are extraneous? You stand a good chance at guessing by visualizing the graphs of the two sides of the original equation... but we will move on to confirmation by calculator/computer graphing... (b) Using a computer or graphing calculator, examine the graphs [ y = 4x + 10 4 ] and [ y = x 10 ] and find decimal approximations of where the curves intersect, if at all. Use this information to confirm your solutions of the equation, and note any extraneous solutions. Discussion Now we re allowed to use a calculator. You should see two root curves risiing through the first quadrant in a concave-down manner. You may have to adjust your zoom window to see that the curves do intersect twice. That means neither solution from part A is extraneous. Now we just need to confirm that the decimal approximations line up. In this case, both solutions look like winners. You can further confirm by asking your calculator to estimate the points of intersection. Typically, you will have to provide the calculator a guess spot to start. You ll learn why in this class. 6. A certain wave is modeled by the function w(t) = 4 sin(3t). What is the first positive value of t for which w(t) =? Answer in radians as a rational multiple of π. Solution First, we must recognize that we are being asked to solve an equation: = 4 sin(3t) Like many trigonometric equations, this one has an infinite number of answers following a pattern. They are all inputs to a triongometric function, so they can be thought of as begin related to angle measures, even when (as appears in this case) the input is actually something like time. Thinking in terms of angles, we ve been asked for the first answer in the first quadrant. Our goal in solving trig equations is to reach a point where either we only have a single trigonometric term, isolated on one side of the equals. Sometimes, we may have a product of multple trigonometric factors, where the product is equal to zero. Sometimes, we have to use a lot of trig identities along the way. This one, however, is tame. We divide by 4 to isolate the sine term: = sin(3t) Now we apply the inverse, sin 1 (AKA arcsin) to both sides of the equation: ( ) sin 1 = 3t We solve for t, which gives us an answer in terms of sin 1 : ( ) t = 1 3 sin 1 You were asked to answer in radians as a multiple of π. Knowing the special values of ( the trig functions on the unit circle (and the standard domain of the inverse trig functions) is key. sin 1 ) = π 4. Thus: t = π 1 Page 5
7. A function p(x) varies in a sinusoidal pattern. It has a range of real numbers between 5 and 50, inclusive, and it has a frequency of 0Hz. Write an equation for p(x). Solution You must know how to assemble a sinusoidal function with amplitude, vertical shift, and frequency/period. Convert the frequency f to the angular velocity ω using [ω = πf] (so 40π = π 0). The amplitude is half the total range ((50 5)/ =.5). The vertical shift is the low end of the range plus the amplitude (5 +.5). p(x) =.5 sin(40πx) + 7.5 8. Determine the exact value of sin( π ) using the sum and difference identities. 1 Solution This calls for the use of trigonometric identities. Easiest to use are the half-angle identities or the sum and difference identities. The following solution will use the difference identities with the observation that π 1 = π 4 π 6 : ( π sin 1) ( π = sin 4 π ) [ ( π ) ( π ) ( π ) ( π )] [ ] 3 = sin cos cos sin = 6 4 6 4 6 1 = 6 4 Page 6
9. Consider the equation: [sin(θ) sin(θ) = 0] (a) Find all valid solutions for θ which lie on the interval [0, π). Provide your solutions in the form of an inverse trig function (e.g., sin 1 ( 3 4 )). Eliminate extraneous solutions, if any. (b) Using your knowledge of exact values on the unit circle, convert solutions from the previous part into angle measures in radians, as multiples of π, where possible. Solution We have nothing but sine terms, but the arguments to the terms are not the same. We have to turn to trig identities to solve this equation, specifically sin(θ) = sin(θ) cos(θ): sin(θ) cos(θ) sin(θ) = 0 Now we can factor the common term out of the difference to form a product: sin(θ)( cos(θ) 1) = 0 This gives us two factors, either of which can equal zero to make the equation true: [sin(θ) = 0] [ cos(θ) 1 = 0] By inspection, the left equation yields θ = sin 1 (0). For the right equation: cos(θ) = 1 ) θ = cos 1 ( 1 We must remember, however, that we were asked for all solutions on the interval [0, π). That means that the left equation yields not just θ = sin 1 (0) = 0 (which is what your calculator would say) but also θ = π 0 = π. That means the right equation yields not just θ = cos ( ) 1 1 = π 3 (also what your calculator would say) but also θ = π π 3 = 5π 3. So the full solution set on the interval is θ = { 0, π 3, π, 5π } 3 Knowing the graphs of sine and cosine and visualizing the functions