Answers to the Practice Problems for Test 2


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1 Answers to the Practice Problems for Test 2 Davi Murphy. Fin f (x) if it is known that x [f(2x)] = x2. By the chain rule, x [f(2x)] = f (2x) 2, so 2f (2x) = x 2. Hence f (2x) = x 2 /2, but the lefthan sie (f (2x)) is written in terms of 2x while the righthan sie (x 2 /2) is written just in terms of x, so we nee to write the righthan sie in terms of 2x as this is the argument of f on the left. So x 2 /2 = (2x) 2 /8 (o some algebra to check this) implies that f (2x) = (2x) 2 /8 for all x, so that f (x) = x 2 /8. 2. (a) Given g(x) = sin x, fin g (42) (x). We re looking for a pattern so we can figure out g (42) (x) without actually taking 42 erivatives! So observe that g () (x) = cos x, g (2) (x) = sin x, g (3) (x) = cos x an g (4) (x) = sin x is equal to g(x) again. So every fourth erivative brings us back to sin x, so g (40) (x) = sin x (0 cycles of 4 erivatives), which implies that g (4) (x) = cos x an g (42) (x) = sin x. (b) Given f(x) = x, fin f (42) (x). Let s first rewrite f(x) = x then use the power rule (without simplifying the coefficient!) to fin the pattern. We have f () (x) = ( )x 2, f (2) (x) = ( )( 2)x 3, f (3) (x) = ( )( 2)( 3)x 4, f (4) (x) = ( )( 2)( 3)( 4)x 5. With this many erivatives compute, let s see if we recognize a pattern yet. Notice that in the FIRST erivative, there is ONE negative sign; in the SECOND erivative, there are TWO negative signs; in the THIRD erivative, there are THREE negative signs; in the FOURTH erivative, there are FOUR negative signs. This is a pattern, which will continue, so in the 42n erivative there will be 42 negative signs, an ( ) 42 =. Next, now that we ve taken care of the negative signs, what s left of the coefficients? In the FIRST erivative, we have ; in the SECOND erivative we have 2; in the THIRD erivative, we have 2 3; in the FOURTH erivative we have What s the pattern? Well, in the nth erivative, we have 2 3 n = n!, so in the 42n erivative our coefficient will be ( ) 42 (42)! = (+)(42)! = 42! (This is a really big number!) Finally, what is the exponent on x in the erivatives of f(x)? In the FIRST erivative, the exponent is 2; in the SECOND, it is 3; in the THIRD, it is 4; in the FOURTH, it is 5. So in the nth erivative, the exponent is n, an thus f (n) (x) = ( ) n n!x n. In particular, f (42) = 42!x 43. (c) If h(x) = 2f(x) + 3g(x), fin h (42) (x). By the sum rule an constant multiple rule, we have h (x) = 2f (x) + 3g (x), h (x) = 2f (x) + 3g (x), an so on. In general, the coefficients 2 an 3 will always be there, so h (42) = 2f (42) (x) + 3g (42) (x) = 2[42!x 43 ] + 3[ sin x] = 2 42!x 43 3 sin x. 3. For what values of c oes the equation ln x = cx 2 have exactly one solution? (Don t forget that c can be either positive or negative.) Whenever c 0, the graph of y = cx 2 is either the xaxis (when c = 0) or a parabola opening own. In both cases, the parabola will have to intersect the graph of y = ln x an o so
2 exactly once. Now, if c > 0, then for c too big the parabola will open an grow faster than ln x so that the two curves will never intersect. For c > 0 but very small, the parabola will be very wie an so will cross the graph of y = ln x twice. The one time that c > 0 an the graph of y = cx 2 intersects y = ln x exactly once must occur in such a way, then, that the two curves have the same tangent line at their point of intersection (if not, then the slope of cx 2 will be less than that of ln x an so the parabola will intersect ln x twice). So this gives us the secon equation 2cx = /x which yiels x 2 = /(2c). Plugging this relationship into the first equation, ln x = cx 2 = c[/(2c)] = /2, which only happens when x = e /2 an c = /(2x 2 ) implies that c = /[2(e /2 ) 2 ] = /(2e). Thus the parabola y = cx 2 intersects the curve y = ln x exactly once when either c 0 or c = /(2e). 