Unit #4 - Inverse Trig, Interpreting Derivatives, Newton s Metho Some problems an solutions selecte or aapte from Hughes-Hallett Calculus. Computing Inverse Trig Derivatives. Starting with the inverse property that sin(arcsin(x)) = x, fin the erivative of arcsin(x). You will nee to use the trig ientity sin (x) + cos (x) =. We begin with: sin(arcsin(x)) = x Take the x erivative of both sies: x [sin(arcsin(x))] = x [x] ( ) cos(arcsin(x)) x arcsin(x) = Solve for x arcsin(x) : This is true, but not helpful. x arcsin(x) = cos(arcsin(x)) Recall: sin (θ) + cos (θ) = so cos(θ) = sin (θ) Using this formula in the arcsin erivative formula, But sin(arcsin(x)) = x, so x arcsin(x) = [sin(arcsin(x))] x arcsin(x) = x We begin with: cos(arccos(x)) = x Take the x erivative of both sies: x [cos(arccos(x))] = x [x] ( ) sin(arccos(x)) x arccos(x) = Solve for x arccos(x) : This is true, but not helpful. Recall: sin (θ) + cos (θ) = so x arccos(x) = sin(arccos(x)) sin(θ) = cos (θ) Using this formula in the arccos erivative formula, But cos(arccos(x)) = x, so x arccos(x) = [cos(arccos(x))] x arccos(x) = x. Starting with the inverse property that cos(arccos(x)) = x, fin the erivative of arccos(x). You will nee to use the trig ientity sin (x) + cos (x) =. 3. Starting with the inverse property that tan(arctan(x)) = x, fin the erivative of arctan(x). You will nee to use the trig ientity sin (x) + cos (x) =, or its relate form, iviing each term by cos (x), tan (x) + = sec (x)
We begin with: tan(arctan(x)) = x Take the x erivative of both sies: x [tan(arctan(x))] = x [x] ( ) sec (arctan(x)) x arctan(x) = Solve for x arctan(x) : This is true, but not helpful. Recall: x arctan(x) = sec (arctan(x)) tan (θ) + = sec (θ) so sec (θ) = + tan (θ) Using this formula in the arctan erivative formula, But tan(arctan(x)) = x, so x arctan(x) = + [tan(arctan(x))] x arctan(x) = + x Interpreting Derivatives 4. The graph of y = x 3 9x 6x + has a slope of 5 at two points. Fin the coorinates of the points. y = x 3 9x 6x + has slopes given by y = 3x 8x 6. The original curve will have a slope of 5 at points where y = 5: Solve for x: 5 = 3x 8x 6 0 = 3x 8x 0 = x 6x 7 0 = (x 7)(x + ) x = 7, The function will have slopes of 5 at x = 7 an x =. The coorinates of these points are (using f(x)), (-, 7) an (7, -09). 6. A ball is thrown up in the air, an its height over time is given by f(t) = 4.9t + 5t + 3 where t is in secons an f(t) is in meters. (a) What is the average velocity of the ball uring the first two secons? Inclue units in your answer. (b) Fin the instantaneous velocity of the ball at t =. (c) Compute the acceleration of the ball at t =. () What is the highest height reache by the ball? (e) How long is the ball in the air? 5. Determine coefficients a an b such that p(x) = x + ax + b satisfies p() = 3 an p () =. Let p(x) = x + ax + b satisfy p() = 3 an p () =. Since p (x) = x + a, this implies 3 = p() = + a + b an = p () = + a; i.e., a = an b = 3. (a) The average velocity is not foun irectly with erivatives, but rather by taking total istance
travele, ivie by time taken. f(0) = 3 meters f() = 33.4 meters = f t so avg spee is = f() f(0) 0 = (33.4 3) meters secons = 5. m/s (b) The instantaneous velocity at t = is given by the erivative. at t =, f (t) = 9.8t + 5 f () = 5.4 m/s (c) The acceleration at t = is given by the secon erivative of position, f (t) = 9.8 so at t = (an all t values), f () = 9.8 m/s () You can use parabola analysis to fin vertex of the parabola. A more calculus-oriente metho is to fin when the velocity of the ball is zero: that will be at the top of its arc. f (t) = 9.8t + 5 Setting f to zero, 0 = 9.8t + 5 t.55 at the top of its arc. The height at this point is f(.55) 34.89 m. (e) Knowing that the height is efine by a parabola, an given that the ball rose for t.55 secons means it will also take.55 secons to return to earth, so it spens (.55) 