on the unit circle are keys to finding all answers (c) Using a calculator, find decimal approximations for your solutions. Graph [y = sin(θ) sin(θ)] using a computer or graphing calculator. Use this information to confirm your solutions. Discussion Calculator work confirms our solutions. Looking at the graph, we immediately see that we have four solutions, which much correspond to the x-intercepts, because we have = 0 in the equation. One x-intercept is definitely 0. π is easy to see as well. We also see one a third of the way between 0 and π. Finally, the last one seems to be one-third of a π short of π, but you d probably want to consult the actual decimal approximation here to be sure it s really 5π 3. (d) Again using a graphing program, compare your graph to the graph of y = [ 1 sin(θ) 1 sin(θ)]. Using proper terminology for sinusoidal functions, explain which properties differ between the two graphs and which are identical. Discussion We see that the x-intercepts don t change, only the height of the humps. Thus, the amplitude changed. The frequency/period did not, nor did we introduce any horizontal or vertical shifts. This is what we d expect, because we only changed the coefficients in front of the trig terms, and those correspond to amplitude. Also important is that we changed both coefficients by the same amount. If we had made different amplitudes for each part, we would change the solutions. Page 7
10. An object is formed from half-sphere with a radius of 5cm which is surmounted by a right circular cone of like radius and a length of 10cm. (a) What is the total volume of this object? (b) What is the length, on the surface of the object, from the tip of the cone to the edge of the hemisphere? Solution for now. : This geometric problem is best diagrammed on the board, but here is a paper explanation Figure : A sketched visualization of the problem (not to scale) The relevant volume formulas are V sphere = 4 3 πr3 where r is radius and h is the height of the cone. V cone = 1 3 πr h Thus, a full sphere of r = 5 would have a volume of 500π 3 cm 3 and so the half-sphere has a volume of 50π 3 cm 3. The cone also has a volume of 50π 3 cm 3. Add the two together and you have a total volume V object = 500 3 π cm3 = 166. 6π cm 3 53.599 cm 3 To envision the length along the surface of the half-sphere, think of a (circular) cross-section. The circumference of this circle would be 10π. The relevant line on the half-sphere is one-quarter of the circumference, or.5π cm. The relevant portion of the length along the cone is the hypotenuse of the cone s right-triangle half-cross-section. The Pythagorean Theorem tells us that this length is 10 + 5 = 5 5 cm. Add these together and you get a length For the record, that s approximately 19.034 cm. l =.5π + 5 5 cm Page 8
11. Flight 453 is 0 miles due south of New Haven. Flight 54 is 30 miles due north of New London. New London is 30 miles due east of New Haven. How far apart are the two flights? What is the compass bearing of New London from Flight 453 s position? Assume 0 is due north and that you do not have to worry about magnetic declination. Figure 3: A visualization of the problem As we can see, using the Pythagorean Theorem, d = 900 + 500 = 3400, therefore For the record, that s approximately 58.309 km. d = 10 34 km We can also see that the unknown angle θ = tan 1 ( 3 ). That s approximately θ = 33.69 Because we are looking for a compass bearing, we actually want the complement of θ, which is Thats approximately 56.31. Bearing = 90 tan 1 ( 3 ) Page 9
1. Simplify the following: 66x 5 3 x 4 + 4 x 1 4(x ) 10 x Solution This is a matter of applying exponent rules, fraction rules, and a touch of factoring. First I choose to combine and simplify exponents in terms, remembering that (x m ) n = x mn and n x m = x m/n. 66x 5 3 x 4 + x 3 4x 0 x Now I factor an x out of the top and bottom and cancel. 66x 1 3 x 3 + x 4x 19 Finally, I d like to convert the fractional exponent back into a root. 66 3 x x 3 + x 4x 19 13. Simplify the following: 4x 16 x + 3 (x + ) 5(x + 6x + 9) Solution More of the same rules as the previous problem, along with factoring a trinomial: 4x 16 x + 3 (x 4)(x + 4) x + 3 5(x + 6x + 9) (x + ) 5(x 4)(x + 3) 5(x + 3) (x + 4) 14. Remove the radical from the denominator of the following by multiplying numerator and denominator by the conjugate of the denominator: + x x Solution All that we need to do here is understand the English. The conjugate of x is + x. We use a clever form of one (1 = + x + ) to stay within the logical rules of algebra. x + x x ( ) + x + x 4 + 4 x + x 4 x Page 10
