Math 0: Calculus I (Math/Sci majors) MWF 3pm, Fulton Hall 30 Homework 4 Solutions Please write neatly, and show all work. Caution: An answer with no work is wrong! Problem A. Use Weierstrass (ɛ,δ)-definition of lim f(x) = L to show that the sum of two continuous functions is continuous. (Recall that we showed rigorously in class that the limit of a sum is the sum of the limits, provided each exist.) Solution A. Let f and g be the two continuous functions in question. We assume that they have a common domain of definition (otherwise the sum of the functions makes no sense who wrote this question?), and we will write f + g for the function x f(x) + g(x). By definition of continuity, the function f + g is continuous if lim(f + g)(x) = (f + g)(a) at each point a where f + g is defined. At each such a, continuity of f and g imply that both lim f(x) = f(a), and g(x) = g(a). lim We ve seen in class (rigorously, using Weierstrass definition) that when two summand limits exist, the limit of a sum is the sum of limits. Thus lim(f + g)(x) f(x) + g(x) f(x) + lim g(x) = f(a) + g(a) = (f + g)(a), so that f + g is continuous at each point where it is defined. Problem B. Ignore Weierstrass definition, and evaluate the following limits: cos x () lim x 0 x sin(h) () lim h 0 h Hints:
cos h () Look at how we evaluated lim h 0 h sin h () Compare with lim h 0 h. Solution B. (): We have cos x lim x 0 x in class. cos x + cos x x 0 x + cos x x 0 cos x x x 0 sin x x sin x x 0 x + cos x + cos x sin x x + cos x. The limit of a product is the product of limits, when the latter exist. sin x We ve seen in class that lim x 0 x =. The last function, ( + cos x), is the reciprocal of a function that has a non-zero limit as x goes to 0, so it exists as well. We have cos x lim = x 0 x + =. (): Let t = h, so that h = t/. As h 0, t does as well, and we have sin(h) lim h 0 h sin t t 0 t/ = lim sin t t 0 t =. Problem C. Find the equation of the line tangent to the graph of y = log(tan x) when x = π/4. Solution C. We ve seen already that (log x) = x, and (tan x) = cos x. In order to compute the tangent line, we compute the derivative using the chain rule: y = tan x cos x = cos x sin x cos x = sin x cos x.
When x = π/4, this evaluates to y x=π/4 = =. At x = π/4, the point on the graph of y = log(tan x) has y-coordinate y = log(tan π) = log = 0. The equation of a line through ( π, 0) with 4 4 slope is given by ( y = x π ). 4 Problem D. Suppose an isosceles triangle has base angle θ, and that the perimeter of the triangle is the fixed length. In this case, the area of the triangle, A, is a function of the angle θ. Find the derivative of the area with respect to θ, da/dθ, when the triangle is equilateral. Solution D. Suppose the length of the two equal length sides is x, and the base is given by y. In this case, the perimeter is x + y, and we have x + y =. On the other hand, drawing the altitude of the triangle that meets the base of the isosceles triangle, we see a right triangle with an angle θ, hypotenuse x, and side adjacent to the angle y/. Trigonometry tells us that cos θ = y/ x, or y = x cos θ. Plugging in to the perimeter equation, we find x + x cos θ =, and solving for x gives x = + cos θ. The area of the triangle is half of its base times its height. The base is given by y, and the same right triangle above tells us the height is given by x sin θ, so we obtain xy sin θ A = = x sin θ cos θ = sin θ cos θ ( + cos θ)
Now that we have A as a function of θ alone, we may compute the derivative directly. (We use primes below as shorthand for d dθ ). da dθ = (sin θ cos θ) ( + cos θ) sin θ cos θ (( + cos θ) ) ( + cos θ) 4 = (cos θ sin θ)( + cos θ) sin θ cos θ ( + cos θ) ( sin θ) ( + cos θ) 4 = (cos θ ( cos θ))( + cos θ) + cos θ( + cos θ)( cos θ) ( + cos θ) 4 = ( cos θ )( + cos θ) + cos θ( cos θ)( + cos θ) ( + cos θ) 4 = cos θ + cos θ( cos θ) ( + cos θ) = cos θ ( + cos θ) When the triangle is equilateral, all of the angles are π/3, so we may compute: since cos π/3 = /. da dθ = cos π 3 θ=π/3 ( + cos π = 0, ) 3 Problem E. If n is an integer, show that cos((n + )x) = cos x cos(nx) cos((n )x). Solution E. We use the sum formula for cos, and the fact that cos is an even function while sin is odd, expanding the right-hand side: cos x cos(nx) cos((n )x) = cos x cos(nx) cos(nx x) = cos x cos(nx) (cos(nx) cos( x) sin(nx) sin( x)) = cos x cos(nx) (cos(nx) cos x + sin(nx) sin x) = cos x cos(nx) cos(nx) cos x sin(nx) sin x = cos x cos(nx) sin(nx) sin x = cos(x + nx) = cos((n + )x). Note that n did not have to be an integer in the above analysis. (Who wrote these questions?)
