Math 19, Homework-1 Solutions

Transcription

1 SSEA Summer 207 Math 9, Homework- Solutions. Consider the graph of function f shown below. Find the following its or eplain why they do not eist: (a) t 2 f(t). = 0. (b) t f(t). =. (c) t 0 f(t). (d) Does not eist. f(t). t 0.5 =.

2 SSEA Math Module Math 9 Homework- Solutions, Page 2 of 7 August 2, Eplain why does not eist. The graph of / jumps at the origin and has a iting value of when approached from the left of = 0 and a iting value of when approached from the right of = 0. The function itself is undefined at = 0. Since no unique it value eists, the it does not eist. 3. Let f() = What can you say about: f() and f()? = ( 2)( 2) ( + 7)( 2) = 2 ( + 7), 2. Note that we cancel the ( 2) term in the numerator and denominator under the assumption that is not equal to 2. The it does not eist because: approaching from left, approaching from right, 2 ( + 7) =, 2 + ( + 7) =, and these values are not the same. The it does not eist because there is no unique iting value as we approach the point = 0 from the left and right = 2 ( 2)( 2) ( + 7)( 2) = 2 2 ( + 7), 2. This it eists since we are not concerned about the behavior of the function at the point of interest, which is = ( + 7) = 2 2 2(2 + 7) = Prove π e < e π by referring to the following figure and choosing appropriate values for a and b. Hint: Slopes of line L a and L b.

3 SSEA Math Module Math 9 Homework- Solutions, Page 3 of 7 August 2, 207 y L a L b y = ln() a b The slope of L a is greater than that of L b. This implies: ln a 0 a 0 > ln b 0 b 0 a ln a > b ln b b ln a > a ln b ln ( a b) > ln (b a ) Choosing a = e and b = π e π > π e. a b > b a, since ln is an increasing function. 5. If f() = 5, must f() eist? If it does, must f() = 5? Can we conclude anything about f()? Eplain your answers. f() may or may not eist. Even if it does, it need not necessarily be equal to f(). The it value of a function does not depend on how the function is defined at the point being approached.

4 SSEA Math Module Math 9 Homework- Solutions, Page 4 of 7 August 2, Compute the following its: (a) t t 2 + 3t + 2 t 2 t 2. t 2 + 3t + 2 t t 2 t 2 = (t + )(t + 2) t (t 2)(t + ) = (t + 2) t (t 2) = 3. (b) = ( ) ( ) ( ) ( ) = = ( + 3) 4 ( ) = ( + 3) + 2 = ( + 3) + 2 = 4. ( ) 2 (c) 4 cos. Since cos() for all real numbers : ( ) 2 cos, 0, ( ) cos 4 ( ) cos 4 ( ) cos 0. ( ) 2 By the squeeze/sandwich theorem, 4 cos = 0. (d) e sin(π/).

5 SSEA Math Module Math 9 Homework- Solutions, Page 5 of 7 August 2, 207 Since sin() for all real numbers : ( π sin, 0 ) e e sin(π/) e e e sin(π/) e e e sin(π/) e 0 e sin(π/) 0. By the squeeze/sandwich theorem, e sin(π/) = Which of the following statements are true, and which are false? If true, say why; if false, give a countereample (that is, an eample confirming the falsehood). (a) If a f() eists but a g() does not eist, then a (f() + g()) does not eist. True- since: a (f()+g()) = a (f()+g())+ a f() = a (f() + g() f()) = a g(), and this does not eist. (b) If neither a f() nor a g() eists, then a (f() + g()) does not eist. False- consider f() = /, and g() = /, with a = 0. Both a f() and a g() do not eist but: which eists. a (c) If f is continuous at a, so is f. (f() + g()) = (/ /) = 0, True- since g() = is continuous g(f()) = f() is continuous (it is the composite of continuous functions). (d) If f is continuous at a, so is f. {, 0 False- for eample, let f() =. f() is a continuous, > 0 functions whereas f() itself is not continuous. 8. If + f() = A and f() = B, find:

