Solutions Of Homework 4

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1 Solutions Of Homework 1. Two parallel sides of a rectangle are being lengthened at the rate of 3 in/sec, while the other two sides are shortened in such a way that the figure remains a rectangle with constant area 60 in 2. i What is the rate of change of the perimeter of the rectangle when the length of an increasing side is 6 in? ii What are the lengths of the sides of the rectangle when the perimeter stops to decrease? Solution: Let A = At be the area, P = P t be the perimeter, = t be the common length of the sides that are being lengthened, and y = yt be the common length of the sides that are being shortened. These functions are related by the equations It is given that dt A = y = 60, P = 2 y. = 3 for any t. Moreover we calculate that 0 = da dt = dt y dt = 3y = dt dt = 3y, i We are asked to find dp/dt when = 6 so that y = 10: dt = 310 = 5 for = 6 and y = 10. Therefore, 6 dp dt = = in/sec. ii We are asked to find and y when dp/dt = 0 : Using dp dt = 2 dt dp, dt dt = 0 dt = dt So the rectangle is the square of the side = y = 60. dp dt = 2 dt. dt 3y = a Write the equations of the tangents and the normals lines to the curve 3 = y 2 2y at the points where the curve crosses the y-ais. b The curves 2 = y and 2y 2 = 12 5 intersect at the points 2, 1 and,. Find the acute angles between the curves at the intersection points. Solution: a The solutions of the equation y 2 2y = 0 is y = 0 and y = 2. So the curve crosses the y-ais at the points 0, 0 and 0, 2. Differentiating both sides of the equation 3 = y 2 2y with respect to we get 3 2 = 2y 2. Thus, = 0 at the point 0, 0 and = 2 at the point 0, 2. That is the slope of the tangent line to the curve at 0, 0 is 0 so that the curve has a vertical normal line at this point and 1

2 the slope of the tangent line to the curve at 0, 2 is 2 so that the slope of the normal line to the curve at this point is 1/2. Therefore, tangent lines are : y = 0, y = 2 2 normal lines are : = 0, y = b Differentiating the equation 2 = y with respect to we get 2 =. Therefore, the slope of the tangent lines to 2 = y at the points 2, 1 and, are 1 and 2. Differentiating the equation 2y 2 = 12 5 with respect to y we get y = 5 implying that = 5 y. Therefore, the slope of the tangent lines to 2y 2 = 12 5 at the points 2, 1 and, are 5 5 and 16. Let us find the angle between the curves at the point 2, 1. The slope at this point of one curve is 1 and of the other curve is 5. Denoting by α and β the angles between the tangent lines of the curves and the -ais, α β is the angle between the curves at this point. As tanα = 1 and tanβ = 5 we see that tanα β = tanα tanβ 1 tanα tanβ = 9. Therefore, the acute angle between the curves at the point 2, 1 is tan 1 9. In a similar way we may find that the acute angle between the curves at the point, is tan a Using an appropriate linearization of the function at a point, find.1 approimately. b Let f = which is an invertible function. Find f 1 0. function. Solution: a Let f =. Then,.1 L.1 where L is the linearization of f at. As f = 2 and f = 1, L = f f = 2 1, and so L.1 = Therefore, b Note that f1 = 0, and f = 3 2 5, and f 1 = 8. So, f 1 0 = 1 f 1 = a Let f = sin 13 cos 22 tan csc. Find f. b Let g = sin cos lnsec 1. Find d g. 2

3 c Find dh where h = sine2 2 cos2. d Find where y = log 3 cos sin. Solution: a ln f = ln sin 1 ln3 cos 2 ln2 tan ln csc 1 f f = cos 3 sin sin 1 3 cos 2 sec2 csc cot 2 tan csc cos f 3 sin = sin 13 cos 22 tan csc sin 1 3 cos 2 sec2 csc cot 2 tan csc b c d g = cos cos lnsec 1 sin lnsec 1 1 sec 1 dh = d = sine 2 cose 2 e cos2 sine 2 d 2 cos2 2 cos2 sine 2 2 cos2 ln 22 cos sin d = 1 cos sin sin sin d ln 3 sin }{{ } u 1 du ln u = sin ln, u = cos ln sin 1 = 1 cos sin sin sin ln 3 sin cos ln sin 1 5. Let f = 3. Find the derivatives f 0, f 0, and f 0. Solution: Note that f is continuous at 0 and so can be written as { f = 3, 0 3, 0 It is clear that f = 3 2 for < 0 and f = 3 2 for > 0. The continuity of f at 0 allow us to see that 3 2 =0 = 0 is the left derivative of f at 0, and 3 2 =0 = 0 is the right derivative of f at 0. Therefore, f 0 = 0 and so Similar arguments show that f 0 = 0 and { f 3 = 2, 0 3 2, 0 f = { 6, 0 6, 0 3

