Solutions to Problem Set 1

Jason Starr Fall 2005 18.01 Calculus Due by 2:00pm sharp Friday, Sept. 16, 2005 Solutions to Problem Set 1 Part I/Part II Part I(20 points) (a) (2 points) (b) (2 points) (c) (2 points) (d) (2 points) (e) (2 points) (f ) (2 points) (g) (2 points) (h) (2 points) (i) (2 points) (j) (2 points) p. p. p. p. p. p. p. p. p. p. 57, 62, 62, 62, 68, 73, 73, 73, 87, 91, Section Section Section Section Section Section Section Section Section Section 2.2, 2.3, 2.3, 2.3, 2.4, 2.5, 2.5, 2.5, 3.1, 3.2, Problem Problem Problem Problem Problem Problem Problem Problem Problem Problem 1 1 28 36 8 4 6 14 5 30 You may assume x > 0. “Speed” is different from “velocity”. Solution (a) By Equation (4) on p. 54, the slope of the tangent line to the parabola at (x0 , y0 ) is 2x0 . Thus the equation of the tangent line is, (y − y0 ) = 2x0 (x − x0 ), or equivalently y = 2x0 x − x20 . (a). At the point (x0 , y0 ) = (−2, 4), the tangent line is, y = −4x − 4. (b). Since the slope is 2x0 , the slope is 8 when x0 equals 4. Plugging in, y0 equals x20 = (4)2 , which is 16. Thus the tangent line with slope 8 is, y = 8x − 16. (c). If x0 is zero, the tangent line has equation y = 0, i.e., the tangent line is the x­axis. Thus the tangent line does not have a well­defined x­intercept. Therefore assume that x0 is nonzero. The x­intercept of the tangent line is the value x1 such that y is 0. Plugging in, this gives the equation, 0 = 2x0 x1 − x20 . 1 18.01 Calculus Due by 2:00pm sharp Friday, Sept. 16, 2005 Jason Starr Fall 2005 Simplifying, this is 2x0 x1 = x20 . Since x0 is nonzero by hypothesis, also 2x0 is nonzero. Dividing both side by 2x0 gives the equation x1 = x0 /2. Thus the x­intercept x1 equals 2 if and only if x0 equals 2x1 = 2 × 2 = 4. As computed in (b), the equation of the tangent line is, y = 8x − 16. Solution (b) Since f (x) equals ax2 + bx + c, substituting x + Δx for x, f (x + Δx) equals, a(x + Δx)2 + b(x + Δx) + c. Expanding the square, this is, a(x2 + 2xΔx + (Δx)2 ) + b(x + Δx) + c. Therefore f (x + Δx) − f (x) equals, [a(x2 + 2xΔx + (Δx)2 ) + b(x + Δx) + c] − [ax2 + bx + c]. Cancelling like terms, namely ax2 , bx and c, this simplifies to, 2axΔx + a(Δx)2 + bΔx. Separating the common factor Δx from these terms, this simplifies to, f (x + Δx) − f (x) = (2ax + aΔx + b)Δx. This complete Step 1. Because of the common factor Δx, the difference quotient is, f (x + Δx) − f (x) = 2ax + aΔx + b, Δx for Δx nonzero. This completes Step 2. Holding a, b and x constant and allowing Δx to vary, the expression 2ax + aΔx + b is a linear function in Δx; perhaps this is easier to see if it is written aΔx + (2ax + b). A linear function is continuous. Thus to compute the limit as Δx approaches 0, it suffices to substitute in Δx equals 0. Therefore, f � (x) = lim [aΔx + (2ax + b)] = a0 + (2ax + b), Δx→0 which simplifies to 2ax + b. Therefore the derivative of ax2 + bx + c is, f � (x) = 2ax + b. Solution (c) The function is defined when 3x+2 is nonzero, i.e., when x is not −2/3. The function is undefined with x equals −2/3. Therefore assume that x is not −2/3. 2 18.01 Calculus Due by 2:00pm sharp Friday, Sept. 16, 2005 Jason Starr Fall 2005 Substituting x + Δx for x gives, f (x + Δx) = 1 . 