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Quiz 2 solution Winter 2022

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School
University of Manitoba**We aren't endorsed by this school
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MATH 1500
Subject
Mathematics
Date
Dec 30, 2024
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3
Uploaded by JusticeMantis4777
Winter 2022Laboratory Quiz #2KeyDue date: March 5, 8:00PM (CDT)Name:Student IDInstructor’s name:GueyeKumarShulyakovaYaoAnswer all questions and show all your work. No calculators allowed. (Total Marks: 25).1. Find the derivativef0(x) of each of the following functions. DO NOT SIMPLIFY YOUR ANSWERAFTER YOU EVALUATE THE DERIVATIVE.(a)[5]f(x) =x-√cotx3xe+ cscx.Solution:f0(x) =0.5z}|{(x-√cotx)0(3xe+ cscx)0.5z}|{-(x-√cotx)0.5z}|{(3xe+ cscx)00.5z}|{(3xe+ cscx)2=1.5z}|{(1--csc2x2√cotx) (3xe+ cscx)-(x-√cotx)1.5z}|{(3exe-1-cscxcotx)(3xe+ cscx)2.TAs: Full credits if the first line in the solution is missing.1
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(b)[3]f(x) =1x·h(x) +ee·5h(x)-7√x4,whereh0(x) exists.Solution:f0(x) =1.5z}|{-1x2·h(x) +1x·h0(x) + 0·5h(x)-7√x40.5z}|{+1x·h(x) +ee·1z}|{(5h0(x)-47x-37.(c)[5]f(x) = sincsc10x+tan(tanx) +π3.Solution:f0(x) =1z}|{coscsc10x+tan(tanx) +π3·2z}|{10 csc9x·(-1) cscxcotx+1.5z}|{sec2(tanx)·sec2x+0.5z}|{0.2.[5]Evaluate the following limit (no credits will be given for L’ Hospital rule).limx→-1tan(x3+x2)2x+ 2.Solution:limx→-1tan(x3+x2)2x+ 2= limx→-1sin(x3+x2)cos(x3+x2)2x+ 2(1 mark)= limx→-1sin(x3+x2)x3+x2·(x3+x2)(2x+ 2)·cos(x3+x2)(1 mark)= limx→-1sin(x3+x2)x3+x2·limx→-1x3+x2(2x+ 2)·cos(x3+x2)(0.5 mark)= 1·limx→-1x22 cos(x3+x2)(1.5 marks)=12 cos 0=12.(1 mark)2
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3. Letf(x) =cosx·k(x)-x22 +k(x).Assume thatk(π) = 2 andk0(π) = 0. Answer the following questions.(a)[5]Findf0(π).Solution:We havef0(x) =1.5z}|{-sinx·k(x) + cosx·k0(x)-2x2 +k(x)0.5z}|{-cosx·k(x)-x21z}|{k0(x)2 +k(x)2|{z}0.5Thenf0(π) =-sinπ·k(π) + cosπ·k0(π)-2π2 +k(π)-cosπ·k(π)-π2k0(π)2 +k(π)2(0.5 mark)=0-2π2 + 2-02 + 22(0.5 mark)=-π2.(0.5 mark)(b)[2]Find the equation of the tangent line to the curvef(x) at the point(π, f(π)).Solution:We havef(π) =-2-π24.(0.5 mark)By the part (a), the slope of the tangent linem=f0(π) =-π2.(0.5 mark)Thus the equation of the tangent line isy--2-π24=-π2(x-π).(1 mark)3
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