M147 Practice Problems for Exam 2


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1 M47 Practice Problems for Exam Exam will cover sections 4., 4.4, 4.5, 4.6, 4.7, 4.8, 5., an 5.. Calculators will not be allowe on the exam. The first ten problems on the exam will be multiple choice. Work will not be checke on these problems, so you will nee to take care in marking your solutions. For the remaining problems unjustifie answers will not receive creit.. Compute the erivative of each of the following functions: a. f(x) = x + x 7. b. c.. f(x) = x sin x. f(x) = ex e x e x + e x. f(x) = (x + x ).. Suppose h(x) = f(x)e g(x), an f() = 4, f () = 7, g() = 0, an g () =. Compute h ().. Compute f (x) if f(x) = sin( x ). 4. Compute x given that sin(xy) = x. Fin an equation for the line that is tangent to this curve at the point (, π 4 ). 5. Fin y x if xy e y = An airplane is flying 6 miles above the groun on a flight path that will take it irectly over a raar tracking station. If the istance between the plane an tracking station is ecreasing at a rate of 400 miles per hour when the istance is 0 miles, what is the spee of the plane? 7. Suppose that two sies of a triangle are 4 cm an 5 cm in length an that the angle between them is increasing at a rate of.06 ra/s. Fin the rate at which the area of the triangle is increasing when the angle between the sies is π. 8. Let f(x) = cosx sin x, π 4 x π 4,
2 an compute f x (). 9. Evaluate the expression tan(sin ( )). 0. Show that. Let x cos x =, < x < +. x y = x tan x, 0 x < π, an compute x.. Use a linear approximation to estimate a value for ln(.99).. Consier a right triangle with hypotenuse length l an sielengths an x. Suppose x is measure as x = 4 ±.05, an use linear approximation to approximate the associate range of error on l. 4. Sketch a graph of the function f(x) = x, on the interval [, ] an etermine all local an global extrema on this interval. 5. Suppose that f(x) is continuous on the interval [, 5] an ifferentiable on the interval (, 5). Show that if f (x) 4 for all x [, 5], then f(5) f(). 6a. For the function f(x) = x + x ; x, fin the intervals on which f is increasing an the intervals on which x is ecreasing. 6b. For the function efine in (6a) fin the intervals on which f is concave up an the intervals on which f is concave own. 7. Suppose that f(x) is twice ifferentiable in an open interval containing the point x = c an has a local minimum at the same point. Show that the function g(x) = e f(x) has a local minimum at x = c. Solutions. a. Applying the power rule to each summan, we fin x (x + x 7 ) = x 7x 8.
3 b. Applying the prouct rule, we fin c. Applying the quotient rule, we fin x sin x = sin x + x cosx. x e x e x x e x + e = (ex + e x )(e x + e x ) (e x e x )(e x e x ), x (e x + e x ) an though consierable simplification is possible, this form is sufficient for the exam.. Proceeing with the chain rule, we set u = x + x an compute x u = u uu x = u( x ) = (x + x )( x ). (This substitution oes not nee to be mae explicitly.). First, an so h (x) = f (x)e g(x) + f(x)e g(x) g (x), h () = f ()e g() + f()e g() g () = 7e 0 + 4e 0 = 7 + = 9.. Metho. Compute irectly an f (x) = cos( x ) ln = ln xx cos( x ) x, f (x) = ln ( sin( x ) ln x + cos( x ) ln x) =( ln ( ) x cos( x ) x sin( x ). Metho. First, observe that x = ( ) x, which eliminates the nee for a neste chain rule. Now, x sin(( ) x ) = cos(( ) x )( ) x ln, an x sin(( ) x ) = sin(( ) x )(( ) x ln ) + cos(( ) x )(( ) x ln ) ln = (ln ) ( cos(( ) x )( ) x sin(( ) x ) x ), which is equivalent to the expression from Metho. 4. We compute implicitly x sin(xy) = x x cos(xy) (xy) = cos(xy)(y + x x x ) =.