5.0 secons in the air. Note: this is not exactly correct, though, because the ball was launche (t = 0) from height f(0) = 3, but the impact occurs when f(t) = 0 later. It will take 5.0 secon to return to height 3 m, but just a little bit longer to actually impact the groun at height 0 m. A more accurate approach to fin the impact time is to use the quaratic formula to fin the two roots of f(t), the larger of which will be the laning time: f(t) = 4.9t + 5t + 3 equals zero when t = 5 ± 5 4( 4.9)(3) ( 4.9) = 5.93 an 0.73 The impact will be at exactly t = 5.93 secons, or just little longer than the estimate of 5.0 we foun using the parabola insight. 7. The height of a san une (in centimeters) is represente by f(t) = 800 5t cm, where t is measure in years since 995. Fin the values f(8) an f (8), incluing units, an etermine what each means in terms of the san une. Since f(t) = 800 5t cm, f(8) = 800 5(8) = 480 cm. Since f (t) = 0t cm/yr, we have f (8) = 0(8) = 80 cm/yr. In the year 003, the san une was 480 cm high an it was eroing at a rate of 80 centimeters per year. 8. With a yearly inflation rate of 5%, prices are given by P (t) = P 0 (.05) t where P 0 is the price in ollars when t = 0 an t is time in years. Suppose P 0 =. How fast (in cents per year) are prices rising when t = 0? P = P 0 (.05) t. If P 0 =, then P (t) = (.05) t ln(.05). At t = 0, P (0) = (.05) 0 ln(.05) = ln(.05) 0.0488 ollars/year. The prices are increasing at a rate of approximately 0.0488 ollars per year initially (at t = 0), or 4.88 cents per year. At t = 0, P (0) = (.05) 0 ln(.05) = 0.0795 ollars/year, or 7.95 cents per year. 9. With t in years since January st, 990, the population P of a small US town has been given by P = 35, 000(0.98) t At what rate was the population changing on January st, 00, in units of people/year? P (t) = 35000(0.98) t, so P (t) = 35000(0.98) t ln(0.98) Since t = 0 is January st 990, an t is in years, Jan st 00 represents t = 0. P (0) = 35000(0.98) 0 ln(0.98) 47 people/year. The rate is negative, inicating that people are leaving the town. 0. The value of an automobile can be approximate by the function V (t) = 5(0.85) t, where t is in years from the ate of purchase, an V (t) is its value, in thousans of ollars. (a) Evaluate an interpret V (4). (b) Fin an expression for V (t). (c) Evaluate an interpret V (4). (a) V (4) 3.05. 4 years after purchase, the car will be worth approximately 3 thousan ollars. 3
(b) V (t) = 5(0.85) t (ln(0.85)) 4.06(0.85) t. (c) V (4) =. means that, 4 years after purchase, the car will be losing value at a rate of roughly thousan ollars per year.. The quantity, q of a certain skateboar sol epens on the selling price, p, in ollars, with q = f(p). For this particular moel of skateboar, f(40) = 5, 000 an f (40) = 00. (a) What o f(40) = 5, 000 an f (40) = 00 tell you about the sales of this skateboar moel? (b) The total revenue, R, earne by the sale of this skateboar is given by. The theory of relativity preicts that an object whose mass is m 0 when it is at rest will appear heavier when moving at spees near the spee of light. When the object is moving at spee v, its mass m is given by m = m 0, where c is the spee of light (v /c ) (a) Fin m v. (b) In terms of physics, what oes m v tell you? R = (price per unit)(quantity sol) = p q R Fin the rate of change of revenue, p when p = 40. This is sometimes written as R p p=40 (c) From the sign of R p, ecie whether p=40 the company woul actually increase revenues by increasing the price of this skateboar from $40 to $4. (a) f(40) = 5, 000 says that at a selling price of $40, 5,000 skateboars will be sol. f (40) = 00 inicates that if the price is increase from $40, for each increase of $ in selling price, sales will rop by roughly 00 skateboars (rop b/c of the negative erivative value). (b) R = pq, so R p = p p q + p q by the prouct p rule. But p q p =, an p = f (p), since f gives the quantity sol q as a function of p. Therefore, at the selling price of $40, R p = p p=40 p q + p q p = f(40) + (40) f (40) = 5000 + 40( 00) = 000 (c) The sign of R is positive. p p=40 This inicates that if p is increase by a small amount (e.g. $40 to $4), revenues will increase. This happens because, even though unit sales ecrease, the loss of sales is offset by the price increase, so the net revenue effect is positive. m = m 0 (v /c ), an m 0 an c are constants. We can rewrite this using powers as m = m 0 [ c v ] / (a) m v = m 0 = m 0 c v [ c v ] 3/ ( [ c v ] 3/ ) c (v) (b) The erivative m represents how the (relativistic) v mass of an object changes as its velocity changes. Note that for small v (v c or v much smaller than c), this rate is almost zero, since relativistic effects are only pronounce when our spee approaches the spee of light. 4
3. A museum has ecie to sell one of its paintings an to invest the procees. If the picture is sol between the years 000 an 00 an the money from the sale is investe in a bank account earning 5% interest per year, compoune annually, then the balance in the year 00 epens on the time t when the painting is sol. Let P (t) be the price the painting will sell for if sol in year t, an let B(t) be the balance from the sale (after interest) in 00, again epening on the year of sale, t. If t = 0 in the year 000, so 0 t 0, then B(t) = P (t)(.05) 0 t (a) Explain why B(t) is given by this formula. (b) Show that the formula for B(t) is equivalent to B(t) = (.05) 0 P (t) (.05) t (c) Fin B (0), given that P (0) = 50, 000 an P (0) = 5000. () Decie whether, if the museum still has the painting in 00 (at t = 0), it shoul continue to hol on to the painting, or sell it immeiately if its goal is to maximize its financial future. (a) The form for B(t) is an exponential growth given 5% interest. The only o part is the exponent, 0 t. Note that the value that goes in the exponent is the length of time for the eposit, not always the variable t. (Usually the two are the same because we start counting the eposit time from t = 0). (b) If the painting is sol at year t, then in the 0-year perio there will be (0-t) years left for the sale price to gain interest in the bank. The sale price is P (t), an it becomes the amount of the initial eposit. (c) B = (.05) 0 P (t)(.05) t P (t) (.05) t ln(.05) ((.05) t ) If P (0) = 50, 000 an P (0) = 5000, B (0) = (.05) 0 5000(.05)0 50000 (.05) 0 ln(.05 ((.05) 0 ) 3776.63 () Since B (0) is negative, if the museum has waite until year 0 to sell the painting, then they shoul sell immeiately, because their final profit is ecreasing at a rate of $3776 per year. This is because the value of the painting is only growing at 5000/5000 = 0.03 or roughly 3% per year, while they coul be making 5% interest if they sol it an investe the procees. 4. Let f(v) be the gas consumption (in liters/km) of a car going at velocity v (in km/hr). In other wors, f(v) tells you how many liters of gas the car uses to go one kilometer, if it is travelling at velocity v. You are tol that f(80) = 0.05 an f (80) = 0.0005 (a) Let g(v) be the istance the same car goes on one liter of gas at velocity v. What is the relationship between f(v) an g(v)? Fin g(80) an g (80). (b) Let h(v) be the car s gas consumption in liters per hour. In other wors, h(v) tells you how many liters of gas the car uses in one hour if it is going at velocity v. What is the relationship between h(v) an f(v)? Fin h(80) an h (80). (c) How woul you explain the practical meaning of the values of these function an their erivatives to a river who knows no calculus? f(v) gives the consumption in l/km at spee v in km/hr. f (80) = 0.05 l/km an f (80) = 0.0005 l/km for each km/hr increase above 80 km/hr. If B(t) = P (t)(.05) 0 t then B(t) = P (t)(.05) 0 (.05) t 0 P (t) = (.05) (.05) t (a) g(v) is the istance the car goes in one km at spee v. By consiering the units, g(v) (km/l) = f(v) l/km. So g(v) = [f(v)]. g(80) = /f(80) = /0.05 = 0 km/l. Fining g requires the chain rule: 5