4. A particle moves on a vertical line so that its coorinate at time t is y = t 3 2t + 3, t 0. (a) Fin the velocity an acceleration functions. The velocity is the rate of change of the position function, so v(t) = y (t) = 3t 2 2. The acceleration is the rate of change of the velocity, so a(t) = v (t) = 6t. (b) When is the particle moving upwar an when is it moving ownwar? The particle is moving upwar when its velocity is positive, which happens when v(t) = 3t 2 2 > 0. Thus this occurs while 3t 2 > 2, i.e., t 2 > 4. This happens when t > 2 or when t < 2, but as we are only observing the particle when t 0, the particle is moving upwar for t > 2 secons. The particle is moving ownwar when v(t) = 3t 2 2 < 0, so whenever t 2 < 4, which is true for 2 < t < 2. Again, as we only observe the particle for t 0, the particle is moving ownwar when 0 t < 2 secons. (c) Fin the total istance that the particle travels in the time interval 0 t 3. The total istance the particle travels is equal to the istance it travels while going own plus the istance it travels while going up again, which is equal to y(2) y(0) + y(3) y(2), since for 0 t < 2 the particle was going own an for 2 < t 3 it is going up. Thus the total istance is ( ) ( ) + ( ) ( ) = ( 3) (3) + ( 6) ( 3) = = 23. () Sketch graphs of the position, velocity an acceleration functions for 0 t 3. Use your calculator an the formulas from part (a) to o this. (e) Where is the graph of the position concave up? Concave own? When is the particle speeing up? When is it slowing own? The graph of the position is concave up when y (t) > 0, but y (t) = a(t) = 6t is > 0 whenever t > 0, so the curve is concave up for all t > 0 an is never concave own for t 0. The answer to when is the particle speeing up an when is it slowing own epens on our interpretation of speeing up an slowing own as referring either to the velocity or to the spee of the particle. If we mean the velocity, then speeing up will occur whenever the rate of change of v(t) is positive, but the rate of change of v(t) is v (t) = a(t) = 6t is positive whenever t > 0 an it is never negative so the velocity in never slowing own for t 0. If, however, by speeing up an slowing own we are referring to the spee of the particle, which is equal to the absolute value of the velocity, v(t) = 3t 2 2, then the spee is increasing when t > 2 an ecreasing for 0 t < 2. 2
3 5. Use[ the] efinition of erivative to prove the Reciprocal Rule: If g is ifferentiable, then x g(x) = g (x). [g(x)] 2 Using the efinition of erivative, [ ] [ ] [ lim g(x) g(x+h) h 0 = g (x) [g(x)] 2. h g(x)g(x+h) 6. Evaluate the following limits: (a) lim x π e 2 sin x x π : x [ ] g(x+h) g(x) = lim g(x) h 0 ] [ h lim h 0 g(x+h) g(x) h lim h 0 g(x)g(x+h) g(x) g(x+h) = lim h 0 ] [ = [ g (x)] hg(x)g(x+h) = e This has the ineterminate form 0/0, so consier the limit lim 2 sin x [2 cos x] x π = e2(0) [2()] e 2. Thus, by l Hôpital s Rule, lim 2 sin x x π x π = 2 as well. (b) lim θ 0 cos θ θ 2 This also has the ineterminate form 0/0, so consier lim θ 0 sin θ 2θ [g(x)] 2 ] = =. Again, this is ine. Therefore, by l Hôpital s, so a secon application of l Hôpital s Rule implies that terminate of the form 0/0, so we consier lim θ 0 cos θ 2 = 2 sin θ Rule, we have lim θ 0 2θ cos θ lim θ 0 = θ 2 2 too. θ+sin θ (c) lim θ 0 tan θ = 2 Once more, we have the ineterminate form 0/0, so consier lim θ 0 +cos θ sec 2 θ = + Hence, applying l Hôpital s Rule, lim θ 0 θ+sin θ tan θ = 2. 2 = Let f be a function such that f(2) = an whose erivative is known to be f (x) = x (a) Use a linear approximation to estimate the value of f(2.03). In general, the linear approximation of a function f(x) near x = a is the equation of the tangent line to the curve y = f(x) when x = a, which is L(x) = f(a) + f (a)[x a]. Thus, for our function, the linear approximating function is L(x) = f(2) + f (2)[x 2] = () + (6)[x 2] (using f (x) = x evaluate when x = 2 to compute f (2) = 6). Therefore, our approximation of f(2.03) L(2.03) = () + (6)[2.03 2] =.8. (b) Will the exact value of f(2.03) be less than or greater than your estimate? Clearly explain why. To etermine whether f(2.03) will be less or greater than L(2.03) epens on the concavity of the curve y = f(x) when x = 2, for if it is concave up, then the tangent line lies below the curve so that L(x) f(x) an our approximation L(2.03) < f(2.03) will be too small. Otherwise, if y = f(x) is concave own when x = 2, the tangent line lies above the curve y = f(x) so that L(x) f(x) an hence L(2.03) f(2.03) will be too big. To ecie whether y = f(x) is concave up or own when x = 2, we check f (x) = x [f (x)] = x [ x 5 + 4] = 2 (x5 +4) /2 [5x 4 ], so f (2) = 20/3 > 0, implies that y = f(x) is concave up when x = 2 so our approximation L(2.03) =.8 will be too small. 8. An airplane is traveling in an elliptical holing pattern escribe by the parametric equations x = 4 cos t y = 3 sin t, 3
