First Midterm Examination

Transcription

1 Çankaya University Department of Mathematics Fall Semester MATH Calculus for Engineering I First Midterm Eamination 1) Find the domain and range of the following functions. Eplain your solution. a)5 pts) f) = 3 ) + 7 b)5 pts) f) = c)5 pts) f) = ln 8 7 ) 15 pts) Find a formula for the piecewise-defined function in the figure. Eplain your solution. y f) 3 6 3) Evaluate the following its if they eist. Eplain your solution. Do NOT use L Hôpital s Rule) a)5 pts) tan 3 ) sin3 ) b)5 pts) 5 cos 3 15 c)5 pts) 5 5

2 4) a)10 pts) By using Intermediate Value Theorem IVT), show that the equation = 0 has four different roots in the interval [ 3, 3]. Eplain your solution. b)10 pts) Find all discontinuities of the following function and classify them as removable or non-removable. Eplain your solution if < 4 6 if = 4 f) = if 4 < < 5 6 if = 5 5 if 5 < < 6 0 if 6 5) Evaluate the following its if they eist. Eplain your solution. Do NOT use L Hôpital s Rule) a)5 pts) b)5 pts) c)5 pts) / ) 1 ln 6) If eists, find all horizontal, vertical and oblique asymptotes for the following functions. Eplain your solution. a)10 pts) f) = b)10 pts) f) =

3 Answers 1) a) f) is a polynomial function so domainf) = R. As we have ) 0 3 ) so range of f is: [7, ) b) As for any real number, f) is defined for all real numbers. So domainf) = R. Clearly, f) > 0 for all R. Minimum of is 4 so maimum of f is 1 4 = 3. Therefore range of f is: 0, 3] c) f) is defined whenever 8 7 > 0 so domainf) = R \ {7/8}. Range of f) is the same as range of ln, therefore the range is R. ) l 1 is a line passing through, 6) and 0, 0). Its slope is m 1 = 0 6) 0 ) = 3 Its equation is: y 0 = 3 0) y = 3 Similarly, l is a line passing through 3, 6) and 0, 0). Its slope is m = and its equation is: y = l 3 is a horizontal line with slope zero and equation y = 6. Putting all these together, we obtain: f) = 3 if < 0 if if > 3

4 3) a) = 3) ) 3 Assuming 3 = ) = 9 b) tan 3 ) sin3 ) 5 cos tan 3 ) = sin3 ) cos = sin 3 ) 3 = = 6 1 cos 3 ) sin3 ) 3 6 cos c) = = = = 3 Right and left its are not equal Limit does NOT eist.

5 4) a) Let s define the function f) = Obviously, this is continuous everywhere, because it is a polynomial. Let s check the sign of the function values on certain points: f) sign Using IVT, we can see that there must be a root to the equation f) = 0 on 3, ), a second root on, 1), a third on 1, ) and a fourth on, 3). b)we have to check f around 3 points: = 4 : f) = = 16 4 f) = = f) = 16 4 f4) = 6 4 f) At = 4, function is discontinuous, because the function value is not equal to the it. Discontinuity is removable because it eists. = 5 : f) = = 5 5 f) = 5 = f) does NOT eist 5 At = 5, function is discontinuous, because it does not eist. Discontinuity is non-removable. = 6 : f) = 5 = 16 6 f) = 0 = f) = 16 6 f6) = 16 = 6 f) At = 6, function is continuous.

6 5) a) = ) = ) = ) = = b) / ) 5/ = / + 4 5/ 1 5/ / ) 7/ ) = 1 13 c) 1 ln 0 1 ln = Limit does NOT eist.

7 6) a) f) = = ) ) y = 155 is a horizontal asymptote. f) = and f) = = 017 is a vertical asymptote. b)degree of numerator is 3, degree of denominator is, so there is an oblique asymptote. Polynomial division gives: f) = = ) ) y = 3 is an oblique asymptote. f) = and f) = f) = and + f) = = 3 and = are vertical asymptotes.

8 Çankaya University Department of Mathematics Fall Semester MATH Calculus for Engineering I Second Midterm Eamination 1) Find the derivatives of the following functions. a)5 pts) f) = sin + ln + e b)5 pts) f) = cos 4) + ln ) + e 3 c)5 pts) f) = ln + e cot d)5 pts) f) = e + ln3) + 1 ) a)10 pts) If y is a function of, find dy d : tan y + 5 y = b)5 pts) Find dy d y = arctan + arccos ) 3) 0 pts) Find the absolute maimum and absolute minimum of the function f) = 3 on [ 3, ].

9 4) 15 pts) Sketch the graph of the function f) = ) Evaluate the following its if they eist. a)4 pts) 1 sin π + ) 3 b)4 pts) ln + e c)7 pts) ) 6) 15 pts) An open top bo has volume 75 cm 3 and is shaped as seen in the figure y Material for base costs 1 $/cm and material for sides costs 10 $/cm. Find the dimensions and y that give the minimum total cost.

