# Judging from the data in the table it appears as if

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Chapter 34 / Exercise 34-13
Principles of Instrumental Analysis
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Judging from the data in the table, itappears as if both limits are 1. This isconfirmed by the graphing calculator.c.Any answer between 0.697 and 0.707is fine as long as you justify it usingvalues in the table.d.The average rate of change isg(0.4)-g(0.1)0.4-0.1=0.693-0.7940.3=-0.337.3.a.This question becomes much simplerif you rewriteFas(a-1-x-1)-1=1a-1x-1=x-aax-1=axx-a.Then we can easily see that the do-mainD={x|x= 0, x=a}and thatthere are no zeros.b.Sincex=ais not in the domain,x=ais the vertical asymptote. Sincethe degree of the numerator is equalto the degree of denominator, we havey=aas the horizontal asymptote.The discontinuities are the infinite dis-continuity atx=aand the removablediscontinuity atx= 0.c.limx0F(x) = 0;limx→∞F(x) =a; andlimxaF(x) does not exist.d.Solve6a6-a= 12 to geta= 4.
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Chapter 34 / Exercise 34-13
Principles of Instrumental Analysis
Skoog Expert Verified
344The AP CALCULUS PROBLEM BOOKDerivatives Test1.D2.D3.A4.C5.A6.C7.D8.B9.C10.D11.D12.D13.D14.D15.E1.a.Taking the derivative implicitly, wehavey-ysiny= 1y(1-siny) = 1y=11-sinyb.Vertical tangents have an undefinedslope.Hence, we set the denomina-tor ofyequal to zero and solve to getsiny= 1, ory=π/2.Now we findthexvalue wheny=π/2:π2+ cosπ2=x+ 1π2=x+ 1x=π2-1Hence, the vertical tangent isx=π2-1.c.We find the second derivative implic-itly.y=-ycosy(1-siny)2Now plug in the expression fory.y=-1(1-siny)cosy(1-siny)2=-cosy(1-siny)32.a.The volume isV=Bh, whereBisthe area of the triangular base. Hence,V=(12(3)(2))(5) = 15.b.By similar triangles, we havebase of triangleheight of triangle=23,orb=23h; so thatV=1223(h)(h)(5) =53h2.When the trough is14full by volume,we have154=53h2, soh=32at thisinstant.Now, we find the implicitderivative with respect tot:dVdt=103hdhdtand plug in our value ofh:-2 =103·32·dhdtdhdt=-25c.The area of the surface isA= 5b=5·23h=103h.Finding the implicitderivative and using the value ofdh/dtfrom part(b), we havedAdt=103·dhdt=103·-25=-433.a.The domain is whatever makesx4-16x20, orx2(x2-16)0; thus,we find have eitherx= 0 orx216.The domain is therefore (-∞,-4){0}(4,).b.We havef(-x) =(-x)4-16(-x)2=x4-16x2=f(x)sofis even.c.Observe:f(x) =12(x4-16x3)-1/2(4x3-32x)=2x3-16xx4-16x3=2x(x2-8)|x|x2-16d.From part(c), we havef(5) =10(25-8)525-16=343so the slope of the normal is-334.
APPENDIX C.ANSWERS345Applications of Derivatives Test1.D2.D3.D4.C5.D6.A7.B8.E9.B10.A11.B12.D13.D14.C15.A1.a.We havev(t) =x(t) = 2π-2πsin 2πt= 2π(1-sin 2πt)b.We havea(t) =v(t) =x(t) =-4π2cos 2πtc.The particle is at rest whenv(t) = 0:2π(1-sin 2πt) = 0sin 2πt= 12πt=π2t=14,54,94d.We find the critical points ofv(t) bysettinga(t) = 0:-4π2cos 2πt= 0cosπt= 0πt=π2,3π2, . . .
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