Math 408CPractice Final Problems Key1 LimitsEvaluate the following limits.1. limx→π2x2− π4√x − πFactoring the numerator, we observe thatlimx→π2x2− π4√x − π= limx→π2(x + π2)(√x + π)(√x − π)√x − π= limx→π2(x + π2)(√x + π)= (π2+ π2)(π + π)= 4π3.2. limx→π−|x − π|x − πFor all x < π we note that |x − π| = π − x, s olimx→π−|x − π|x − π= limx→π−π − xx − π= limx→π−−1= −1.3. limx→π+|x − π|x − πFor all x > π we note that |x − π| = x − π, solimx→π+|x − π|x − π= limx→π+x − πx − π= limx→π+1= 1.14. limx→πx3(x − π)5We observe thatlimx→π−x3(x − π)5= −∞andlimx→π+x3(x − π)5= +∞.Therefore the limit does not exist.5. limx→0sin(πx)xlimx→0sin(πx)x= limx→0πsin(πx)πx= πlimx→0sin(πx)πx= π.6. limx→∞sin(πx)xSince −1/x ≤ sin(πx)/x ≤ 1/x for x > 0, the Squeeze Theorem gives thatlimx→∞−1x≤ limx→∞sin(πx)x≤ limx→∞1x.Sincelimx→∞−1x= limx→∞1x= 0,it follows thatlimx→∞sin(πx)x= 0.7. limx→∞xπ−3Because π − 3 > 0,limx→∞xπ−3= ∞.28. limx→∞xπ−4Because π − 4 < 0,limx→∞xπ−4= 0.9. limx→∞πx3x3+ x2+ x + 1Dividing numerator and denominator by x3, we findlimx→∞πx3x3+ x2+ x + 1= limx→∞π1 + 1/x + 1/x2+ 1/x3=limx→∞πlimx→∞1 + limx→∞(1/x) + limx→∞(1/x2) + limx→∞(1/x3)=π1 + 0 + 0 + 0= π.10. limx→−∞cos(πx)As x → −∞, πx→ 0. Since cos is a continuous function, it follows thatlimx→−∞cos(πx) = lima→0cos(a)= cos(0)= 1.32 Differentiation11. Ifddxf(x) = 0, what can be said about f(x)?Ifddxf(x) = 0, this means the s lope of f(x) is zero everywhere. Since this means f neverincreases nor decreases, f must be constant.Compute dy/dx for the following function using the limit definition of the derivative.12. y = x3+ x.ddx[x3+ x] = limh→0[(x + h)3+ (x + h)] − [x3+ x]h= limh→0x3+ 3x2h + 3xh2+ h3+ x + h − x3− xh= limh→03x2h + 3xh2+ h3+ hh= limh→0(3x2+ 3xh + h2+ 1)= 3x2+ 1.Compute dy/dx for the following relations.13. y =sec xln xBy the quotient rule,dydx=(ln x)(tan x sec x) − (sec x)(1/x)(ln x)2414. y = πxxπBy the product rule,dydx= (πxln π)(xπ) + (πx)(πxπ−1)15. y =p1/xdydx=ddxx−1/2= −12x−3/2.16. y = (tan x − sec x)(tan x + sec x)Solution 1:We could do this with the product rule:dydx= (tan x − sec x)(sec2x + tan x sec x) + (sec2x − tan x sec x)(tan x + sec x)= (tan x − sec x)(sec x + tan x) sec x + sec x(sec x − tan x)(tan x + sec x)= 0.Solution 2:Or we could remember a trig identity:dydx=ddx[tan2x − sec2x]=ddx[−1]= 0.17. y = csc(sec x)By the chain rule,dydx= −csc(sec x) cot(sec x) tan x sec x.518. y = exln xBy the product rule,dydx= (ex)(1/x) + (ex)(ln x).19. y = tan(x) tan−1(x)By the product rule,dydx= (tan x)11 + x2+ (sec2x)(tan−1x).20. y = sin−1(x) + sec−1(1/x)Solution 1:We could brute force it:dydx=ddxsin−1(x) +ddxsec−1(1/x)=1√1 − x2+1(1/x)p(1/x)2− 1(−x−2)=1√1 − x2−1xp(1/x)2− 1=1√1 − x2−1√1 − x2= 0.Solution 2:Or we could note thatdydx=ddx[sin−1(x) + sec−1(1/x)]=ddx[sin−1(x) + cos−1(x)]=ddx[π/2]= 0.621. y2+ 2y= sin xWe differentiate implicitly with respect to x:y2+ 2y= sin x⇒ddx[y2] +ddx[2y] =ddx[sin x]⇒ 2ydydx+ (ln 2)2ydydx= cos x⇒dydx=cos x2y + (ln 2)2y.22. y = x√xTaking the natural log of both sides, we observe thaty = x√x⇒ ln y = ln(x√x)⇒ddx[ln y] =ddx[√x ln x]⇒1ydydx= (√x)1x+12√x(ln x)⇒dydx= x√x2 + ln x2√x.23. y2= x√xThis is similar to the previous problem, except we differentiate implicitly:y2= x√x⇒ ln(y2) = ln(x√x)⇒ddx[2 ln y] =ddx[√x ln x]⇒2ydydx= (√x)1x+12√x(ln x)⇒dydx=y22 + ln x2√x.