University of Toronto Mississauga

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1 Surname: First Name: Student Number: Tutorial: Universit of Toronto Mississauga Mathematical and Computational Sciences MATY5Y Term Test Duration - 0 minutes No Aids Permitted This eam contains pages (including this cover page) and 4 problems. Check to see if an pages are missing and ensure that all required information at the top of this page has been filled in. No aids are permitted on this eamination. Eamples of illegal aids include, but are not limited to tetbooks, notes, calculators, or an electronic device. Unless otherwise indicated, ou are required to show our work on each problem on this eam. The following rules appl: Organize our work in a reasonabl neat and coherent wa, in the space provided. Work scattered all over the page without a clear ordering will receive ver little credit. Msterious or unsupported answers will not receive full credit. A correct answer, unsupported b calculations, eplanation, or algebraic work, will receive no credit; an incorrect answer supported b substantiall correct calculations and eplanations might still receive partial credit. Problem Points Score Total: 0 If ou need more space, use the back of the pages; clearl indicate when ou have done this.

2 MATY5Y Term Test - Januar 6, 08. Each question is worth mark. Write our answer in the given bo. No part marks will be awarded for this question. (i) ( point) Calculate the derivative d d + Solution: Let = + so that ln() = ( + ) ln(). Differentiating implicitl ield d d = ln() + + d [ d = + ln() + + ]. Your answer: [ ] + ln() + + (ii) ( point) Determine the derivative of the function f() = ln(t)dt. Solution: Let F () = ln(t)dt, which b the FTC satisfies F () = ln(). The given epression is F ( ), which we differentiate b the chain rule to get d d F ( ) = F ( ) = ln( ) = 4 ln(). Your answer: 4 ln() (iii) ( point) Which function F is an anti-derivative of f() = A. F () = ( + ) B. F () = ln( + ) 0π C. F () = D. F () = + + / E. None of the above. +? Solution: You can differentiate each function to see which ields f as its derivative. The onl such function is B. Your answer: B Page of

3 MATY5Y Term Test - Januar 6, 08 (iv) ( point) Let f() = 0 on [0, 5]. Determine the right Riemann sum of f if [0, 5] is partitioned into five subintervals of equal length. Solution: You can write out the Riemann sum if ou like, but since f is the constant function, the height of each rectangle is identicall 0. Thus the Riemann sum is the area under the curve, which is a rectangle of height 0 and width 5, ielding R(f) = 50. Your answer: 50 (v) ( point) Write down the second order polnomial approimation to f() = e at a = 0. Solution: Differentiating gives f () = e and f () = e + 4 e so that f(0) =, f (0) = 0, f (0) =. Substituting this into the equation for the polnomial approimation, we get p,0 () = f(0) + f (0) + f (0) = +. Your answer: p,0 () = + (vi) ( point) Consider a metal clinder of radius r and height h. When heated, both the height and radius increase in length. In particular, the radius increases at a rate of 0.5 cm/s, while the height increases at a rate of 0.5 cm/s. Determine the rate of change of the volume of the clinder when the radius is cm and the height is cm. [The volume of a clinder is V = πr h.] Solution: Both r and h change in time. Differentiating implicitl ields Inputting r =, h =, dr dt =, dh dt = 4 dv dt = πrhdr dh + πr dt dt. we get dv dt = 6π + π = 7π cm /s. Your answer: 7π cm /s Page of

4 MATY5Y Term Test - Januar 6, 08 Graph Graph Graph Graph 4 Figure : The graphs for questions (vii) - (). These graphs are also on our formula sheet, which ou can tear from the eam booklet. (vii) ( point) Indicate which graph(s) satisf f () > 0 when > 0. Solution: A positive derivative means that f is strictl increasing. The applies onl to Graphs and. Your answer:, Page 4 of

5 MATY5Y Term Test - Januar 6, 08 (viii) ( point) Indicate which graphs change concavit for < 0. Solution: The onl graph whose concavit does not change for < 0 is Graph, which is consistentl concave up. Hence the answer is Graphs,,4. Your answer:,, 4 (i) ( point) If f is the function illustrated in Graph, which graph corresponds to f? Solution: There are multiple was to approach this problem. From Graph, we see that f is decreasing when < 0 and increasing when > 0. This means that f () < 0 when < 0 and f () > 0 when > 0. The onl graph which satisfies this is Graph. Your answer: () ( point) If f is the function illustrated in Graph, which graph corresponds to f? Solution: There are multiple was to approach this problem. We know Graph is f, which has two critical points at roughl = ±/. In addition, f is decreasing for (, /) (/, ) suggesting that f () < 0 on this same interval. This corresponds to Graph 4. Even if we did not know that Graph corresponds to f, we could still guess that Graph 4 is the correct answer. Indeed, looking at Graph we see that f is almost alwas concave down, ecept for on a small interval corresponding roughl to ( /, /). Hence f () > 0 on ( /, /) and f () < 0 everwhere else. The onl graph which does this is Graph 4. Your answer: 4 Page 5 of