4 where x an y have units of miles. The control tower is 5 miles east of the origin. At what point will the airplane be flying irectly towar the control tower? We want to fin where the tangent line to the ellipse passes through the point (5, 0), which correspons to the location of the control tower, an such that the plane is flying towar rather than away from this point. As t increases, the plane goes aroun the ellipse counterclockwise, so the time when it is flying towar the control tower will be when it is in the IVth quarant rather than when it is in the Ist. Now the equation for the tangent line to the ellipse at time t is given by y (3 sin t) = y x t[x (4 cos t)], where y x t = y/ x/ = 3 cos t 4 sin t. Hence the equation of the tangent line is y 3 sin t = 3 cos t 4 sin t [x 4 cos t], an we nee this line to pass through the point (5, 0), which is the location of the control tower. Thus, replace y by 0 an x by 5 an let s see what we get: 0 3 sin t = 3 cos t 4 sin t [5 4 cos t]. Multiplying both sies by 4 sin t an istributing the 3 cos t on the righthan sie we obtain 2 sin 2 t = 5 cos t 2 cos 2 t. Aing 2 cos 2 t to both sies an recognizing the Pythagorean ientity, we have 2 sin 2 t + 2 cos 2 t = 2 which will equal 5 cos t, so cos t = 2/5 = 4/5. Thus t is the angle such that the ajacent sie has length 4 an the hypotenuse has length 5, but we want t to be an angle in the IVth quarant, so the opposite sie must be negative, so it is 3. Hence sin t = 3/5. Therefore, the plane will be flying irectly towar the control tower when x = 4 cos t = 4(4/5) = 6/5 an y = 3 sin t = 3( 3/5) = 9/5, so when the plane is at the location (6/5, 9/5). 9. Fin points P an Q on the parabola y = x 2 so that the triangle ABC forme by the xaxis an the tangent lines at P an Q is an equilateral triangle. The triangle forme by the three lines will be equilateral when the angles insie at the base vertices are both 60. Thus, if P is the point on the left sie of the origin an Q is the point on the right sie of the origin on the parabola y = x 2, we must have the slope of y = x 2 at P is equal to tan(60 ) = 3 an the slope of y = x 2 at Q is tan(60 ) = 3. Yet the slope of y = x 2 for any x is y/x = 2x, so the value of x at P satisfies 2x = 3, so x = 3/2 an y = ( 3/2) 2 = /4 at P. At Q, 2x = 3, so x = 3/2 an y = /4. Hence P = ( 3/2, /4) an Q = ( 3/2, /4). 0. Implicitly ifferentiate the equation x 4 + 4x 2 y 2 + y 4 = x 3 y with respect to x an solve for y x. Differentiating with respect to x, we have 4x 3 + (4x 2 )[2y y x ] + (y2 )[8x] + 4y 3 y x = (x3 )[ y (y)[3x 2 ]. Put everything involving y x on the left an everything else on right to get 8x2 y y 4y 3 y y x x3 x = [8x2 y + 4y 3 x 3 ] y x = 3x2 y 4x 3 8xy 2, so y x = 3x2 y 4x 3 8xy 2. 8x 2 y+4y 3 x 3 x ] +. Water is flowing at a constant rate into a spherical tank. Let V (t) be the volume of water in the tank an H(t) be the height of the water in the tank at time t. (a) What are the meanings of V (t) an H (t)? Are these erivatives positive, negative or zero? V (t) refers to the rate at which the volume is changing, an thus is the constant rate at which water is flowing into the tank. H (t) means the rate at which the water level in the tank is changing. (b) Is V (t) positive, negative or zero? Explain. As V (t) is constant, its erivative V (t) is zero. x + 4