[Note that we cannot substitute√x√xin for y, since y may be negative. This is whyimplicit differentiation was necessary.]73 Applications of DifferentiationSolve the following related rates problems.24. Three children begin stretching out a large piece of latex in the shap e of an equilateraltriangle by each holding a corner as they walk away from its center (do not give childrenlatex). If one of the sides is expanding at a rate of 1 m/s, how fast is the area expandingwhen one of the sides has length 10 m?The area of a triangle is given by A = hs/2, where h is its height and b the length of sideperpendicular to the height. In an equilateral triangle, h = b√3/2, so A = b2√3/4. Byimplicitly differentiating with respect to t,A = b2√3/4⇒dAdt=√32bdbdt.We are given that db/dt = 1 and we wish to find dA/dt when b = 10. We see thatdAdtb=10m=√32(10m)(1m/s) = 5√3m2/s.25. After the children tire of the latex, they notice a bucket of rocks sitting on the porch anddecide to dump them out on the hood of their dad’s Dodge Stratus (do not keep yourbucket of rocks where children can find it). If the rocks form a conical shape with height10 cm that is increasing in volume at a rate of 5 cm3/s, how fast is the radius of the pileincreasing when the volume is 20 cm3?The volume of a cone is given by V = πr2h/3, where h is its height and r is the radiusof the base. Noting that h is constant, by implicitly differentiating with respect to t weobserveV = πr2h/3⇒dVdt=2πhr3drdt⇒dVdt32πhr=drdtWe are given that dV/dt = 5cm3/s and h = 10cm. We wish to find dr/dt when V = 20cm3.Since r =p3V/(hπ), we havedrdtV =20cm3= (5cm3/s)32π(10cm)p3(20cm3)/((10cm)π)=√34√2πcm/s.826. Clearing the rocks off the hood, the children notice that the car’s cadet blue finish hasbeen horribly scratched, so they resolve to fix it with some paint left over from when thehouse was painted last July (do not leave paint where children may find it). As they pourthe brown paint onto the hood of the Stratus, it forms a circular puddle. If the area isincreasing at a rate of 3 cm2/s, how fast is the puddle’s radius increasing when the radiusis 9 cm?The area of a circle is given by A = πr2, where r is the radius. Implicit differentiationyieldsA = πr2⇒dAdt= 2πrdrdt⇒dAdt12πr=drdtWe are given that dA/dt = 3cm2/s. We wish to find dr/dt when r = 9cm. Thus,drdtr=9cm=12π(9cm)(3cm2/s) =16πcm/s.Solve the following optimization problems.27. Suppose we have two numbers x, y ∈ R such that x + 4y = 9. What is the largest theproduct xy can possibly be?Given y, we may write x = 9 − 4y. Thus, xy = y(9 − 4y). We wish to find the valueof y