4 Now, by the continuity of f at 0 we see that the left derivative at 0 is d 6 =0 = 6 while the right one is 6. Hence f 0 does not eist. 6. a Let f be a differentiable function that satisfies the equation f 2 sin f = for all in some open interval containing 1. If f1 = 0, find f 1. b Let f be a differentiable function satisfying the equation d f3 = 27 2 for all. Find f. c Assume that there is a differentiable function f defined on an open interval containing 1 and assume that f satisfies the equation f = f for all in this interval. If f1 = 1, find f 1. Solution: a Taking the derivative of the equation f 2 sin f = with respect to we get f 2 2 sin f cos f f = 1. Substituting = 1 we obtain that f 1 = 1 3. b d f3 = 3f 3. So d f3 = 27 2 implies that f 3 = 9 2. Hence f = 2. c f = f = f ln = ln f Taking the derivatives of the last equation with respect to we get: f ln 1 f = ln f 1 f f d We need to know the derivative d. Let u =. Then, ln u = ln and taking the derivative of this equation with respect to we obtain that 1 du du = ln 1 = u = ln 1. Substituting the value of du we just found in the equation we get the equation f ln 1 f = ln f 1 f f ln 1 If we put = 1 in the last equation we obtain the equation f 1 ln f1 = ln f f 1 1f ln As 1 1 = 1, ln 1 = 0, f1 = 1, the latest equation becomes 1 = f 1. 7.

5 a Is there a real number r satisfying r r 2 sin 2 cos r = 111? b Let f be a continuous function on [0, 1]. Assume that f0 = f1. Show that there is a real number a such that fa = f a 1 3. Hint: Let g be a continuous function on a closed interval. If g has no zero values in that interval, it must be either positive or negative throughout that interval. Why? Solution: a Yes, there is. Let g := sin 2. We will show that cos there is a real number r such that gr = 111 answering the question affirmatively. Now let us evaluate the function g, for instance, at 1 and 10 : g10 = g1 = sin = 57 < 111, 1 }{{} 2 sin sin 2 cos 10 }{{} = > 111. sin The conclusion is that 111 is between g1 and g10. As the function g is continuous everywhere, in particular it is continuous on the closed interval [1, 10]. Consequently, we may apply the intermediate value theorem to deduce that there is a real number r in [1, 10] such that gr = 111, as desired. b Define the function g := f f 1/3. It suffices to show that there is an a such that ga = 0. Suppose that there is no a such that ga = 0. Then, as g is continuous on the closed interval [0, 2/3], it follows by the intermediate value theorem that either g > 0 or g < 0 for all in the interval [0, 2/3]. Consider, for instance, the case g > 0 for all in [0, 2/3]. Then, f > f 1/3 for all in [0, 2/3]. In this case, letting be 0, 1/3, and 2/3, we see that f0 > f1/3, f1/3 > f2/3, f2/3 > f1. The above inequalities imply that f0 > f1. But this contradicts the given information f0 = f1 in the problem. By the similar reasons we may obtain in the case g < 0 for all that f0 < f1, which again contradicts the given information f0 = f1 in the problem. Therefore, there must be a real number a in the interval [0, 2/3] such that ga = 0 and so fa = fa 1/3. 8. a Let f be a function that satisfies the equation f y = f fy 2y for all real numbers and y. Assume that 0 f = 1 3. If possible, find f. b Suppose that f is a differentiable at 0 and that f0 = 0. Let g be the function defined { f by g =, 0 where a is a real number. If g is continuous at 0, what is a? a, = 0 5

6 Solution: a f h = f fh 2h and so Taking its as h 0 we obtain that f h f h = fh h 2. f f h f fh = = 2 h 0 h h 0 h = 2 1 h h 0 fh }{{} 1/3 = 2 3. b g is continuous at 0 if and only if g = g0 = a. As g = f when 0, we see 0 f that g =. As f is differentiable at 0 and as f0 = 0, we have that 0 0 f f0 h f0 fh 0 = = h 0 h h 0 h. Therefore, if g is continuous at 0 then a = f 0 9. a Find the it 7 3. b Find the it ne 1/n n. n Solution: a 7 3 = = }{{} u=1/ = u 0 1 7u 1/7u 21 = e 21 3/u 21/7u 1 7u = 1 7u u 0 u 0 = }{{} 21 is continuous 1/7u 1 7u u 0 } {{} e 21 b 10. ne 1/n n = n n e 1/n 1 1/n = }{{} h=1/n = d e =0 = e 0 = 1 e h 1 e 0h e 0 = h 0 h }{{} h 0 h a Let f be a function satisfying f 3/2 in an open interval containing 0. Show that f is differentiable at 0. 6 e 0 =1

7 b Let g be a function satisfying g 1/3 in an open interval containing 0. Assume also that g0 = 0. Show that g is not differentiable at 0. Solution: a We see by putting = 0 in the given inequality that f = 0. Multiplying the inequality f 3/2 by 1/ we obtain that f 1/2 for all nonzero values of in an open interval containing 0. Now, as f0 = 0, f f f0 1/2 1/2 1/2 f f0 1/2 Note that 0 1/2 = 0 = 0 1/2, and so the sandwich theorem implies that f f0 = 0. 0 Thus, f is differentiable at 0 and its derivative at 0 is 0. b We see in this case that g g0 2/3. As 0 2/3 =, the derivative of g at 0, equivalently the it of the function as goes to 0, cannot be a finite a real number. Hence, g has no derivative at 0. g g0 7