YORK UNIVERSITY. Faculty of Science Department of Mathematics and Statistics. MATH A Test #2. June 25, 2014 SOLUTIONS

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1 YORK UNIVERSITY Faculty of Science Department of Mathematics an Statistics MATH A Test # June 5, 04 SOLUTIONS Family Name (print): Given Name: Stuent No: Signature: INSTRUCTIONS:. Please write your name, stuent number an final answers in ink.. This is a close-book test, uration 80 minutes. 3. Calculator of the moel Sharp EL-50 RN is permitte. NO INTERNET CONNECTED DEVICES OR OTHER AIDS ARE PERMITTED. 4. There are eight questions on ten pages. Answer all questions. Fill in answers in esignate spaces. Your work must justify the answer you give. Show your work on the space provie. If you nee more space, use the back of a page an clearly inicate this fact on an original page each time when you use the back of a page for your work. 5. Remain seate until we collect all the test papers. 6. Do the easiest questions first, GOOD LUCK! Question Points Score Total: 70

2 Name: Stuent No:. ( pts) Fin each of the its in parts (a), (b) an (c), if it exists; if oes not exist, state so. Justify each answer you give. (a) (b) x x x x 6 x + 3x + 3 x(x ) (c) x 0 x Now, while x x x 6 x + 3x + = (x + )(x 3) x (x + )(x + ) = x 3 x x + = x (x 3) x (x + ) = 3 + = 5 = 5. x = 3 ex x = Hence, the it oes not exist. 3 = x x 3 = 0 =. 3 x(x ) x x 0 x = = (x x 0 x x 0 x ). x 0 +(x x ) = x 0 +( x ) =, x 0 (x x ) = x 0 ( x ) =. () Fin the value of the constant a that makes the function { x + if x f(x) = 5 ax if x > continuous for all x. f(x) is continuous at x =, if Now, x f(x) = f(), that is f(x) = f(x) = f(). x x + x f(x) = + ) = + = = f(). (x x On the other han, x 5 a =, which yiels a = 5 = 3. f(x) = if an only if + x +(5 ax ) =, that is

3 cos. (a) (4 + 4 pts) Show that = 0. Hint: You may nee to use the fact that =. cos = ( = ( cos + cos + cos ) = + cos ) = cos ( + cos ) = + cos = = 0. (b) Fin the it ( tan ). Hint: You may nee to use the its from part (a). ( tan ) = ( cos ) = 3. (4 + 3 pts) = cos = cos (a) Use the Sanwich Theorem to evaluate the it x sin π x 0 x = 0. = 0 = 0. cos sin ( + cos ) Note that x = x. sin π x ( x 0) x x sin π x x ( x 0). But ( x ) = x = 0. x 0 x 0 Hence, by the Sanwich Theorem, x sin π x 0 x = 0. (b) Use the Intermeiate Value Theorem to show that there is a root of the equation x 5 + x 3 x + = x + 8 on the interval (0, ). Consier the function f(x) = x 5 + x 3 x + x + 8. f(x) is continuous on [0, ] an f(0) = 8 < 0, while f() = = 4 3 = > 0. Hence, by the Intermeiate Value Theorem there exists c (0, ) such that f(c) = 0, that is c 5 + c 3 c + c + 8 = 0, i.e., c 5 + c 3 c + = c + 8.

4 4. (5 + pts) (a) Use the formal efinition to fin the erivative of f(x) =, for x. x Note: No creits will be given for any other solution. f f(x + h) f(x) (x) = = h = x+h x h h h(x )(x + h ) = = (x )(x + h ) = (x ). = x (x + h ) h(x )(x + h ) (x )(x + h ) (b) For each of the following statements etermine if it is always true or sometimes false: i. If f(x) is continuous, then f(x) is ifferentiable. False. ii. If f(x) is not continuous, then f(x) is not ifferentiable. True. 5. ( pts) x (a) If y =, then + x = By the Chain Rule, = x ( + x ) x = ( + x ) = + x x( + x ) ( x )x x ( + x ) = tan x, then = (b) If y = = (tan x ) = (c) If y = x sin x, then = Use logarithmic ifferentiation, So, (tan x ) = (tan x ) sec x ln y = ln x sin x = sin x ln x x ( + x ) y y = sin x (sin x ln x) = cos x ln x + x. + x x 4x ( + x ) = x ( + x ) x 4. (x ) = sin x = y(cos x ln x + x ) = xsin x (cos x ln x + sin x x ). 3 sec x 4 tan x.

5 () If y = x + ln x, for all x > 0, then = y=e+ Note: Please rea the question carefully. Note that y = x + ln x is one-to-one an y = e + x = e. Hence, Therefore, 6. (4 + 4 pts) = (x + ln x) = ( + x ) = x + x. = = x+ x = x x +. = x y=e+ x + = x=e e e +. (a) Fin, if the function is given implicitly by ey + xy = e. Differentiate both sies implicitly, (ey + xy) = e e y + y + x = 0 (ey + x) = y = y e y + x. (b) Fin equations for the tangent an normal lines to the curve e y + xy = e at x = 0. x = 0 = e y + 0 y = e = e y = e = y =. So, we have to fin equations of the tangent an normal lines at the point (0, ). From part (a) the slope m of the tangent line at that point is efine as m = = (0,) e + 0 = e. Hence, the equation of the tangent line will be y = e (x 0) or y + e x = 0. The slope of the normal line at the same point is efine as m = m = consequently, the equation of the normal line will be e = e. An y = e(x 0) or y ex = 0. 4

6 7. (6 pts) A lake is pollute by waste from a plant locate on its shore. Ecologists etermine that when the level of pollutant is x parts per million (ppm), there will be left F fish of a certain species in the lake, where F = 4 + x. When there are fish left in the lake, the pollution is increasing at the rate of. ppm/year. At what rate is the fish population changing at this time? We have to fin F t = By the Chain Rule, F t = F when F = an t = x x = = 6 x = x = 4. F t F = ( ) = x (4 + x) ( x) = (4 + x) x = x(4 + x).. So, substituting x = 4 an =., we obtain t t = x=4 4(4 + 4). = Hence, the fish population is ecreasing by 300 fish per year.. = (6 pts) During a meical proceure, the size of a roughly spherical tumor is estimate by measuring its raius an using the formula V = 4 3 πr3 to compute its volume. Estimate the percentage error that can be allowe in the measurement of raius r to ensure that there will be no more than 0% error in the calculation of volume V. Let r 0 be the true value of r. r V = r (4 3 πr3 ) = 4πr. We have to ensure that the percentage error in the measurement of V is no more than 0%, that is 00 V V (r 0 ) 0. Using linear approximation, 00 V V (r 0 ) 0 00 V (r 0 ) r V (r 0 ) πr0 r 4 3 πr r r 0 0 3(00 r r 0 ) 0 00 r r = 3 3. Therefore, we must measure raius within an error of ± 3 3 %. The en. 5