Review Sheet 2 Solutions

Transcription

1 Review Sheet Solutions. A bacteria culture initially contains 00 cells and grows at a rate proportional to its size. After an hour the population has increased to 40 cells. (a) Find an expression for the number of cells in the colony after t hours. (b) Find the number of bacteria after 3 hours. (c) Find the rate of growth at t 3 hours. (d) When will the population reach 0,000 cells? If the rate of growth of the population is proportional to the population, we know that the growth will be exponential, so the number of cells after t hours have elapsed will have the form P t A b t. In this formula, we know that A P 0, the initial population, so A 00. To find b, we plug in our other data point: P b b 4. b ln 4. b Log P t_ : 00 b t The number of bacteria after three hours is P 3 00 b 3 P Note that we could also write P t in the form P t 00 b t 00 b t t. To find when the population will reach 0,000, we solve the equation P t 0 000: We get t t t ln 00 ln 4. N Log 00 Log We reach 0000 bacteria at t hours, or approximately 3 hours and minutes after the colony begins with 00 members. Here is the graphical solution: Plot P t, 0 000, t, 0,

2 review answers.nb. If y x 3 x and dx dt 5, find dy dt when x. We have that dy dt 3 x dx dt dx dt 3 x 5 5, and this is equal to when x. 3. A spherical balloon is being inflated so that its volume increases at a rate of ft/min. How fast is the radius increasing when the diameter of the balloon is 4 ft across? Let V t be the volume of the balloon at time t. We are given that V ' t in units of ft/min. Let r be the radius of the balloon (also a function of time). We want to know r' dr dt at the instant of time that the diameter (which is twice the radius) is 4 ft, or the radius is ft. The formula for the volume of a sphere is V 4 3 Π r3, so we have our relation between V and r. Taking the derivative with respect to time, we have that V ' 4 Π r dr dt. Since V ' (a constant), we can solve for dr dt to get dr dt 4 Π r ; the units are ft3 min. At the instant in time when r, we obtain that dr dt 4 Π 8 Π ft3 min. 4. Find the critical points of the following functions: a) g t 3 t 4 b) h x x x 4. Recall that the critical points (or critical numbers) of a function are the values x=a such that the derivative either equals 0 or fails to exist at a. We know that the function g t 3 t 4 whenever 3 t 4 0, and equals 3 t 4 whenever 3 t 4 0. In other words, the graph of g consists of two lines that meet in a V-shape at the point where 3 t 4 0, which is the point t 4 3. That point is a critical point, because g ' doesn't exist at a "corner" point, and it is the only critical point, because everywhere else the slope of the curve is either 3 or -3. Here is a graph of g: Plot Abs 3 t 4, t,, The function h is defined everywhere, because the denominator is always positive. Furthermore, the derivative exists everywhere, so the only critical points of h are of the form h' x 0. We can find the derivative using the quotient rule: h' x x 4 x x. This fraction is 0 wherever its numerator is 0 (again, the denominator is never 0). So the critical points x 4 of h are the solutions of x x 4 0 x ± are visible. 4 4 ± 5. Here is a graph of h in which the two critical points

3 review answers.nb 3 Plot x, x, 3, 5 x Find global max and min values of f x x ln x on the interval [/, ]. The extreme values will occur either at critical points or endpoints. The derivative is f ' x x, which is defined everywhere on the interval. f ' x 0 at x. Therefore, to find the extremes, we need to find the largest and smallest values among f, f, and f. f x_ : x Log x ; N f.935 N f. N f We see the global max is , which occurs at x, and the global min is, which occurs at x. Plot f x, x,,, PlotRange.,

4 4 review answers.nb 6. Repeat # 4 for f x x 4 x on [-, ]. The function is defined and continuous on the given interval. The extreme values can occur only at critical points or endpoints. The derivative is f ' x 4 x x x. Note that f ' is not defined (due to division by 0), so x is a critical 4 x point (and also an endpoint). The derivative exists everywhere else on the given interval. Is the derivative ever equal to 0 on this interval? f ' x 4 x x 4 x 0 4 x x x x ±. So there is one critical point in the interior of the interval, namely. So, to find the max and min, we need to check the values f, f, and f. f x_ : x 4 x f 3 f f 0 Therefore, we see that the global max is, which occurs at x, and the global min is 3, which occurs at x. Plot f x, x,, Find the point the MVT guarantees will exist for the function f x x on [, 9]. Illustrate. f x_ : x

5 review answers.nb 5 The secant slope is 9 9. Since f is continuous on [, 9] and differentiable for all x 0 (in particular, on (, 9), the 8 4 MVT guarantees that there is at least one point c between and 9 such that f ' c 4. f ' x x 4. So the slope of the tangent line at x 4 is equal to the slope of the secant line. Plot f x, 4 x 4, 4 x, x, 0, x, and this equals 4 when The graph of f '' is shown. What are the x-coordinates of the points of inflection of f? On which intervals is f concave up/down? Points where f '' x 0 are candidates for points of inflection. To be a point of inflection, the second derivative has to change sign through the point, so the original function f actually changes concavity at the point. Therefore, x and x 7 are the points of inflection. On,, we have f '' 0 f is concave down. On, 7, we have f '' 0 (except at x 4), so f is concave up on, 7. Finally, on 7,, f '' 0 f is concave down.

6 6 review answers.nb 9. Let f x x x 4. Find (a) the interval(s) on which f is increasing and decreasing; (b) the local maximum and minimum value(s) of f, and (c) the intervals of concavity and the point(s) of infection. The domain of f is all real x. The derivative is f ' x x 4 x 3 x x x x x, which exists everywhere and is 0 at x, 0, and /, the three critical points. By using test points, one finds that f ' 0 for x and 0 x, and f ' 0 for x 0 and x. Therefore, x is a local maximum, since f increases as x approaches -/ from below, and then decreases as x moves away from -/ ; similarly, x is a global maximum. On the other hand, f decreases to the left of x 0, and increases to the right of it, so x 0 is a local minimum point. The local maximum outputs are f f 9 4.5, and the local minimum output is f 0. We use the second derivative to locate intervals of concavity and points of inflection: f '' x x. The graph of the second derivative is a parabola centered on the y-axis and opening down, with x-intercepts ± 6. We see that f '' 0 for x and x, and f '' 0 for x. Therefore, f is concave up on the interval, and concave down on the intervals, 6 and,. The points ± 6 6 are the points of inflection (where the concavity changes). f x_ : x x 4 inflex pt local max inflex pt local min 4 local max The graph of the first derivative g' of a function g is shown (on the review sheet). Identify the local extreme points and sketch a graph of g. Local maxima occur where g ' x 0 and goes from positive to negative as you move to the right. Thus the local maxima are at x and x 5. The point x 3 is a local minimum, since the derivative is 0 at this point and g ' goes from negative to positive as we move to the right. A sketch will be omitted here.

7 review answers.nb 7. Evaluate the following limits: sin 4 x a) lim x 0. Since the numerator and denominator each tend to 0, we can use L'Hospital's Rule: lim 4 cos 4 x tan 5 x x sec 5 x 5 5 cos x b) lim x 0. Since the numerator and denominator each tend to 0, we cana try L'Hospital's Rule: x lim x 0 sin x x lim x 0 cos x. (We used L'Hospital's Rule twice.) c) L lim x x x. We first take logs: ln L lim x x ln x lim x ln x x lim x x =0. So ln L 0 L 0.