4 Solving for, we fin x At the point (, π), we have 4 x = y cos(xy). x x = π cos( π 4 ) 4 = π. The equation for the tangent line is π (y 4 ) = ( π )(x ). 5. We begin by computing the xerivative of the entire equation, y + x ey x x = 0. Solving for, we obtain x x = y x e. y At this stage we can either compute a secon erivative irectly from this final expression or take another erivative of our original equation. The former approach is the most irect, an we fin y x = (x ey ) y( ey ) x x = (x ey )( y ) y + ye y ( y x e y (x e y ) (x e y ) = y(x ey ) + y e y (x e y ). Note. In case you re curious, the secon metho woul look like this: so that Finally, we substitute our expression for x y y x = ( ) e y ( y ) x e y x e y (x e y ) giving the same result. x + x + x y x ey ( x ) e y y x = 0, y = x ey ( x ). x x e y to fin x e y ) y = (x e y ) + y ey (x e y ) = y(x ey ) + y e y, (x e y ) 6. In this case, we are given that r = 400, where r enotes the istance between the plane t an the tracking station. If we let x enote the horizontal istance between the plane an 4
5 the tracking station, then what we are looking for is x, the plane s spee. In orer to fin t a relation between x r an, we begin by relating x an r. We have t t x + 6 = r. Upon ifferentiation of this equation with respect to t, we fin x x t = rr t. When r = 0, we have x = 00 6 = 8, an therefore (8) x t = (0)( 400) x t = = 500. The negative sign inicates that the plane is moving towar the tracking station, but since the problem asks for spee, the correct answer is First, observe that what we know is θ A an what we woul like to know is, where A t t is the area of the triangle an θ is the angle between the sies of lengths 4 an 5. In orer to get a relationship between A an θ, we recall that the area of a triangle is A = bh, where b is the length of the base of the triangle an h is the height of the triangle. If we raw the triangle with the 5 cm sie as the base (i.e., b = 5) an let h enote the triangle s height, then we immeiately fin the relation We can now write the area as In orer to get a relationship between A last expression with respect to t. We fin sin θ = h 4 h = 4 sinθ. A = (5)4 sinθ = 0 sinθ. an θ t t, we take the erivative of each sie of this A t = 0 cosθθ t. At θ = π (60o ), we have cos( π ) = an consequently A t = 0( )(.06) =. cm /s. 8. First, f (x) = sin x cosx. 5
6 Also, f(0) = f () = 0. We have, then, f x () = f (f ()) = f (0) = =. 9. For calculations like this it s often convenient to set θ = sin (using θ because this is an angle), so that sin θ =. (See Figure.) θ Figure : Figure for Problem 9 solution. Accoring to the Pythagorean Theorem, the ajacent sielength is b = 9 4 = 5. In this way, tan(sin ) = tanθ = Set f(x) = cosx an use the formula f x (x) = f (f (x)) = sin(cos x). In orer to evaluate cos x, set θ = cos x (we use θ because this is an angle) an note that consequently cosθ = x sin θ = cos θ = x. 6
7 Notice here that since the range of cos x is [0, π], we know that θ [0, π], an so we know sin θ 0. This chooses the sign in front of cos θ. We finally have sin(cos x) =. x. If we take the natural logarithm of both sies, we have ln y = lnx tan x = (tanx)(ln x). Now ifferentiate each sie with respect to x to obtain y x = (sec x)(ln x) + tan x x. Multiplying this last expression by y = x tan x, we conclue x = xtan x ((sec x)(ln x) + tan x x ).. We start with f(x) = ln x an use the linear approximation f(x) = f(a) + f (x)(x a), where it is reasonable here to take a =. We fin We can now compute f(x) ln + (x ) = x. f(.99).99 =.0. (The exact value, to four ecimal places, is .00.). First, the length l is given by the Pythagorean Theorem, By linear approximation, we have l = + x. l(x + x) l(x) l (x) x, where l(x + x) l(x) is the absolute error, x = 4, x =.05 an l (x) = x x + 9. We have, then, We conclue l (4)(.05) = 4 (.05) = l(4 ±.05) = 5 ±.04. 7
8 4. First, the graph is given in Figure. The easiest way to graph a function like this is to expan it out as follows: { + x, x 0 f(x) = x, 0 x. Each iniviual piece is easy to graph. The local minimizers are x =, an the local minima are 0,. The global minimizer is x = an the global minimum is 0. The local an global maximizer is x = 0 an the local an global maximum is. Figure : Figure for Problem 4 solution. 5. By the Mean Value Theorem, we know that there exists some value c (, 5) so that f (c) = f(5) f(). Since the largest possible value for f (c) on this interval is 4 an since the smallest possible value for f (c) on this interval is, we have the inequality Multiplying this last inequality by, we fin f(5) f() 4. f(5) f(). 6a. First, f (x) = ( + x)x x ( + x) = x + x ( + x), 8
9 an from this we can ientify the critical points are x =,, 0. Plotting this on a number line, we fin that f is ecreasing on [, ) (, 0] (note that the point where f is unefine is exclue, but the other enpoints are inclue), an increasing on (, ] [0, + ). 6b. We compute f (x) = ( + x) (x + ) (x + x)( + x) ( + x) 4 = (x + ) (x + x) ( + x) = ( + x). The only possible point of inflection is x =, an plotting this on a number line we fin f is concave own on (, ) an concave up on (, + ). 7. By our assumptions on f, we know f (c) = 0 an (by the first erivative test) f (x) < 0 for x < c (an close to c; i.e., f ecreases own to the minimum) an f (x) > 0 for x > c (an close to c). We nee to show that precisely the same two conitions hol for g(x) = e f(x). We have g (x) = e f(x) f (x) = 0, from which we see that g (c) = e f(c) f (c) = 0, an also that g (x) always has the same sign as g (x). This means that for x near c we have g (x) < 0 for x < c an g (x) > 0 for x > c, an so by the first erivative test g has a minimum at x = c. Note. There is a subtle problem in irectly applying the secon erivative test because f can have a minimum with f (c) = 0. (E.g., f(x) = x 4 with c = 0.) In that case we woul fin g (c) = 0, an this is not enough to conclue that x = c is a minimizer of g. 9
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