3(x + Δx) + 2 To compute the difference, f (x + Δx) − f (x) = 1 1 − , 3(x + Δx) + 2 3x + 2 we express both fractions with the common denominator (3(x + Δx) + 2)(3x + 2), � � � � 1 3x + 2 1 3(x + Δx) + 2 × − = × 3(x + Δx) + 2 3x + 2 3x + 2 3(x + Δx) + 2 3x + 2 3(x + Δx) + 2 − . (3(x + Δx) + 2)(3x + 2) (3(x + Δx) + 2)(3x + 2) This simplifies to, (3x + 2) − (3(x + Δx) + 2) . (3(x + Δx) + 2)(3x + 2) Cancelling the like terms 3x and 2, this simplifies to, f (x + Δx) − f (x) = −3Δx . (3(x + Δx) + 2)(3x + 2) This completes Step 1. Because of the factor Δx in the numerator of f (x + Δx) − f (x), the difference quotient is, f (x + Δx) − f (x) −3 = , Δx (3(x + Δx) + 2)(3x + 2) for Δx nonzero. This completes Step 2. Considered as a function of Δx, is the expression (−3)/(3x+2+3Δx)(3x+2) defined and continuous at Δx = 0? The only values of Δx where the expression is undefined or discontinuous are the values where the denominator equals 0. By hypothesis, x is not −2/3, and thus 3x + 2 is not zero. Therefore the denominator is 0 if and only if 3x + 2 + 3Δx is 0. Thus the function (−3)/(3x + 2 + 3Δx)(3x + 2) has a single infinite discontinuity when Δx equals −x − 2/3. Again using the hypothesis that x is not −2/3, −x − 2/3 does not equal 0. In other words, there is a single point where the function is undefined and discontinuous, but this point is different from Δx = 0. Therefore (−3)/(3x + 2 + 3Δx)(3x + 2) is defined and continuous at Δx = 0. So the limit can be computed by substituting in 0 for Δx. Therefore the derivative of f (x) = 1/(3x + 2) is, −3 = −3/(3x + 2)2 . Δx→0 (3x + 2 + 3Δx)(3x + 2) f � (x) = lim 3 18.01 Calculus Due by 2:00pm sharp Friday, Sept. 16, 2005 Jason Starr Fall 2005 Solution (d) The function is defined when x is nonnegative and undefined when x is negative. Therefore assume that x is nonnegative, x ≥ 0. Substituting x + Δx for x gives, f (x + Δx) = � 2(x + Δx). Please note this is defined if and only if x + Δx is nonnegative, i.e., Δx ≥ −x. Also, as always, Δx is nonzero. To compute the difference, � √ f (x + Δx) − f (x) = 2(x + Δx) − 2x, � √ multiply and divide by the sum 2(x + Δx) + 2x, � √ � √ 2(x + Δx) + 2x √ . f (x + Δx) − f (x) = ( 2(x + Δx) − 2x) × � 2(x + Δx) + 2x Although at first glance this seems to make the expression more complicated, in fact the expression � √ now simplifies. The numerator is of the form (a − b)(a + b) for a = 2(x + Δx) and b = 2x. But (a − b)(a + b) simplifies to a “difference of squares”, a2 − b2 . Thus the expression simplifies to, � √ ( 2(x + Δx))2 − ( 2x)2 2(x + Δx) − (2x) � √ √ . =� 2(x + Δx) + 2x 2(x + Δx) + 2x Cancelling the like term 2x, this simplifies to, 2Δx √ . f (x + Δx) − f (x) = � 2(x + Δx) + 2x This completes Step 1. Because of the factor Δx in the numerator of f (x + Δx) − f (x), the difference quotient is, f (x + Δx) − f (x) 2 √ , =� Δx 2(x + Δx) + 2x for Δx nonzero and satisfying Δx ≥ −x. This completes Step 2. There are 2 cases depending on whether x is positive or zero. First consider the case that x is zero. Then the difference quotient is, 2 √ . 2Δx This expression has an infinite discontinuity as Δx approaches 0. Therefore the limit is undefined. Since the derivative is the limit of the difference quotient, f � (x) is undefined for x = 0. 4 18.01 Calculus Due by 2:00pm sharp Friday, Sept. 16, 2005 Jason Starr Fall 2005 Next consider� the case that √ x is positive. Considered as a function of Δx, for Δx ≥ −x, the expression 2/( 2(x + Δx) + is nonzero. � 2x) is defined and continuous as long�as the denominator √ √ Since 2x is positive and 2(x + Δx) is nonnegative, the sum 2(x + Δx) + 2x is positive. Therefore the expression is defined and continuous at Δx =√0. So the limit can be computed by substituting in 0 for Δx. Therefore the derivative of f (x) = 2x for x > 0 is, f � (x) = lim � Δx→0 2 2(x + Δx) + √ 2 1 = √ =√ . 