Final Exam Review Sheet Solutions

Transcription

1 Final Exam Review Sheet Solutions. Find the derivatives of the following functions: a) f x x 3 tan x 3. f ' x x 3 tan x 3 x 3 sec x 3 3 x. Product rule and chain rule used. b) g x x 6 5 x ln x. g ' x 6 x 5 5 ln x x 6 5 x ln 3 x x. Product rule and chain rule used.. Find the equation of the tangent line to the curve x y x y 3 at the point,. We first find the derivative y', where y f x is the implicitly defined function of x whose graph is the "function-graph" portion of the curve that passes through the point (, ). We find y' by implicit differentiation: y x y y' x y 3 3 x y y' 0 y' x y 3 x y y x y 3 y' y x y 3 x y 3 x y. Therefore, the slope of the tangent line at (, ) is y' x,y The equation of the tangent line is easily given in point-slope form: y 5 x, or y 5 x. Here is a graph of the curve and the tangent line at (, ): 6 5 Out[]= 3 3 3

2 final review answers.nb 3. Find the derivative of the function h x x sin x 3 using logarithmic differentiation. x3 5 We begin by taking ln of both sides, and expanding on the RHS by using properties of logs: ln h x ln x sin x 3 ln x sin x 3 ln x sin x x3 5 x3 5 3 x3 5 3 ln x sin x ln x3 5 3 ln x ln sin x ln x3 5. Next, we differentiate both sides: h' x cos x h x 3 x sin x x3 5 x3 3 x Finally, we multiply by h x to solve for the desired h' x : h' x h x 3 x cos x sin x x3 5 x3 3 x x sin x 3 3 x x3 5 cos x sin x x3 5 x3 3 x.. Suppose that a sample of a radioactive element decays to 95% of its original amount after years. Find the half-life of the element, and find the time required for the sample to decay to 0% of its original amount. Let A t be the amont of the element remaining in the sample after t years have elapsed. Let A 0 be the initial amount, that is, the amount present at time t 0. Then we know that A t A 0 k t, since the substance decays exponentially. To determine the value of k, we plug in t, when the amount will have decayed to 95% of its original value: A A 0 k.95 A 0 k.95 k ln.95, which is equal to In[3]:= k Log.95 Out[3]= To find the half-life, we solve the equation A t.5 A 0 for t: A 0 k t.5 A 0 k t.5 t ln.5, which equals k In[]:= Log.5 k Out[]= The half-life is therefore approximately 7 years. To find the time required for the sample to decay to 0% of its original size, we solve the equation A t.0 A 0 : A 0 k t.0 A 0 k t.0 t ln (.0)/k, which has the value In[5]:= Log.0 k Out[5]= 89.78

3 final review answers.nb 3 So it takes almost 90 years for the sample to decay to 0% of its original size. Here is a graph of the function A t and the horizontal lines at 50% and 0% of the original amount, which is set to 00. In[6]:= Plot 00 k t, 50, 0, t, 0, Out[6]= The volume of a cube is increasing at a rate of 0 cm 3 min. How fast is the surface area increasing when the length of an edge is 30 cm? Let the side of the cube be s. Then the volume is V s s 3. We are given that d V d t V ' 0 cm 3 min. Therefore, 3 s s' 0, so s' 0 3 s cm min. The surface area of the cube is S 6 s, since the cube has six square faces of side length s. We want to find d S d t S ' when s 30. We see that S ' s s', and when s 30 and s' , we obtain S ' ^ cm min. 6. Find the linear approximation to f x 5 x near x 3. Use it to estimate f If Annie were to approximate f 3.08 crudely by f 3.08 f 3, what percentage error would she make? The linear approximation "formula" is f x L x f a f ' a x a, for x near to a. In our case, a 3 and f a f 3. Moreover, f ' x x, so f ' a f ' 3 5 x Therefore, L x 3 x 3. Our approximate value of f Mathematica gives the "exact" value In[7]:= Out[7]= We see that 3.9 is an excellent approximation to the exact value. Now suppose that Annie approximates f 3.08 by f 3. What is her percentage error? Using our approximation of 3.9 as the exact value, we see that the absolute size of her error is.06, so her relative error is.06 expressed as a percent. 3.9 In[8]:= Out[8]= 0.058