g = ( )(f(v)) f (v) so g (80) = ( )(f(80)) f (80) = ( ) 0.05 0.0005 0. km/l per km/h spee above 80 (b) If h(v) is the consumption in l/hr, then we nee a combination of the spee an consumption rate: h(v) l/hr = [f(v) (l/km)] [v (km/hr)] h(v) = f(v) v so h(80) = f(80) 80 = 0.05 80 = 4 l/hr Careful with the erivative of h: our variable is v, so v f = f, an v v =. ( ) ( ) v h (v) = v f(v) v + f(v) v = f (v)v + f(v) so h (80) = f (80) 80 + f(80) =.0005 80 +.05 = 0.09 l/hr for each km/h above 80 km/h (c) Part (a) tells us that the car can go 0 km on liter. Since the first erivative is negative, it means going faster reuces the istance the car can travel on each liter. Part (b) tells us that at 80 km/hr, the car uses 4 l/hr. Since the erivative is positive, going at higher spees will use up more gas per hour. 5. If P (x) is a polynomial, an it can be factore so that P (x) = (x a) Q(x), where Q(x) is also a polynomial, we call x = a a ouble zero of the polynomial P (x). (a) If x = a is a ouble zero (i.e. P (x) can be written as (x a) Q(x)), show that both P (a) = 0 an P (a) = 0. (b) Show that if P (x) is a polynomial, an both P (a) = 0 an P (a) = 0, then we can factor P (x) into the form (x a) Q(x). (a) Clearly P (a) = 0: P (a) = (a a) Q(a) = 0 To fin the erivative, we use the prouct rule & chain rule, so at x = a, P (x) = (x a)q(x) + (x a) Q (x) P (a) = (a a)q(a) + (a a) Q (a) so P (a) = 0 We can be sure of this because Q is a polynomial, so Q (a) has a nice finite value (the only way P (a) coul not equal zero woul be if Q (a) i not exist). (b) If both P (a) an P (a) equal zero, an P (x) is a polynomial, then (x a) must be a factor in both P an P (this is how you fin & check factors by han). Since x = a is a root of P (x), we can write P (x) = (x a)r(x), where R(x) is another polynomial. From this, P (x) = R(x) + (x a)r (x) but P (a) = 0, so P (a) = R(a) + (a a)r (a) = 0 so R(a) = 0 This means R(a) also has an (x a) factor in it: R(x) = (x a)t (x), where T (x) is yet another polynomial. Substituting this form back into P, we get P (x) = (x a)r(x) = (x a)[(x a)t (x)] = (x a) T (x) so P (x) must have a ouble root at x = a if both P (a) = P (a) = 0. 6. If g(x) is a polynomial, it is sai to have a zero of multiplicity m at x = a if it can be factore so that g(x) = (x a) m h(x) where h(x) is a polynomial such that h(a) 0. Explain why having a polynomial having a zero of multiplicity m at x = a must then satisfy g(a) = 0, g (a) = 0,..., an g (m) (a) = 0. (Note: g (m) inicates the m-th erivative of g(x).) If g(x) = (x a) m h(x), then repeate erivatives will give g (x) = m(x a) m h(x) + (x a) m h (x) g (x) = m(m )(x a) m h(x) +... [terms with highe. g (m ) (x) = m(m )... (x a) h(x) +... [terms with hig Note that, since every term has an (x a) factor, when we evaluate the function or its erivatives at x = a, we get zero. 7. (a) Fin the eighth erivative of f(x) = x 7 + 5x 5 4x 3 + 6x 7. Look for patterns as you go... (b) Fin the seventh erivative of f(x). 6
(a) Since each erivative removes one power from a polynomial, the eighth erivative of any seventhegree polynomial will be zero. (b) Only the erivative of the x 7 term will survive seven erivatives. Looking at the pattern, f (x) = 7x 6 +... f (x) = 7 6x 5 +.... f (7) = 7 6 5... x 0 = 7! (7 factorial) = 5040 8. (a) Use the formula for the area of a circle of raius r, A = πr, to fin A r. (b) The result from part (a) shoul look familiar. What oes A represent geometrically? r (c) Use the ifference quotient efinition of the erivative to explain the observation you mae in part (b). very small r. The erivative relationship tells us that the resulting change in area of the circle will be approximate by A r A r = πr so A (πr) r We can see that this prouct of the circumference an the change in raius provies a goo approximation to the total change in area of the circle with a iagram: (a) If A = πr, then A r = πr. (b) This rate looks like the formula for the circumference of a circle, πr. (c) Imagine a circle with raius r. Then we enlarge it by just the tiniest bit, increasing the raius by a Linear Approximations an Tangent Lines 9. Fin the equation of the tangent line to the graph of f at (,), where f is given by f(x) = x 3 x +. f(x) = x 3 x +. In general, the slopes of the function are given by f (x) = 6x 4x At the point (, ) (which you shoul check is actually on the graph of f(x)!), the slope is f () = 6 4 = Using the point/slope formula for a line (or the tangent line formula), a line tangent to the graph of f(x) at the point (, ) is 0. (a) Fin the equation of the tangent line to f(x) = x 3 at x =. (b) Sketch the curve an the tangent line on the same axes, an ecie whether using the tangent line to approximate f(x) = x 3 woul prouce over- or uner-estimates of f(x) near x =. (a) f(x) = x 3, so f (x) = 3x. At x =, f() = 8 an f () =, so the tangent line to f(x) at x = is (b) y = (x ) + 8 y = f ()(x ) + f() = (x ) + or y = x 7
From the graph of y = x 3, it is clear that the tangent line at x = will lie below the actual curve. This means that using the tangent line to estimate f(x) values will prouce unerestimates of f(x).. Fin the equation of the line tangent to the graph of f at (3, 57), where f is given by f(x) = 4x 3 7x +. Differentiating gives f (x) = x 4x, so f (3) = 66. Thus the equation of the tangent line is y = y 0 + f (x 0 )(x x 0 ) = 57 + 66(x 3).. Given a power function of the form f(x) = ax n, with f (3) = 6 an f (6) = 8, fin n an a. Since f(x) = ax n, f (x) = anx n. We know that f (3) = (an)3 n = 6, an f (6) = (an)6 n = 8. Therefore, f (6) f (3) = 8 6 = 8. But so n = 8, an so n = 4. f (6) f (3) = (an)6n (an)3 n = n, Substituting n = 4 into the expression for f (3), we get 4a3 3 = 6, so a = 4 7. 3. Fin the equation of the line tangent to the graph of f at (, ), where f is given by f(x) = x 3 5x + 5. Differentiating gives f (x) = 6x 0x, so f () = 4. Thus the equation of the tangent line is y = 4(x ), or y = + 4(x ). 4. Fin all values of x where the tangent lines to y = x 8 an y = x 9 are parallel. Let f(x) = x 8 an let g(x) = x 9. The two graphs have parallel tangent lines at all x where f (x) = g (x). hence, x = 0 or x = 8 9. f (x) = g (x) 8x 7 = 9x 8 8x 7 9x 8 = 0 x 7 (8 9x) = 0 The point at x = 0 is easy to visualize (both graphs are flat there). Here is a graph showing the parallel tangents at x = 8/9. 5. Consier the function f(x) = 9 e x. (a) Fin the slope of the graph of f(x) at the point where the graph crosses the x-axis. (b) Fin the equation of the tangent line to the curve at this point. (c) Fin the equation of the line perpenicular to the tangent line at this point. (This is the normal line.) (a) f(x) = 9 e x crosses the x-axis where 0 = 9 e x, which happens when e x = 9, so x = ln 9. Since f (x) = e x, f (ln 9) = 9. (b) y = 9(x ln(9)). (c) The slope of the normal line is the negative reciprocal of the slope of the tangent, so y = 9 (x ln(9)). 6. Consier the function y = x. (a) Fin the tangent line base at x =, an fin where the tangent line will intersect the x axis. (b) Fin the point on the graph x = a where the tangent line will pass through the origin. (a) We fin the linearization using f(x) = x, so f (x) = x ln() (non-e exponential erivative rule). At the point x =, f() = = an f () = ln(), so the linear approximation to f(x) is L(x) = + ( ln())(x ). Solving for where this line intersects the x axis (or the y = 0 line), we fin the x intercept is approximately -0.447. (b) This question is more general. Instea of asking for a linearization at a specific point, it is asking at what point woul the linearization pass through the origin? Let us give the point a name: x = a (as oppose to x = use in part (a)). From the function an the erivatives, the linearization at the point x = a is given by: L a (x) = a f(a) + a ln() (x a) f (a) 8