15. Sketch a graph of each of the following. Find and label foci: Solution You should recognize these as conic sections in stanrd form. The first is an ellipse. The second is a hyperbola, due to the addition of a minus in front of one of the fraction terms. (a) (x 4) (y + ) + = 1 5 9 One way to notate the standard form of an ellipse is: (x x c ) (r x ) + (y y c) (r y ) = 1 This is an ellipse centered at (x c, y c ) with a horitzontal semi-axis (half the whole width) of r x and a vertical semi-axis of r y. Thus, this ellipse is centered at (4, ). Because r x = 5 and r y = 3, the ellipse is 10 units wide and 6 units high. Don t forget to invert the signs of the horizontal and vertical shifts, and don t forget to square root the denominators! The foci of the ellipse must lie on the major axis (the longer axis). In this case, r x > r y, so r x is the semi-major axis. Many books refer to the semi-major axis as a and the semi-minor axis as b. They then refer to the distance to a focus (measured from the center) as c. The relationship governing this is c = a b or c = a b. In this ellipse, that means c = 6.4, so the foci are at (4 ± 6, ). Don t forget to subtract rather than add to find c, and don t forget to calculate the foci along the major axis! Figure 4: The Ellipse Page 11
(b) (x + 1) y 16 9 = 1 If one and only one of the fractions has a minus in front of it, the conic section is a hyperbola. The axis which has no minus is the transverse axis and the one with the minus is the conjugate axis. The prefix trans- means across, so this is the axis the hyperbola s curves will cross. The term conjugate should make you think of a companion, like a complex conjugate, and thus you should think of the minus. Hyperbolas have no major and minor axis. Rather, r x = 4 and r y = 3 form half the width and height of a box that guides the slope of the asymptotes. The distance from the center to the foci ( c ) is found using c = a + b AKA c = (r x ) + (r y ), so c = (r x ) + (r y ). The foci always lie along the transverse axis, outside the central box. For this hyerbola, the center is at ( 1, 0) and c = 5, so the foci lie at ( 1 ± 5, 0). Figure 5: The Hyperbola Page 1
16. Sketch a graph of each of the following. Where applicable, label all roots/zeroes, holes/point discontinuities, poles/asymptotes, y-intercepts, and end behaviors: Solution Important rules for rational functions before we get into our calculations: The factored binomials (x n) show us the roots, holes, and poles: Binomials that appear only on the top give us roots at x = n Binomials that appear only on the bottom give us poles at x = n Binomials that appear equally on the top and bottom give us holes at the point (x n, y n ), where x n = n and y n is found by eliminating the paired binomials and inserting x = n for other x s in the equation for y. End behavior depends on the overall degree (# of x s) of the rational expression: If the degree on the bottom is higher than the top, the end behavior is y 0 If the degree of the top and bottom are equal, the end behavior is y r, where r is the ratio of the leading coefficients of the numerator and denominator when each is expressed as a polynomial in standard form. If the degree on the top is higher than the bottom, then y ±, where the infinity approached in each direction depends on whether the degree is even or odd and whether r is positive or negative. x + 3 (a) y = (x 3)(x + 4) This one has a root at x = 3 and poles at x = 3 and x = 4. The degree on the bottom () is higher than the top (1), so this curve approaches y = 0 for large x. The y-intercept (0, 1 4 ) helps us see that the curve zig-zags from positive to negative between the two poles. At the far right, the curve should be positive, because a large positive x will yield a large positive y. At the far left, the curve should be negative, because a large negative x will yield a large negative y. (3x + 6)(x ) (b) y = (x )(x + 1) This one has a root at x = (solve 3x + 6 = 0), a pole at x = 1, and a hole at x =. How do we find the y-coordinate for the hole? Eliminate the x terms: (3x + 6)(x ) y = (x )(x + 1) 3x + 6 x + 1 Substitute x = : y = 3 +6 +1 = 4 Thus, the hole is at (, 4). The degree of the top () and bottom () are equal, so the end behavior depends on the leading coefficients: y 3 1. x(x + )(x ) (c) y = x 1 This one has roots at x =, x = 0, x = and a pole at x = 1. The degree of the top (3) is larger than the degree of the bottom (1). The overall degree is 3 1 =. This is an even degree, so the end behavior should be like a parabola: either both ends head to or both head to. How do we figure out which? First, connecting the features in the curve can give us an idea. Second, what if we subsitute a very large positive number for x? We will get a very large positive number for y. Therefore, we can reason that the end behavior for large positive x is y. The same will be true for large negative x because the function is even. If the function was odd, we d see the other end head to. Page 13
Figure 6: Rational Graphs A & B Page 14
Figure 7: Rational Graph C Page 15