Problem F. Find the values of x at which the tangent line to the graph of y = cos x + sin x is horizontal. Solution F. The tangent line is horizontal when its slope is 0. The slope of the tangent at x is also the value of the derivative at x. We compute directly that the derivative is given by y = sin x + cos x = cos x sin x. Thus the tangent line is horizontal exactly when cos x = sin x. Squaring both sides, and replacing cos x with sin x, we find 5 sin x = ±/. This occurs for x = π/4 + πn/, for any integer n. However, cos x = sin x in particular implies that cos and sin have the same sign, so we only look at the first and third quadrants. Thus when x = π/4 + πn, for any integer n, we have cos x = sin x, and the tangent line to y = sin x + cos x is horizontal. Problem G. Let f(x) = sin(x). What is f (0) (x)? Solution G. We compute directly, using the chain rule repeatedly: f (x) = cos(x) f () (x) = 4 sin(x) f (3) (x) = 8 cos(x) f (4) (x) = 6 sin(x) Thus f (4) (x) = 6f(x), and f (4n) (x) = 6 n f(x), for any integer n. This implies that f (00) (x) = 6 5 f(x), so that f (0) (x) = (6 5 f(x)) = 6 5 f (x) = 6 5 cos(x) = 0 cos(x). Problem H. Let f(x) = e x sin x. What is f (95) (x)?
Solution H. We compute directly, using the product rule repeatedly: f (x) = e x sin x + e x cos x f () (x) = e x sin x + e x cos x + e x cos x e x sin x = e x cos x f (3) (x) = (e x cos x e x sin x) f (4) (x) = (e x cos x e x sin x e x sin x e x cos x) = ( e x sin x) = 4e x sin x. Thus f (4) (x) = 4f(x), and f (4n) = ( 4) n f(x), for any integer n. This implies that f (9) (x) = ( 4) 3 f(x), so that f (95) (x) = ( ( 4) 3 f(x) ) (3) = ( 4) 3 f (3) (x) = ( 4) 3 (e x cos x e x sin x) = 47 e x (cos x sin x) Problem I. Suppose that cos x is defined so that its range is in the interval [π, π]. Compute the derivative d ( cos x ). dx Solution I. By definition, we have cos(cos x) = x. Taking the derivative of each side, we obtain so that sin(cos x) (cos x) =, (cos x) = sin(cos x). Since sin x + cos x =, we have sin x = ± cos x. Thus sin(cos x) = ± cos (cos x) = ± x. Since the range of cos is in the interval [π, π], the value of sin(cos x) is less than or equal to 0, so that sin(cos x) = x. We conclude (cos x) = sin(cos x) =. x Problem J. Define the one-sided limit lim + f(x).
Solution J. Suppose there exists L so that the following holds: For any ɛ > 0, there exists δ > 0 so that, for all 0 < x a < δ, we have f(x) L < ɛ. In that case, lim f(x) exists and is equal to L. + Problem K. A particle moves along the x-axis so that its position at time t is given by x(t) = t t3. Find the intervals on which the x-coordinate is increasing/decreasing. Find the acceleration of the particle at time t = /. Solution K. We have the derivative x (t) = t 3t, and the second derivative x (t) = 3t. Factoring x (t), we see x (t) = t (4 3t). Thus x (t) is equal to 0 when t = 0 or 4/3. For t < 0, t/ < 0 and 4 3t > 0, so that x (t) < 0. When 0 < t < 4/3, t/ > 0 and 4 3t > 0, so that x (t) > 0. When t > 4/3, t/ > 0 and 4 3t < 0, so that x (t) < 0. Thus x(t) is increasing for t between 0 and 4/3, while x(t) is decreasing for t < 0 or t > 4/3. Finally, the acceleration is given by x (/) = 3 =.