That is true in general, but we want the point x = a where the linearization will go through (0, 0), i.e. for which L a (0) = 0: Solving for a, 0 = a + a ln()(0 a) 0 = a ( a ln()) 0 = a ln() a ln() = a = ln().44 At that x point, the graph of y = x s tangent line will pass exactly through the origin. 7. (a) Fin the tangent line approximation to f(x) = e x at x = 0. (b) Use a sketch of f(x) an the tangent line to etermine whether the tangent line prouces over- or uner-estimates of f(x). (c) Use your answer from part (b) to ecie whether the statement e x + x is always true or not. (a) f(x) = e x, so f (x) = e x as well. To buil the tangent line at x = 0, we use a = 0 as our reference point: f(0) = e 0 =, an f (0) = e 0 =. The tangent line is therefore l(x) = (x 0) + = x + (b) 8. The spee of soun in ry air is f(t ) = 33.3 + T 73.5 m/s where T is the temperature in egrees Celsius. Fin a linear function that approximates the spee of soun for temperatures near 0 o C. f(t ) = 33.3 f (T ) = 33.3 + T 73.5 ) ( + T 73.5 so at T = 0 o C, f(0) = 33.3 f (0) = 73.5 33.3 ()(73.5) 0.606 Thus the spee of soun for air temperatures aroun 0 o C is f(t ) 0.606(T 0)+33.3, or f(t) 0.606T +33.3 m/s 9. Fin the equations of the tangent lines to the graph of y = sin(x) at x = 0, an at x = π/3. (a) Use each tangent line to approximate sin(π/6). (b) Woul you expect these results to be equally accurate, given that they are taken at equal istances on either sie of π/6? If there is a ifference in accuracy, can you explain it? (a) At x = 0, the tangent line is efine by f(0) = 0 an f (0) =, so y = (x 0) + 0 = x is the tangent line to f(x) at x = π 3. Since the exponential graph is concave up, it curves upwars away from the graph. This means that the linear approximation will always be an unerestimate of the original function. (c) Since the linear function will always unerestimate the value of e x, we can conclue that + x e x, an they will be equal only at the tangent point, x = 0. At x = π, the tangent line is efine by (π/3) = 3 3 an f (π/3) =, so y = ( x π ) 3 + 3 is the tangent line to f(x) at x = 0. The estimates of each tangent line at the point x = π/6 woul be Base on x = 0 tangent line, f(x) x, so f(π/6) π/6 0.536. Base on x = π/3 tangent line, f(x) ( x π 3 ) + 3, ( π/6 π 3 ) 3 + 0.604 so f(π/6) The actual value of f(x) = sin(x) at x = π/6 is sin(π/6) = 0.5. 9
(b) From these calculations, the estimate obtaine by using the tangent line base at x = 0 gives the more accurate preiction for f(x) at x = π/6 A sketch might help explain these results. In the interval x [0, π/6], the function stays very close to linear (i.e. oes not curve much), which means that the tangent line stays a goo approximation for a relatively long time. The function is most curve/least linear aroun its peak, so the linear approximation aroun x = π/3 is less accurate even over the same x. 30. Fin the quaratic polynomial g(x) = ax + bx + c which best fits the function f(x) = e x at x = 0, in the sense that g(0) = f(0), g (0) = f (0), an g (0) = f (0) so f (x) = e x an f (x) = e x Evaluating at x = 0, f(x) = e x f(0) = f (0) = f (0) = Comparing with the erivatives of the quaratic, so g (x) = ax + b an g (x) = a Evaluating at x = 0, g(x) = ax + bx + c g(0) = c, g (0) = b, an g (0) = a For g(x) to fit the shape of f(x) near x = 0, we woul then pick c =, b = an a = or a = 0.5, so g(x) = 0.5x + x + woul be the best fit quaratic to f(x) = e x near x = 0. Here is a graph of the two functions, f(x) = e x in black, g(x) = 0.5x + x + in re. Notice how similar they look near their intersection. 