4 final review answers.nb We see that Annie's relative error is about.5%. Notice that if we compare.06 to her approximate value of, we get essentially the same answer for the percentage error. In[9]:=.06 Out[9]= Find the critical points of the function f x x x 3, and determine which correspond to local extremes of f. The function is defined and differentiable on,, so the only critical points are of the type x c, where f ' c 0. We have that f ' x x 3 x 3 x 3 x x 3 x x 3 x x 3 x 7 x, so the values of x at which f ' x 0 are 0, 7, and. We use test values to determine the sign of f ' on the open intervals determined by the critical points: On, 0, we use x, and find f ' 0, so f is increasing on, 0. On 0, 7, we use x 7, and see that f ' 7 0, so f is decreasing on 0, 7. This implies that x 0 is a local maximum point. On 7,, we use x 5 7, and find f ' 5 7 0, so f is increasing on 7,, which implies that x 7 is a local minimum point. Finally, on,, we use x, and find f ' 0, so f is increasing on,. Therefore, x is a "terrace point" of the function f. Here is a graph confirming our analysis: In[]:= Plot x ^ x ^3, x,, Out[]=

5 final review answers.nb 5 8. For the function f x cos x sin x on the interval 0, Π, find the sub-intervals on which f is increasing and decreasing, and the sub-intervals on which f is concave up and concave down. Also find the global maximum and minimum values of f on the interval. f ' x cos x sin x cos x cos x sin x. Therefore, f ' x 0 when cos x 0 and when sin x. On the interval 0, Π, cos x 0 at x Π and x 3 Π, and sin x at x 3 Π. Therefore, we see that there are two critical points on the interval: Π and 3 Π. On the interval 0, Π, we use test value x Π and find that f ' Π 0, so f is decreasing on 0, Π. On Π, 3 Π, we use test value x Π, and find f ' Π 0, so f is increasing on Π, 3 Π. On 3 Π, Π, we use test value x 7 Π, and we find f ' 7 Π , so f is decreasing on 3 Π, Π. This implies that x Π is a local minimum, and x 3 Π is a local maximum, of f. We can now determine the global maximum and global minimum values by evaluating f at the critical points and the endpoints: In[]:= In[9]:= Out[9]= f x_ : Cos x ^ Sin x f 0, f Π, f 3 Π, f Π,,, From this we see that the global minimum (output) value is - and the global maximum value is +, and that these occur at the local minimum and local maximum points found earlier. Finally, we study the concavity by investigating the second derivative: f '' x sin x sin x cos x cos x. f '' x sin x sin x cos x cos x sin x sin x sin x sin x sin x. Therefore, the second derivative is 0 when the factor ( sin x sin x sin x sin x 0, or sin x sin x 0 We see that the second derivative is 0 when sin x, which happens at x Π 6 and 5 Π 6 (on our interval), and when sin x, which occurs at x 3 Π. By checking the sign of the second derivative, one checks that f is concave down on 0, Π 6), concave up on Π 6, 5 Π 6, and concave down on 5 Π 6, Π. -- The concavity doesn't change at 3 Π. The following plot of the function f confirms this analysis. NOTE: A problem such as this is definitely too hard to be included on the exam! However, the tools of the first and second derivatives are used just as they would be in an easier case; one just has to work harder to handle the trig functions, etc.

6 6 final review answers.nb In[3]:= Plot Cos x ^ Sin x, x, 0, Pi Out[3]= Find the following limits: x a) lim 9 x lim 9 x 8 x 5 x 9. 5 x 5 L'Hospital's Rule used once; the original limit is of the form "0/0." b) lim x x 3 x lim x x 3 x 0 Either remember x dominates any power of x, or use L'Hospital's rule to reduce the power of x in the numerator until the numerator is a constant. c) lim x 0 x x : Let the limit be L, and take logs to get ln L lim x 0 x ln x lim x 0 Rule, we find that ln L lim x 0 x. Therefore, L, which equals ln x, which is of the form "0/0." Using L'Hospital's x In[3]:= N Out[3]= A plot confirms this result:

7 final review answers.nb 7 In[30]:= Plot x ^ x, x,., Out[30]= A poster is to have an area of 80 in, with one-inch margins along the bottom and the two sides, and a two-inch margin at the top. What dimensions will give the largest printed area (the total area less the margins)? Let w be the width and h the height of the overall poster. Then w h 80. The printed area will have a width of w and a height of h 3, to accommodate the margins as described in the problem. Therefore, the area of the printed region is w h 3 w h h 3 w 6 80 h 3 w 6. From the constraint that w h 80, or h 80 w, we can write the area of the printed region as a function of w: p w w 3 w 6. We need to find the global maximum of this function on 0,. (Note that as w 0 and w, we have that p w, so there will be a finite interval of "reasonable" values of w for which the printed area is positive.) To find the maximum, we find the critical points of p: p' w 360 w 3, and p' w 0 when w, or w 0, or w ± 0. The maximum we seek can only occur at the positive critical point w 0, which is approximately equal to So the dimensions of the optimal poster are w 0 and h , which is approximately equal to 6.37.

8 8 final review answers.nb. If f ' x 8 x 3 x 3 and f 6, find f x. The general antiderivative of 8 x 3 x 3 is x 6 x 3 x C. To determine the value of C, we use the initial condition f C C 5. Therefore, f x x 6 x 3 x 5.. The graph of f ' is shown. Sketch the graph of f, assuming that f A stone dropped off a cliff hits the ground with a speed of 0 ft/sec. What is the height of the cliff? (Recall that the acceleration due to gravity is a constant a t 3 (ft/sec)/sec near the surface of the earth.) Let us call the height of the cliff h. We know the initial position of the stone is s 0 h, where the position is the height of the stone above the canyon floor. We also know that the initial velocity of the stone is v 0 0, since the stone is "dropped" -- it is not thrown up or down. Finally, we know that the acceleration is a constant: a t 3 ft/sec. Taking the antiderivative of the acceeleration to get the velocity, we find that v t 3 t C, and C 0 since the initial velocity of the stone is 0. Taking the antiderivative of velocity to get the position, we find that s t 6 t C. Plugging in t 0, we find that s 0 h 6 0 C C h, so s t 6 t h. Now, at what time t is the instantaneous velocity equal to -0 ft/sec (this will be the moment of impact)? Solve 3 t 0 t sec. So the height of the stone is 0 at time t 5 sec. Therefore, we can now solve for h: 0 s h h ft. The cliff is 5 feet high.. Compute the left, right, and midpoint Riemann Sums of the function f x x 6 x on the interval 0, 6 using 3 subintervals. Then compute the same sums using 6 subintervals. We first compute a table of values for the function f, that includes all of the values we will need to compute the requested Riemann Sums:

9 final review answers.nb 9 x 0 f x If we use three subintervals, the width of each subinterval is x 6 0. If we use a left sum, the three rectangle heights are 3 f 0, f, and f, so the left sum is f 0 x f x f x which is equal to 0* + (-8)* + (-8)* = -3. Here is a picture: LeftSum f, 0, 6, 3 The sum The right sum with three subintervals is equal to f x f x f 6 x 3. The midpoint sum will involve three rectangles where the heights are computed at the midpoints of the subintervals, so it equals f f 3 x f 5 x Here is a picture: MidpointSum f, 0, 6, 3 The sum The three Riemann Sums using six subintervals are computed in the same way; we record the answers here without further comment: The left sum = -35, the right sum = -35, and the midpoint sum =

10 0 final review answers.nb 5. Find the exact value of 0 6 x 6 x x using the Fundamental Theorem of Calculus. An antiderivative of the integrand is F x x3 3 3 x 6, so 0 x 6 x x F x 6 0 F 6 F Use geometry to find 0 5 f x x, where f is the function whose graph is shown We add the areas of the following regions, read from left to right: triangle + square + rectangle with semicircular dome on top + triangle = / + + ( + Π/) + / = + Π/. Let's compare this to a Riemann Sum approximation (the "internal" name of this function is f). N Π MidpointSum f, 0, 5, 0 The sum Compute the exact values of the following integrals. In each case, we use the Fundamental Theorem of Calculus, and compare with a midpoint sum: a) 3 x x ln x 3 ln 3 ln =.0986.

11 final review answers.nb f x_ : x MidpointSum f,, 3, 0 The sum Π b) Π cos x x sin x Π Π sin Π sin Π 0. MidpointSum Cos, Π, Π, 50 The sum c) a x x x x 3 x f x_ : x

12 final review answers.nb MidpointSum f,,, 50 The sum