6 SSEA Math Module Math 9 Homework- Solutions, Page 6 of 7 August 2, 207 (a) + f( 3 ) < 3 < < ( 3 ) 0 f(3 ) = f(y) = B, + y 0 where y = 3. (b) f( 3 ). 0 < < 3 < 0 ( 3 ) 0 + f(3 ) = f(y) = A, y 0 + where y = 3. (c) + f( 2 4 ) < 4 < 2 < ( 2 4 ) 0 + f(2 4 ) = f(y) = A, + y 0 + where y = 2 4. (d) f( 2 4 ). 0 < < 0 0 < 4 < 2 < ( 2 4 ) 0 + where y = 2 4. f(2 4 ) = f(y) = A, + y 0 9. A line y = b is a horizontal asymptote of the graph of a function y = f() if either: f() = b, or f() = b. (a) Find the horizontal asymptote of the curve y = 2 + sin(). y = 2 + sin() = 2 + sin().

7 SSEA Math Module Math 9 Homework- Solutions, Page 7 of 7 August 2, 207 sin() To evaluate, note that: sin() sin() 0 sin() 0 sin() 0 sin() 0. Therefore, by the squeeze/sandwich theorem, sin() = 0. Therefore: y = 2 + sin() = 2, which implies that the horizontal asymptote is the line y = 2. (b) Does the graph of the function f cross its horizontal asymptote at all? The horizontal asymptote y = 2 crosses the graph of y = f() at several points as shown in the following figure: 0. For what value of a is: f() = { 2, < 3 2a, 3

8 SSEA Math Module Math 9 Homework- Solutions, Page 8 of 7 August 2, 207 continuous at every? f is continuous if: 3 f() = 3 f() }{{} =3 2 Therefore, 6a = 3 2 a = 4/3. = 3+ f() }{{} =2a 3 = f(3). }{{} =2a 3. A continuous function y = f() is known to be negative at = 0 and positive at =. Given this fact, why does the equation f() = 0 has at least one solution between = 0 and =? Illustrate with a sketch. Hint: Intermediate value theorem. By the Intermediate Value Theorem, f must take all values between f(0) (-ve) and f() (+ve) since it is continuous. Since f(0) < 0 < f(), there is at least one point between 0 and as shown in the following figure such that f() = 0: y f() y = f() 0 f(0) 2. If the product function h() = f() g() is continuous at = 0, must f() and g() be continuous at = 0? Give reasons for your answer. No. For instance, f() = 0 and g() can be any discontinuous function at = 0. Then, f()g() = 0, which is a constant function that is continuous at = 0.

9 SSEA Math Module Math 9 Homework- Solutions, Page 9 of 7 August 2, 207 Figure : 3. Refer to Figure () which shows the graphs of several functions over a closed interval D. At what domain points does each of the function appear to be: At what domain points does each of the function appear to be: (a) Differentiable?

10 SSEA Math Module Math 9 Homework- Solutions, Page 0 of 7 August 2, 207 Top left: The function is differentiable on its domain 3 2. Top right: The function is differentiable on its domain 2 3. Middle left: The function is differentiable on 3 < 0 and 0 < 3. Middle Right: The function is differentiable on 2 <, < < 0, 0 < < 2, and 2 < < 3. Bottom left: The function is differentiable on < 0, and 0 < 2. Bottom right: The function is differentiable on 3 < 2, 2 < < 2, and 2 < 3. (b) Continuous but not differentiable? Top left: None. Top right: None. Middle left: None. Middle Right: The function is continuous but not differentiable at = since it has a corner there. Bottom left: The function is continuous but not differentiable at = 0 since it has a cusp there. Bottom right: The function is continuous but not differentiable at = 2 and = 2 since there are corners at those points. (c) Neither continuous nor differentiable? Top left: None. Top right: None. Middle left: The function is neither continuous nor differentiable at = 0 since f() + f().