5 (c) Let t, t 2 an t 3 be times when the tank is onequarter full, half full, an threequarters full, respectively. Are the values H (t ), H (t 2 ), H (t 3 ) positive, negative or zero? Why? When the tank is onequarter full, as water continues to flow in at a constant rate the wih is increasing so the neee increase in epth is ecreasing, so H (t ) < 0. When the tank is threequarters full, the wihs of the tank are ecreasing as we continue to pour water in at a constant rate, so to accommoate the steay increase in volume while the wih ecreases, the change in height must be increasing, so H (t 3 ) > 0. When the tank if half full, we re changing from the case where wihs were increasing so that H (t) < 0 to where wihs are ecreasing so that H (t) > 0. Hence, when the tank is half full H (t) = 0 must be the transition from H (t) < 0 to H (t) > (a) Explain why x = x 2 for all real numbers x. By x 2, we mean the positive square root of the positive number x 2, which will therefore be x. (b) Use part (a) an an the Chain Rule to show that x x = x x. Employing part (a) an the chain rule, x [ x ] = x [ x 2 ] = 2 (x2 ) /2 [2x] = x = x x 2 x. (c) If f(x) = sin x, fin f (x) an sketch the graph of f an f. Where is f not ifferentiable? Using the chain rule an the result of part (b), we have f (x) = sin x sin x [cos x], which is efine so long as sin x 0. Thus f(x) is not ifferentiable when x = 0, ±π, ±2π, ±3π,.... () If g(x) = sin x, fin g (x) an sketch the graph of g an g. Where is g not ifferentiable? Using (b) an the chain rule, we have g x (x) = cos x x, which is efine so long as x 0. Thus g(x) = sin x is not ifferentiable only when x = A hockey team plays in an arena with a seating capacity of 5,000 spectators. With ticket prices at $2, average attenance at a game has been,000. A market survey inicates that for each ollar that ticket prices are lowere, the average attenance will increase by 000. How shoul the owners of the team set ticket prices to maximize their revenue from ticket sales? Currently eman satisfies p = 2, q = 000 but our market survey tells us that when p =, we ll have q = Base on this, assuming eman is linear, using the pointslope equation for the line we have q 000 = [p 2] = 000[p 2]. Hence q = [p 2] = p+2000 = p. Therefore the owner s revenue is R(p) = pq = p[ p] = 23000p 000p 2. To fin when this is maximize, consier R (p) = p, which is always efine but is zero when p = 23000/2000 =.50. The revenue is inee maximize (in fact we have an absolute maximum) when p =.50 as R (p) = 2000 is always negative so the revenue function is always concave own. Therefore, the owner s shoul lower ticket prices to $.50 in orer to maximize their revenue. 4. Let f(x) = x 3 + 3x 2 + 9x 8. (a) Fin the relative extrema point(s) of f. First, consier f (x) = 3x 2 + 6x + 9 = 3(x 2 2x 3) = 3(x 3)(x + ). This is efine for all x, so our only critical numbers are when it is zero, which is when x = 3 or x =. To etermine if we have relative max or min at these points, consier the Secon Derivative Test: f (x) = 6x + 6, so f (3) = 2 < 0 implies f has a local max when x = 3 an f ( ) = +2 > 0 implies that f has a local min when x =. Thus, f has a local max point at (3, f(3)) = (3, 9) an a local min point at (, f( )) = (, 23). 5
6 (b) Determine the inflection point(s) of f. The inflection point(s) of f occur where the concavity changes, which it etermine by the secon erivative changing sign. Now f (x) = 6x + 6 = 0 only when x = an for x <, f (x) > 0 (e.g., f (0) = 6 > 0) so f is concave up while for x >, f (x) < 0 (e.g., f (2) = 6 < 0) so f is concave own. Thus f changes from being concave up to concave own at x =, so f has an inflection point at (, f()) = (, 7). (c) When is the graph of y = f(x) concave up? When is it concave own? We just answere this question above: f is concave own when f < 0, on the interval (, ) an f is concave up when f > 0 on the interval (, ). () When is the graph of y = f(x) increasing? When is it ecreasing? The function is increasing when f > 0 an ecreasing when f < 0. Our critical numbers were x = an x = 3, so checking f ( 2) = 5 < 0 implies f is ecreasing on the interval (, ); f (0) = 9 > 0 implies f is increasing on the interval (, 3); an f (4) = 5 < 0 implies that f is ecreasing on the interval (3, ). 