2 2x 2x 2x √ To summarize, f � (x) is undefined for x = 0 and f � (x) = 1/ 2x for x > 0. Solution (e) The velocity of the first particle is, v1 (t) = s�1 (t) = 2t − 6, and the velocity of the second particle is, v2 (t) = s�2 (t) = −4t + 9, using the solution of Problem 1 of Section 2.3. Therefore the speed of the first particle is, � 6 − 2t, t ≤ 3 |v1 (t)| = |2t − 6| = 2t − 6, t > 3. and the speed of the second particle is, � |v2 (t)| = | − 4t + 9| = 9 − 4t, t ≤ 9/4 4t − 9, t > 9/4. (a). There are 3 cases to consider: 0 ≤ t ≤ 9/4, 9/4 < t ≤ 3 and t > 3. In the first case, |v1 (t)| equals |v2 (t)| if and only if, (6 − 2t = 9 − 4t) if and only if (2t = 3) if and only if (t = 3/2). So for 0 ≤ t ≤ 9/4, the speeds are equal for precisely one moment, t = 3/2. At this time, both speeds equal 3. Note, however, the velocity of the first particle is −3 and the velocity of the second particle is +3, i.e., the velocities are not equal. In the second case, |v1 (t)| equals |v2 (t)| if and only if, (6 − 2t = 4t − 9) if and only if (6t = 15)if and only if (t = 15/6). Note that 15/6 = 5/2 = 2 12 is between 9/4 = 2 41 and 3. So, for 9/4 < t ≤ 3, the speeds are equal for precisely one moment, t = 5/2. At this time, both speeds equal 1. 5 18.01 Calculus Due by 2:00pm sharp Friday, Sept. 16, 2005 Jason Starr Fall 2005 In the third case, |v1 (t)| equals |v2 (t)| if and only if, (2t − 6 = 4t − 9) if and only if (2t = 3) if and only if (t = 3/2). However, 3/2 = 1 21 is less than 3. So for t > 9/4, the particles never have the same speed. In sum­ mary, for t ≥ 0 the two particles have equal speed at precisely two moments, t = 3/2 and t = 5/2. (b). The moment when the two particles have the same position is the solution of the equation, s1 (t) = s2 (t), or equivalently t2 − 6t = −2t2 + 9t, or equivalently 3t2 = 15t. The two solutions of this quadratic equation are t = 0 and t = 5. Now, v1 (0) equals 2 ×0− 6 = −6 and v2 (0) equals −4× 0+9 = 9. Also, v1 (5) equals 2 ×5− 6 = 4 and v2 (5) equals −4 × 5 + 9 = −11. Thus, for t = 0, the particles have velocities, v1 = −6 and v2 = 9. And, for t = 5, the particles have velocities v1 = 4 and v2 = −11. Solution (f ) The expression 6/(2x − 4) has an infinite discontinuity when the denominator equals 0. The denominator is 0 when 2x − 4 = 0, or equivalently, x = 2. Therefore the limit, lim x→2 6 , 2x − 4 is undefined. Solution (g) The expression (x2 + 3x)/(x2 − x + 3) is defined and continuous so long as the denominator is nonzero. Plugging in 3 for x, the denominator equals, (3)2 − (3) + 3 = 9, when x = 3. Since the denominator is nonzero, the limit exists and equals, (3)2 + 3(3) 18 x2 + 3x = = = 2. 2 2 x→3 x − x + 3 (3) − (3) + 3 9 lim √ Solution (h) The expression (x − 4)/( x − 2) is undefined when x = 4, since the denominator is 0. However, by the same “difference of squares” technique from Solution (d), √ √ 1 1 x+2 x+2 √ =√ = , ×√ x−4 x−2 x−2 x+2 for x nonnegative and not 4. Therefore, for x nonnegative and not 4, √ x−4 √ = x + 2. x−2 6 18.01 Calculus Due by 2:00pm sharp Friday, Sept. 16, 2005 Jason Starr Fall 2005 √ The expression x + 2 is defined and continuous for all nonnegative x. Therefore the limit is obtained by plugging in 4 for x; √ √ x−4 lim √ = lim( x + 2) = 4 + 2 = 4. x→4 x − 2 x→4 Solution (i) By Rules 1–4 on pp. 84–85, the derivative is, y � = (3x2 )� + (−5x)� + (2)� (Rule 4) = 3(x2 )� + (−5)(x)� + 2(1)� (Rule 3) = 3(2x) + (−5)(1) + 2(0) (Rule 2 and