11 SSEA Math Module Math 9 Homework- Solutions, Page of 7 August 2, 207 Middle Right: The function is neither continuous nor differentiable at = 0 and = 2. Bottom left: None. Bottom right: None. 4. Find values of constants m and c for which the function: { sin(), < π, y = m + c, π is: (a) Continuous at = π. The function y = f() is continuous at = π if: We have π π f() = f() = f(π). + f() = sin(π) = 0, π f() = mπ + c = 0 π + mπ + c = 0 m = c π. So all values of m that satisfy the condition m = c/π will result in f being continuous at = π. (b) Differentiable at = π. { We have f cos(), < π, () ==. The function y = f() is m, π differentiable at = π if the left and right derivatives at = π eists and are equal: f( + h) f() h 0+ h cos(π) = m f( + h) f() = h 0 h m = and c = mπ = π. So m = and c = π will result in f being differentiable at = π.

12 SSEA Math Module Math 9 Homework- Solutions, Page 2 of 7 August 2, Suppose that a function f satisfies the following conditions for all real values of and y: f( + y) = f() f(y), Show that f () = f() for all values of. f() = + g(), where g() =. Hint: Use the it definition of the derivative. f () = h 0 f( + h) f() h f( + h) f() }{{} =f() f(h) = h 0 h = h 0 f() f(h) f() h f(h) }{{} =+hg(h) = f() h 0 h = f() h 0 + hg(h) h = f() h 0 g(h) }{{} = = f(). (note that is not a variable in the it) 6. What is d9999 cos()? d9999 The third derivative of cos() is: d 3 d2 d cos() = ( sin()) = ( cos()) = sin(). d3 d2 d

13 SSEA Math Module Math 9 Homework- Solutions, Page 3 of 7 August 2, 207 Computing the third derivative 9999/3 = 3333 times ( )) d 3 d 3 ( d 3 d d3 3 d (cos()) = sin() Use chain rule to compute the derivative of: ( y = 4 sin + ). ( dy d = 4 cos + ) d d ( = 4 cos + ) ( = 2 cos + ) = ( ) cos +. + ( + ) d ( ) + d 2 8. Use the implicit rule of differentiation to compute dy/d if: ( ) y sin = y. y Differentiating w.r.t. on both sides ( ) ( ) ( ) dy d d sin + y cos y y d y dy ( ) ( ) ( d sin + y cos ) dy y y y 2 d [ ( ) dy sin ( ) ] d y y cos + = y y dy d = = y dy d = y dy d sin y ( ) ( y y cos y ). +

14 SSEA Math Module Math 9 Homework- Solutions, Page 4 of 7 August 2, Implicit Rule: Find the two points where the curve 2 + y + y 2 = 7 crosses the -ais, and show that the tangents to the curve at these points are parallel. What is the common slope of these tangents? The curve f(, y) = 2 + y + y 2 7 = 0 crosses the -ais when the y-coordinate is zero; that is: f(, 0) = = 0 = ± 7. We will now compute the slopes at the points ( 7, 0) and ( 7, 0): d ( 2 + y + y 2) = d d d (7) 2 + y + dy dy + 2y d d = 0 dy ( + 2y) = 2 y d dy d = 2 + y + 2y dy d = = 2, ( 7,0) dy d = ( 7,0) = 2. So the slopes at these two points are equal. 20. Chain Rule: Suppose that f() = 2 and g() =. Then the composites: (f g)() = 2 = 2 and (g f)() = 2 = 2 are both differentiable at = 0 even though g itself is not differentiable at = 0. Does this contradict the chain rule? Eplain? The assumption in the chain rule requires g to be differentiable only at f(0) and not at = 0 itself. Clearly, in this case, g() = is differentiable at f(0) = 0. Therefore this does not contradict the assumptions made in the chain rule. Optional Challenge Problems.