5. A metal storage tank with volume 000 m 3 is to be constructe in the shape of a right circular cyliner surmounte by a hemisphere. What imensions will require the least amount of metal? Let r be the raius of the cyliner (an hence also of the hemisphere) an h be the height of the cyliner. Then the volume of the tank is V = πr 2 h + 2 ( 4 3 πr3 ) = πr 2 h πr3. Hence 000 = πr 2 h πr3, so πr 2 h = πr3 which means h = πr3 πr 2. In our first interpretation of this problem, let s assume the tank has no bottom but only the sies an top. Then the amount of metal use to make the tank is equal to the lateral surface area of the cyliner plus the surface area of the hemisphere, which is A = 2πrh + 2 (4πr2 ) = 2πr πr3 πr 2 + 2πr 2 = 2000r 4 3 πr2 + 2πr 2 = 2000r πr2. To minimize this A, consier A (r) = 2000r πr = 4 3 πr r 2, which is not efine when r = 0 (but r = 0 is not in the omain of A(r) since the volume is 000 an if r = 0 then the volume woul be 0 also) an is zero when 4 3 πr3 = 2000, so r = (500/π) /3 an h = 0 (using the formula relating h to r above). This means that our tank is really just the hemisphere without the cyliner base, which is fine. To confirm that this is a minimum, look at A (r) = 4000r π, which is > 0 for all r > 0, so A(r) is concave up on the entire omain {r > 0} an hence we have an absolute minimum surface area when r = (500/π) /3, h = 0. A secon interpretation of the problem says the tank shoul inclue a circular base, so that the area is now A(r) = 2000r πr2 + πr 2 = 2000r πr2 (the same as the area formula above except that we now a another πr 2 for the circular bottom). Then this area is minimize when A (r) = 2000r πr = 0 3 πr r 2 6
7 is not efine (when r = 0, which is still not in the omain so this isn t a critical number) or when it is zero. But A (r) = 0 when 0 3 πr3 = 2000, so r = (600/π) /3 = an h = as well (again using the formula for h above). To ensure that the area is minimize for this value of r, check that A (r) = 4000r π is > 0 for all r > 0, which is the omain of A(r), so that the function is always concave up an hence our critical point is at an absolute minumum. 6. Suppose that the price p an the number of copies q (measure in 00s) of an ol vieo game satisfy the eman equation 6p + q + qp = 94. Determine the rate at which the quantity is changing when q = 4, p = 9 an the price is falling $.50 per week. This is a relate rates problem, so ifferentiate with respect to time, t: 6 p + q (p)[ q ] = 0. Now evaluate when q = 4, p = 9,a n p (4)[.5] + (9) q = q = 0 implies that q =.5 to fin q = 5/0 =.5. p + (q)[ ] + q : 6(.5) + 7. Suppose that the cost function of proucing R raios is given by the formula C(R) = 00+ R. Fin the marginal cost at prouction level R = 00. The marginal cost is the erivative of the cost function, so is C (R) = 2 R /2. Hence the marginal cost when proucing R = 00 raios is C (00) = 2 = /20 = A water trough is 0 m long an a crosssection has the shape of an isosceles trapezoi that is 30 cm wie at the bottom an 80 cm wie at the top, an has a height 50 cm. If the trough is being fille with water at the rate of 0.2 ft 3 /min, how fast is the water level rising when the water is 30 cm eep? The volume of the trough is going to be the area of the trapezoi times 0, since the trough is 0 m long. Let h be the epth of water in the trough an w be the wih of the water s surface in the trough when the epth is h. We nee to convert all of the cm measurements to m, since the length an the change in volume, V/ = 0.2 are expresse in terms of meters rather than centimeters, so the base of the trapezoi is 30 cm = 0.3 m, the top is 80 cm = 0.8 m an the height is 50 cm = 0.5 m. This isoceles trapezoi can be broken up into a rectangle of wih 0.3 an height 0.5 an two congruent triangles (because the trapezoi is isoceles) of height 0.5 m. As the top of the trapezoi has length 0.8 an 0.3 m of that is use up by the rectangle, the remaining 0.5 m must be evenly ivie between the two triangles on either en, so the tops of the triangles have length 0.25 m. This is half of their height, so when the water is h meters eep in the trough, the wih of the water s surface within one of these triangles (using a similar triangle arguement) is half the epth. Thus w = 0.5h h = h+0.3. Therefore the area of the trapezoi when the water is h meters eep is A = 2 [(0.3) + (h + 0.3)]h an the volume is thus V = 0A = 5[0.6+h]h = 3h+5h 2. Therefore V = 3 h h h +0h = (3+0h). Thus 0.2 = [3 + 0(0.3)] h = 6 h h, so = 0.2/6 = m/min, which is 3.33 cm/min. + 7
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