6 MATY5Y Term Test - Januar 6, 08. (5 points) Find the point on the line = + 5 which is closest to the origin (0, 0), and determine the corresponding distance. Be sure to show our answer is a minimum. Your solution must use calculus. Solution: The distance R between two points ( 0, 0 ) and (, ) is given b R = ( 0 ) + ( 0 ). One of our points is the origin ( 0, 0 ) = (0, 0) and the other must lie on the line = + 5, so (, ) = (, + 5), hence R = + ( + 5). From here there are two was to proceed. Differentiating and equating to zero gives dr d = + 6( + 5) + ( + 5) = ( + 5) = 0 from which we infer that there is a single critical point at = /. The denominator is alwas positive, so the sign of f is determine entirel b This is negative when < / and positive when > /, impling the same about f. Hence b the First Derivative Test, = / is a minimum. An easier approach is to realize that is minimized when is minimized, hence it suffices to minimize R = + ( + 5), avoiding the complication of the square root sign. The critical points satisf dr = = 0 d giving = / as a critical point. The Second Derivative Test is now eas to appl, since dr d = 0 > 0, showing that = / is a minimum. There are no boundar cases, since can be arbitraril large in both the positive and negative directions. This is thus the global minimum, and has a corresponding distance of ( R = ) ( ( + ) 9 + 5) = =. Your answer: =, R = 0 Page 6 of

7 MATY5Y Term Test - Januar 6, 08. (5 points) Consider the equation + + =. Find d d if there are more than one. when =. Provide all possible solutions Solution: Let s begin b finding the values of which correspond to =. Substituting = into our equation gives + + = or + = 0. This is factored as ( + )( ) = 0, so = and =. Now we differentiate: d d + + d d + = 0 d d = +. Our first solution corresponds to (, ) = (, ), which we substitute to get d ( ) d = (,) + ( ) = =. The other solution is (, ) = (, ) which gives d d = (,) + () =. Your answer: d d =, Page 7 of

8 MATY5Y Term Test - Januar 6, (0 points) Consider the function f() =. Use the graph paper to plot the corresponding function. + You should include as part of our calculations the domain, - and -intercepts, vertical/horizontal/oblique asmptotes critical points, maima and minima, intervals of increasing/decreasing, inflection points intervals of concavit. Be sure to clearl show all work used to determine the plot. The net page is also available if ou need more space Page 8 of

9 MATY5Y Term Test - Januar 6, 08 Additional space for Question 4. Solution: The domain of the function is R \ { }. The - and - intercept agree and occurs at the origin. There is a single vertical asmptote at =. It satisfies lim f() = lim + f() =. + As the degree of the numerator is strictl larger than the denominator, there are no horizontal asmptotes. Polnomial long division ields so that = is an oblique asmptote. The derivative of f is f () = = ( ) ( + ) ( + ) ( + ) = ( + ) so there are critical points at =,, 0. The factor of ( + ) in the denominator does not affect the sign of f, so the table of increasing and decreasing can be found as follows: < < < < < 0 > f () + + This also shows that = is a local maimum with critical value f( ) = 4, and = 0 is a local minimum with critical value f(0) = 0. The second derivative is f () = ( + )( + ) ( + )( + ) ( + )( + ) ( + ) ( + ) 4 = ( + ) = ( + ) which is never zero, but there is a potential inflection point at =. Indeed, if < then f () < 0, showing that f is concave down here, and if > then f () > 0, showing that f is concave up. Page 9 of

10 MATY5Y Term Test - Januar 6, 08 This page is for additional work and will not be marked. Page 0 of

11 MATY5Y Term Test - Januar 6, 08 Graph Graph Graph Graph 4 Formula Sheet Point-Slope Formula: 0 = m( 0 ) Product Rule: Quotient Rule: Chain Rule: Left Riemann Sum Right Riemann Sum d d f()g() = f ()g() + f()g () d f() d g() = f ()g() f()g () g() d d f(g()) = f (g())g () n f (a + i ) where = (b a)/n i=0 n f (a + (i + ) ) where = (b a)/n i=0 Polnomial Approimation p n,a () = n = n, k= n k = k= n k=0 n(n + ), f (k) (a) ( a) k k! n k = k= n(n + )(n + ) 6 Page of