Rule 1) = 6x − 5. Of course this is also a special case of Problem 1 from Section 2.3. At x = 2, the derivative is y � = 6(2) − 5 = 7. Therefore the equation of the tangent line is, (y − 4) = 7(x − 2) or equivalently y = 7x − 10. Solution (j) In the first method, the fraction is simplified to, f (x) = 2x + 6x4 − 2x6 = 2x−4 + 6x−1 − 2x. x5 Using Equation (3) on p. 90, the derivative is, f � (x) = 2(x−4 )� + 6(x−1 )� − 2(x)� = 2(−4x−5 ) + 6(−x−2 ) − 2(1) = −8x−5 − 6x−2 − 2. Clearing denominators, the derivative is, f � (x) = (−8 − 6x3 − 2x5 )/x5 . In the second method, expressing f (x) as a quotient, f (x) = g(x)/h(x), g(x) = 2x + 6x4 − 2x6 , h(x) = x5 , the quotient rule gives, f � (x) = g � (x)h(x) − g(x)h� (x) . h(x)2 Using Section 3.1, g � (x) = 2 + 24x3 − 12x5 , h� (x) = 5x4 . Therefore the quotient rule gives, f � (x) = (2 + 24x3 − 12x5 )(x5 ) − (2x + 6x4 − 2x6 )(5x4 ) . x10 7 18.01 Calculus Due by 2:00pm sharp Friday, Sept. 16, 2005 Jason Starr Fall 2005 Expanding and simplifying, the numerator equals, (2x5 + 24x8 − 12x10 ) − (10x5 + 30x8 − 10x10 ) = −8x5 − 6x8 − 2x10 . Thus the quotient rule gives, −8x5 − 6x8 − 2x10 . x10 Factoring x5 from numerator and denominator, this is, f � (x) = f � (x) = (−8 − 6x3 − 2x5 )/x5 , just as in the first method. Part II(30 points) Problem 1(15 points) The derivative of f (x) = 1/x is f � (x) = −1/x2 (for x nonzero). (a)(5 points) Show that for the tangent line to the graph of f (x) at (x0 , y0 ), the x­intercept of the line is 2x0 and the y­intercept of the line is 2y0 . Solution to (a) The equation of the tangent line is, (y − y0 ) = −1 (x − x0 ). x20 The x­intercept is the unique value x = x1 for which y = 0. Plugging in x = x1 and y = 0 gives the equation, −1 −y0 = 2 (x1 − x0 ). x0 Simplifying, this gives, x1 = x0 + x20 y0 . Since x0 y0 equals 1, this simplifies to, x1 = x0 + x0 (x0 y0 ) = x0 + x0 (1) = 2x0 . Similarly, the y­intercept is the unique value y = y1 for which x = 0. Plugging in x = 0 and y = y1 gives the equation, 1 −1 y1 − y0 = 2 (−x0 ) = . x0 x0 Since 1/x0 equals y0 , the equation is, y1 − y0 = y0 . Solving, this gives, y1 = 2y0 . 8 18.01 Calculus Due by 2:00pm sharp Friday, Sept. 16, 2005 Jason Starr Fall 2005 (b)(5 points) Part (a) implies the following: For every pair of real numbers (x1 , y1 ), there is a tangent line to the graph of f (x) with x­intercept x1 and y­intercept y1 if and only if x1 y1 equals 4. You may use this fact freely. Let (a, b) be a point such that ab is nonzero and less than 1 (possibly negative). Show that a line L with x­intercept x1 and y­intercept y1 is a tangent line to the graph of f (x) containing (a, b) if and only if x1 satisfies, bx21 − 4x1 + 4a = 0, and y1 satisfies, ay12 − 4y1 + 4b = 0. Hint. Using Equations 1–5 on pp. 11–12 of the textbook, deduce that L contains (a, b) if and only if bx1 + ay1 equals x1 y1 . Then use the fact above to eliminate one of x1 or y1 from this equation, and simplify. Solution to (b) The equation of the line L with x­intercept x1 and y­intercept y1 is, x1 y + y1 x = x1 y1 . Therefore (a, b) is on L if and only if, bx1 + ay1 = x1 y1 . (1) First, substituting in y1 = 4/x1 to Equation 1 gives, bx1 + 4a = 4. x1 Multiplying both sides by x1 and simplifying gives, bx21 − 4x1 + 4a = 0. Next, substituting in x1 = 4/y1 to Equation 1 gives, 4b + ay1 = 4. y1 Multiplying both sides by y1 and simplifying gives, ay12 − 4y1 + 4b = 0. (c)(5 points) Using the quadratic formula and Part (b), which you may now use freely, write down the equations of the 2 tangent lines to the graph of f (x) containing the point (a, b) = (5, −3). Solution to (c) Plugging in a = 5 and b = −3 to the equation bx21 − 4x1 + 4a = 0 gives, −3x21 − 4x1 + 20 = 0. 