15 SSEA Math Module Math 9 Homework- Solutions, Page 5 of 7 August 2, What is 0 0? If we try to apply familiar rules of eponents, we get both 0 and as possible answers. What value would you like 0 0 have? (a) Calculate for = 0, 0 2, 0 3 etc., as long as possible with your calculator. What pattern do you see for the values of? We see a pattern like [0.7943, , 0.993, 0.999, ,...]. Clearly the values approach, suggesting that 0 0 =. (b) Calculate (/) / ln() for = 0, 0 2, 0 3 etc., as long as possible with your calculator. What pattern do you see now? Does the it value involve a well known irrational number? The values quickly approach which is /e suggesting that 0 0 = /e. 22. If c (f() + g()) = 3 and c (f() g()) =, find c f() g(). Subtracting these two equations [f() + g()] 2 = f() 2 + g() 2 + 2f() g(), [f() g()] 2 = f() 2 + g() 2 2f() g(). 4f() g() = [f() + g()] 2 [f() g()] 2 f() g() = ( [f() + g()] 2 [f() g()] 2) 4 ( f() g() = [f() + g()] 2 [f() g()] 2) c c 4 = ( 4 [f() + c g()]2 [f() g()] 2) c = ( [ ] 2 [ ] ) 2 4 f() + g() f() g() c c = ( 3 2 ( ) 2) = Antipodal points Is there any reason to believe that there is always a pair of antipodal (diametrically opposite) points on the earth s equator where the temperatures are the same? Eplain. Hint: Intermediate value theorem

16 SSEA Math Module Math 9 Homework- Solutions, Page 6 of 7 August 2, 207 Yes. Let R be the radius of the equator (earth) and suppose at a fied instant of time we label noon as the zero point, 0, on the equator 0+πR represents the midnight point (at the same eact time). Suppose is a point on the equator just after noon +πr is simultaneously just after midnight. It seems reasonable that the temperature T at a point just after noon is hotter than it would be at the diametrically opposite point just after midnight: That is, T ( ) T ( + πr) > 0. At eactly the same moment in time, pick 2 to be a point just before midnight 2 +πr is just before noon. Then T ( 2 ) T ( 2 +πr) < 0. Assuming the temperature function T is continuous along the equator (which is reasonable), the Intermediate Value Theorem says there is a point c between 0 (noon) and πr (simultaneously midnight) such that T (c) T (c + πr) = 0; i.e., there is always a pair of antipodal points on the earth s equator where the temperatures are the same. 24. Prove that: ( + )/ = e, by computing the derivative of f() = ln() at = using the it definition of the derivative. That is, use: f f( + h) f() () =. h 0 h f f( + h) f() () = h 0 h = h 0 ln( + h) ln() }{{} =0 h ln( + h) = h 0 h = ln ( + h) /h h 0 = ln But f () = / f () =. This implies: [ h 0 ( + h)/h]. ln ( + h)/h h 0 =. }{{} inside argument

17 SSEA Math Module Math 9 Homework- Solutions, Page 7 of 7 August 2, 207 But this is possible only if the inside argument is equal to e since ln(e) =. Therefore: ( + h 0 h)/h = e ( + ) / = e. 25. The period of a clock pendulum The period T of a clock pendulum (time for one full swing and back) is given by the formula: T 2 = 4π2 L, g where T is measured in seconds, g = 9.8 m/s 2, and L, the length of the pendulum is measured in meters. Find approimately: (a) The length of a clock pendulum whose period is T = sec. L = gt 2 4π 2 = g 0.25 m. 4π2 (b) The change dt in T if the pendulum with a period of T = sec is lengthened by 0.0 m. T 2 = 4π2 L g Therefore, if dl = 0.0 m, then dt = T = 2π g L dt dl = 2π g 2 L dt = π Lg dl. π Lg dl = sec. (c) The resulting amount the clock gains or loses in a day as a result of the change in period by the amount dt that was found in the previous part. Since there are 86, 400 sec in a day, the pendulum loses about , 400 sec/day sec/day, or about 29 min/day;