10 Answers 1) a) f ) = cos e b) f ) = 8 cos4) sin4) e 3 c) f ) = ln + + e cot e sin d) f ) = e + e + 1 ) ) + 1) e + ln3) + 1) ) a) Using implicit differentiation, we obtain: sec y y y + 5 yy = ) sec y + 5 y y = y y = y sec y + 5 y b) y =

11 3) f ) = 3 ln ) 1 = 3 ln 3 + ) Assuming ln 3 1 = 3 + = 3 + ) 1) Critical points are: f = 0 =, = 1 f does NOT eist = ± We have to check all candidate points, in other words, critical points inside the interval [ 3, ] and endpoints: 3 f) Using the fact that 3 > 7, we obtain: Absolute maimum is: f ) = 9 Absolute minimum is: f ) = 0

12 4) Domain: R \ { } intercept: y = 0 = 3 y intercept: = 0 y = 3 4 f) = 1 and f) = 1 y = 1 is Horizontal Asymptote. f) = and f) = = is Vertical Asymptote. + f = f = + 4). The only critical point is = because f is never zero ) 3 f f + f y y = 1 =

13 5) a) This is an indeterminacy of the form 0 0 Use L Hôpital: 1 sin π + ) 3 = cos π + ) 6 1 = 0 b) This is an indeterminacy of the form Use L Hôpital: ln + e = 1 + e + e = 1 + e + e + e = 0 c) This is an indeterminacy of the form 1 Find the logarithm and use L Hôpital: L = ) ln L = ln ) ln ) = = ) ln = = 0 ln L = 0 L = 1

14 6) Volume is 75: y = 75 y = 75 Cost function is: c) = y = = Find the Derivative: c ) = = = 4 ) 3 15 c ) = 0 3 = 15 = 5 y = 3 Using the fact that c ) < 0 for < 5 and c ) > 0 for > 5, we conclude that = 5 gives the local and absolute minimum value.

15 Çankaya University Department of Mathematics Fall Semester MATH Calculus for Engineering I Final Eamination 1) Evaluate the following its if they eist. a)7 pts) b)8 pts) tan3) sin ) e ) a)9 pts) Find the derivative of the function: y = f) = + e ) ln b)6 pts) Find the slope of the tangent line to the curve y + y 8 = 6 at the point 1, 1). 3) Evaluate the integrals: a)7 pts) ) 7 + 4)d b)8 pts) 1 3e 3 + e 3 + d

16 4) a)10 pts) Find the area between the curve y = and the line y = 18. b)10 pts) Evaluate the integral 3 + 1)e d 5) Evaluate the integrals. a)10 pts) π π sin 3 cos 1 d b)10 pts) d 7 + 6) a)9 pts) Evaluate the integral ) 3) d b)6 pts) Evaluate the integral if eists 0 e d

17 Answers 1) a) Using the change of variable u = we obtain: u 0 tan3u ) u sinu) Now multiply and divide by 3u and rearrange to obtain: tan3u ) 3u = u 0 3u sinu) Similarly, multiply and divide by and rearrange: 3 = u 0 tan3u ) u 3u sinu) = = 3 b)this it is in the form 0, so using L Hôpital s rule we obtain: 0 e = e = e = e At this point, the it is NOT in the form 0, so we can NOT use L Hôpital. 0 Just insert = 0 to obtain: = 1 6

18 ) a) ln y = ln ln + e ) ln y) = 1 ln + e ) e + e ln y y = ln + e ) e + e ln y = + e ) ln ln + e ) ) e + e ln b) Using implicit differentiation we obtain: y + 8 yy + 8y 7 y = 0 y = 7 y y + y 7 At 1, 1) the slope is: y = = 1

19 3) a) The substitution u = gives du = 3 + 1) d 1 3 du = + 4) d Rewriting the integral in terms of u, we obtain: ) 7 + 4) d = 1 3 u 7 du = u8 4 + c = ) c b) Use the substitution u = e 3 + du = 3e 3 + ) d The new integral its are: = 1 u = e 3 + = u = e Rewriting the integral in terms of u, we obtain: 1 3e 3 + e 3 + d = e 6 +4 e 3 + du u = ln u e 6 +4 e 3 + = lne 6 + 4) lne 3 + ) = ln e6 + 4 e 3 +

20 4) a) 0 y y = y = The area is: = = 0 7) + ) = 0 = 7, = A = 7 ) ) d = ) d = = ) 3 = ) 3

21 b) We have to use integration by parts. u = du = 3 d dv = e d v = e 3 + 1)e d = 3 + 1) e e 3 d = 3 + 1) e + 3 e d = 3 + 1) e 3 4 e + c = 6 + 5) e 4 + c 5) a) π π sin 3 cos 1 d = = π π π π sin cos 1 sin d 1 cos ) cos 1 sin d Using the substitution: u = cos, du = sin d = = u )u 1 du u 14 u 1 ) du = u15 15 u13 13 =

22 b) We need the trigonometric substitution = 7 tan θ, d = 7 sec θ dθ d 7 sec θ dθ = tan θ = sec θ dθ 1 + tan θ = sec θ dθ sec θ = sec θ dθ = ln sec θ + tan θ + c = ln sec arctan ) + + c 7 7 OR: = ln c = ln ln 7 + c OR: = ln k

23 6) a)using partial fractions epansion, we find: ) 3) = A + B C 3) 4 + = A + B) 3) + C + 1) 4 + = A + C) + 3A + B) + 3B + C) A + C = 4 3A + B = 3B + C = A = 6 13, B = 13, C = ) 3) d = 6 13 d d d 3 = 3 13 ln arctan ln 3 + c b) 0 e d = t t 0 e d = t e t 0 = t e t + 1 = 1 The integral is convergent.