3. Consier the graphs of y = sin(x) (regular sine graph), an y = ke x (exponential ecay, but scale vertically by k). If k, the two graphs will intersect. What is the smallest value of k for which two graphs will be tangent at that intersection point? Let f(x) = sin(x) an g(x) = ke x. They intersect when f(x) = g(x), an they are tangent at that intersection if f (x) = g (x) as well. Thus we must have sin(x) = ke x an cos(x) = ke x We can t solve either equation on its own, but we can ivie one by the other: sin(x) cos(x) = tan(x) = ke x ke x x = 3π 4, 7π 4,... Since we only nee one value of k, we try the first value, x = 3π/4. sin(3π/4) = ke 3π/4 e 3π/4 = k k 7.46 We confirm our answer by verifying both the values an erivatives are equal at x = 3π/4, sin(3π/4) = 7.46e 3π/4 0.707 (same y: intersection) an cos(3π/4) = 7.46e 3π/4 0.707 (same erivative) The actual point of tangency is at (x, y) = A sketch is shown below. ( 3π 4, ). 0
so a linearization of g(x) = e x y = (x 0) + or near x = 0 is y = x + 3. (a) Show that + kx is the local linearization of ( + x) k near x = 0. (b) Someone claims that the square root of. is about.05. Without using a calculator, is this estimate about right, an how can you ecie using part (a)? Note that his is the same as the approximation we obtaine before, except that our prouct version ha an aitional term, x. These approximations give the same straight-line estimate of the function, but I woul expect the first (multiplication) version to be more accurate because it contains more information (the square term that the pure linear approximation was missing). We will see more of this iea in Taylor polynomials an Taylor series. (a) f(x) = ( + x) k f (x) = k( + x) k so at x = 0, f(0) = k = f (0) = k( k ) = k so the tangent line at x = 0 will be y = k(x 0) + or y = + kx 34. (a) Show that x is the local linearization of near x = 0. + x (b) From your answer to part (a), show that near x = 0, + x x. (b) As an estimate for the square root of., we coul note that. = ( + 0.) /. This matches exactly the form of f(0.) if we choose k =. From our linearization above, f(0.) + (0.) =.05 so yes, a goo approximation for. is.05. (Calculator gives the value of.0488. 33. (a) Fin the local linearlization of e x near x = 0. (b) Square your answer to part (a) to fin an approximation to e x. (c) Compare your answer in part (b) to the actual linearization to e x near x = 0, an iscuss which is more accurate. (a) e x has a tangent line/local linearization near x = 0 of y = x + (slope, point (0, )). (b) Multiplying this approximation by itself, we get (e x )(e x ) or e x (x + )(x + ) = x + x + (c) To compare with the actual linearization of g(x) = e x, we fin its erivative an value at x = 0, g(x) = e x g(0) = g (x) = e x g (0) = (c) Without ifferentiating, what o you think the erivative of is at x = 0? + x (+x). (a) Let f(x) = /( + x). Then f (x) = At x = 0, f(0) = an f (0) =. So near x = 0, f(x) x + = x + (b) For small x values (i.e. x near zero), we can approximate /(+x) with x. Replace the variable x with y (because the name oesn t matter), /( + y) y If we choose y small but equal to x, then + x x (c) The linearization of /( + x ) is the linear part of x, or just. Since the erivative at x = 0 is the coefficient for x in the linear part, this means x at x = 0 must equal zero. + x
35. (a) Fin the local linearization of near x = 0. f(x) = + x (b) Using your answer to (a), what quaratic function woul you expect to approximate g(x) = + x? (c) Using your answer to (b), what woul you expect the erivative of +x at x = 0 to be even without oing any ifferentiation? (a) We know f(0) = an f (0) = (+(0)) =, so the local linearization is f(x) x. (b) Next, g(x) = f(x ), so we expect that g(x) x. (c) Noting that the approximation we foun in (b) is ownwar opening parabola with its vertex at x = 0 (i.e. its local maximum or a critical point), we expect that the erivative of g(x) at x = 0 will be zero.