9 18.01 Calculus Due by 2:00pm sharp Friday, Sept. 16, 2005 Jason Starr Fall 2005 By the quadratic formula, the solutions are, x1 = 4 1 � ± (−4)2 − 4(−3)(20). 2(−3) 2(−3) x1 = −2 1 √ −2 16 −2 ± 8 ± ± = 256 = . 3 −6 6 3 3 Simplifying, this gives, Thus the solutions are x1 = 6/3 = 2 and x1 = −10/3. Since y1 = 4/x2 , the corresponding solutions of y1 are y1 = 2 and y1 = −6/5. Since the equation of L is y1 x + x1 y = x1 y1 , the equations of the 2 tangent lines containing (5, −3) are, 2x + 2y = 4 −6 x + −10 y = 4. 5 3 Simplifying, the equations of the 2 tangent lines containing (5, −3) are, x + y = 2 and 9x + 25y = −30. Problem 2(10 points) For a mass moving vertically under constant acceleration −g, the displace­ ment function is, x(t) = −gt2 /2 + v0 t + x0 , where x0 is the displacement and v0 is the instantaneous velocity at time t = 0. A scientist uses a magnetic field to conduct an experiment simulating zero gravity. At time t = 0, the scientist drops a mass from a height of 10m with instantaneous velocity v0 = 0 under constant acceleration −10m/s2 . When the mass drops below height 5m, the field is switched on and the particle continues to move with new acceleration 0m/s2 . Assuming the displacement and velocity are continuous, determine the height and instantaneous velocity of the mass at time t = 1.2s. Show your work. Solution to Problem 2 Before the field is switched on, the displacement function is, x(t) = −10t2 /2 + 0t + 10 = −5t2 + 10. Differentiating, the velocity function is, v(t) = x� (t) = −10t. The time t1 at which the field is switched on is the positive solution of the equation x(t1 ) = 5. Plugging in and solving gives, 5 = −5t21 + 10 or equivalently 5t21 = 5 or equivalently t21 = 1. 10 18.01 Calculus Due by 2:00pm sharp Friday, Sept. 16, 2005 Jason Starr Fall 2005 Therefore the field is activated at time t1 = 1s. The displacement function for t > t1 is, x(t) = g1 (t − t1 )2 /2 + v1 (t − t1 ) + x1 , where g1 is the new acceleration, v1 is the instantaneous velocity at time t = t1 , and x1 is the displacement at time t = t1 . At time t1 , the displacement is x1 = 5 and the instantaneous velocity is v(t1 ) = −10(1) = −10. For t > t1 , the particle moves with acceleration g1 = 0. Thus the displacement function is, x(t) = 0(t − 1)2 + (−10)(t − 1) + 5 = −10t + 15. Plugging in t = 1.2s gives, x(1.2) = −10(1.2) + 15 = −12 + 15 = 3m. Problem 3(5 points) For a differentiable function f (x) and a real number a, using the difference quotient definition of the derivative, show that the function g(x) = f (ax) has derivative, g � (x) = af � (ax). Solution to Problem 3 Plugging in, g(x + Δx) − g(x) = f (a(x + Δx)) − f (ax) = f (ax + aΔx) − f (ax). Thus the difference quotient for g(x) is, g(x + Δx) − g(x) f (ax + aΔx) − f (ax) = . Δx Δx By definition of the derivative, g(x + Δx) − g(x) g � (x) = lim . Δx→0 Δx Now, for an expression E(Δx) involving Δx, lim Δx→Δx0 E(Δx) = lim aΔx→aΔx0 E(Δx). Applying this to the limit above, and using that a(0) equals 0, f (ax + aΔx) − f (ax) f (ax + aΔx) − f (ax) a × = = lim aΔx→0 aΔx→0 Δx Δx a f (ax + aΔx) − f (ax) a lim . aΔx→0 aΔx Substituting h = aΔx, this is, f (ax + h) − f (ax) g � (x) = a lim . h→0 h This limit is precisely the derivative of f (x) at the point ax. Therefore, g � (x) = lim g � (x) = af � (ax). 11