Newton s Metho 36. Consier the equation e x + x =. This equation has a solution near x = 0. By replacing the left sie of the quation by its linearization near x = 0, fin an approximate value for the solution. (In other wors, perform one step of Newton s metho, starting at x = 0.) Our equation is e x + x = f(x) To fin the linearization of f(x) near x = 0, we nee f an its erivative f, both evaluate at x = 0. f(x) = e x + x so f(0) = e 0 + 0 = f (x) = e x + so f (0) = e 0 + = The linearization is then e x + x () (x 0) + f (0) f(0) = x + We now replace the original (unsolvable) equation e x + x = with the simpler approximation: x + = Solving this secon version is straightforwar, yieling x = 0.5. This is actually a fair approximation to the solution, since e 0.5 + 0.5.49 which is close to the RHS value of. (If we continue our linearizations an their approximations, we woul get values even closer to the real solution, which is (to 4 ecimal places) 0.449.) 37. Use Newton s Metho with the equation x = an initial value x 0 = 3 to calculate x, x, x 3 (the next three solution estimates generate by Newton s metho). Moving everything to the left sie, we get x = 0 f(x) Differentiating, we have f (x) = x. Therefore: x = x 0 f(x 0) f (x 0 ) = x 0 x 0 x 0 = 3 3 3 =.83333 x = x f(x ) f (x ) =.83333.83333.83333.46 x 3 = x f(x ) f (x ) =.46.46.46.45 This sequence provies successive approximations to the exactly solution,which woul equal.44 38. Use Newton s Metho with the function x 3 = 5 an initial value x 0 =.5 to calculate x, x, x 3 (the next three solution estimates generate by Newton s metho). Moving everything to the left sie, we get x 3 5 = 0 f(x) Differentiating, we have f (x) = 3x. Therefore: x = x 0 f(x 0) f (x 0 ) = x 0 x3 0 5 3x 0.74074 =.5.53 5 3.5 = x = x f(x ) f (x ) =.74074.740743 5 3.74074.705 x 3 = x f(x ) f (x ) =.705.7053 5 3.705.70998 This sequence provies successive approximations to the thir root of 5, 3 5.70997 39. Use Newton s Metho to approximate 4 3 an compare with the value obtaine from a calculator. (Hint: write out a simple equation that 4 3 woul satisfy, an use Newton s metho to solve that.) We nee to fin an approximation to x = 4 3 using Newton s Metho. Since we are assuming we on t have a cube-root function, we nee to translate this into an equivalent form. Cubing both sies we get x 3 = 4 To use Newton s metho, we have to move everying to one sie to get the form f(x) = 0: x 3 4 = 0 f(x) With f(x) = x 3 4, we then get f (x) = 3x. A goo initial guess coul be x 0 = 8 /3 =. Thus: x = x 0 f(x0) f (x 0) = 3 4 3 =.66667 x = x f(x) f (x ) =.66667.666673 4 3.66667.59 x 3 = x f(x) f (x ) =.59.593 4 3.59.5874 x 4 = x 3 f(x3) f (x 3) =.5874.58743 4 3.5874.5874 Using a calculator to fin 4 3, we get: 4 3.58740. 3