Sheet1DIFFERENTIAL CALCULUSQUESTIONSCHOICESANSWERDISCUSSION1A Type of derivative obtained from a function which depends upon a single independent variable?a. Partial Derivativeb. Ordinary Derivativec. Derivative of a functiond. All of the aboveb. Ordinary DerivativeThere are two types of Derivative:

Ordinary Derivative - the derivative obtained from a function which depends upon a single independent variable.

Partial Derivative - the derivative obtained from a function which depends upon two or more independent variable.2If a function is defined by the equation y = f(x), then y is said to be an?a. Explicit function of xb. Odd function of xc. Periodic function of x d. Even function of xa. Explicit function of xThere are five types of Function:

If f(-x) = f(x), then it is an Even function of x If f(-x) = -f(x), then it is an Odd function of x If f(x+p)= f(x),then it is a Periodic function of x If y=f(x), then it is an Explicit function of x If f(x,y) = 0, then it is an Implicit function of x3Find the second derivative of y by implicit differentiation from the equation 4x+8y=36a. 64xb. -9/4y c. 32xyd.16/9 yb. -9/4y4x+8y=36 x+2y=9 --->Eq.1differentiate both sides: 2x+4yy'=0 y'=-2x/4y y'=-x/2y --->Eq.2 Take second derivative: y"=-1/2[(y-xy')/y] --->Eq.3 substitute Eq.2 in Eq.3: y"=-1/2[y-x(-x/2y)/y] y"=-1/2[y+(x/2y)/y] y"=-1/2y[(2y+x)/2y] substitute Eq.1 in y": y"=-9/4y4The derivative with respect to x of 2 cos (x+2) isa. -8x cos(x+2) sin(x+2)b. 2 sin (x+2) cos (x+2)c. -2 sin (x+2) cos (x+2)d. 8x cos(x+2) sin(x+2)a. -8x cos(x+2) sin(x+2)note: cos =(1+cos 2)/2 simplifying: y=2cos(x+2) y=1+cos2(x+2) y=cos(2x+4) note: d(cos u)=-sin du where u=2x+4; du=4x y'=-sin(2x+4)(4x) y'=-4x sin2(x^2+2) note sin2=2 cos sin y'=-4x[2cos(x+2)sin(x+2)] y'= -8x cos(x+2) sin(x+2)5Find the slope of the ellipse x+4y-10x-16y+5=0 at the point where y = 2+8^ and x = 7.a. -0.1463 b. -0.1538c. -0.1768d. -1654c. -0.1768x+4y-10x-16y+5=0 note slope= y' as given: y=2+8^ = 4.828, x=7 differentiate and substitute x=7 and y=4.28 2x+8yy'-10-16y'=0 y'(8y-16)=0 y'=(10-2x)/(8y-16) y'=-0.17686If y = 4cosx+sin2x, what is the slope of the curve when x = 2 radians?a. -2.21b. -3.25c. -4.94d. 2.21c.-4.94y'=4 (-sinx)+cos2x(2) y'=2 cos2x-4sinx at x=2rad y'=2 cos2[2(180/)] - 4sin[2(180/)] y'=-4.947What is the derivative with respect to x of (x+1) - x?a. 3x+6b. 6x+3c. 3x+3d. 6x-3b 6x+3y= (x+1) - x note: d(u)^n= nu^n-1 du y'= 3(x+1)-3x y'=3(x+1)-3x y'=3(x+2x+1)-3x y'=6x+38Differentiate y = log (x + 1)a. 4x (x+1)b. log e(x)(x+1)c. 4x log e/ (x+1)d. 2x (x+1)c. 4x log e/ (x+1)y = log (x + 1) note d(log u)= log e du/u where: u=(x+1) du= 2(x+1)(2x)= 4x(x+1) y'=log e [4x(x+1)/(x+1)] y'=4x log e/ (x+1)9What is the first derivative of the expression (xy) = e?a. 0b. x/yc. -y/x (1+ln xy)d. -y/x (1+ln xy)d.-y/x(1+lnxy)(xy)=e take in on both sides: ln(xy)= ln e x ln (xy)= 1 note d(uv)= udv+vdu differentiatingboth sides: x[(xy'+y)/xy]+ln xy= 0 xy'+y+y ln xy=0 xy'=-y - y ln xy xy'=-y(1+ln xy) y'=-y/x(1+ln xy)10Find the derivative with respect to x the function 2-3xa. -2x / 2 - 3xb. -3x / 2 - 3xc. 3x / 2 - 3xd. -3x / 2-3xd. -3x / 2-3x y=2-3x note: du= du/2u where u=2-3x ; du=-6x y'=-6x/22-3x y'=-3x / 2-3x11Find the derivative of (x + 1) / xa. (x+1) / x - (x+1) / xb. 3(x+1)/ x - (x+1)/ xc. 4(x+1) / x - 2(x+1) / xd. 2(x+1) / x - (x+1) / xb. 3(x+1)/ x - (x+1)/ xy' = (x+1) / x note d(u/v)= (vdu-udv)/v where: u=(x+1): du=3(x+1) v=x; v=x; dv=1 y'=[x(3)(x+1) - (x+1)]/2 y'= 3(x+1)/ x - (x+1)/ x12Evaluate: Lim x1 [x-1 / x+3x-4]a. 1/5b. 2/5c. 3/5d. 4/5b. 2/5Lim x1 [x-1 / x+3x-4] substitute x=1; Lim= (1)-1/(1)+3(1)-4 is indeterminate note use L'Hospitals rule; Lim=2x / 2x+3 Lim=2(1) / 2(1)+3=2/513If f(-x)=-f(x),then f(x)is called ana. Odd function of xb. Periodic function of xc. Implicit function of xd. Even function of xa. Odd function of xThere are five types of Function: If f(-x) = f(x), then it is an Even function of x If f(-x) = -f(x), then it is an Odd function of x If f(x+p)= f(x),then it is a Periodic function of x If y=f(x), then it is an Explicit function of x If f(x,y) = 0, then it is an Implicit function of x14Evaluate: Lim x0 (1-cosx) / xa. 0b. 2c. 1/2d. -1/2c. 1/2Lim x0 (1-cosx) / x Lim= 1-cos0 / 0=indeterminate Apply L' Hospitals rule: Lim=sinx / 2x lim=sin2(0) / 2(0) = indeterminate Apply L' Hospitals rule again: Lim=cosx / 2 Lim=cos0 / 2 = 1/215Compute the following limit: Limit: x [x+4 / x-4]a. 1b. 0c. 2d. infinitea. 1Lim x [x+4 / x-4] Limit =(+4) / (-4) = indeterminate Apply L'Hositals rule: Lim= 1/1= 116Find the slope of xy = 8 at the point (2,2)a. 2b. -1c. -1/2d. -2d. -2xy = 8 y=8 / x differentiate and substitute x=2; y'8(-2x)/x^4=-16/x y'=-16/(2) y'=-217This method is used to approximate the root of an equation which cannot be obtained by the ordinary methods of algebra?a. Partial approximationb. Direct approximationc. Newton's method of approximationd. None of the abovec. Newton's method of approximationNewton's method of approximation is a method used to approximate the root of an equqtion which cannot be obtained by the ordinary methods of algebra. The equation generally involves transcendental functions.18Find the slope of the curve x + y - 6x + 10y + 5=0 at point (1,0).a. 1/5b. 2/5c. 1/4d. 2b. 2/5x + y - 6x + 10y + 5 = 0 Note slope = y' differentiate and substitute x =1 and y = 0 2x+2yy'-6+10y'+0 = 0 y'(2y+10) = 6 - 2x y' = 6-2x / 2y+10 y' = 6 - 2(1) / 2(0) + 10 = 2/519Find the equation of the normal to the curve x + y = 5 at the point (2,1)a. x= 2yb. y= 2xc. 2x+3y =3d. x +y =1a. x=2yx + y = 1 let m1=slope of the given curve differentiate and substitute x=2 and y=1; 2x+2yy'=0 y'=-x/y =-2 m1=-2 note since the line is nominal to the curveat the given point the slope m2 of the line is equal the negative reciprocal of the slope of the given curve. m2=-1 / m1 = 1/2 using point slope form: y-y1= m(x-x1) y-1= 1/2(x-2) x=2y20Locate the points of the inflection of the curve y = f(x)=xea. -2 3b. 2 2c. -2 2d. 2 3c. -2 2y=xey'=x(e)+e(2x) y"=x(e)+e(2)+2x(e) y"=xe+4xe+2e at point of inflection y"=0: 0=x+4x+2 by quadratic formula: x=-4((4)-4*1*2) / 2*1 x=-2 221In the curve 2+12x-x , find the critical points.a. (2,18) and (2,-14)b. (-2,18) and (2,-14)c. (2,18) and (-2,-14) d. (-2,18) and (-2,14)c. (2,18) and (-2,-14)y=2+12x-x note: critical points are points wherein the slope of the curve is zero. y'=12-3x 0=12-3x x=4 x=2 y=2+12(2)-(2)=-18 or y=2+12(-2)-(-2)=-1422A body which moves in a straight line is said to be moving in . If the motion along the line is given by the equation s=f(t).a. Free motion b. Rectilinear motionc. Straight motion d. Vertical motionb. Rectilinear motionA body which moves in a straight line is said to be moving in rectilinear motion . If the motion along the line is given by the equation s=f(t) where s =directed distance traveled by the body at any time t.23Find the radius of curvature of a parabola y-4x =0 at point (4,4).a. 25.78 unitsb. 20.33 unitsc. 15.42 unitsd. 22.36 unitsd. 22.36 unitsR=[1+(y')]^3/2 / |y"| y-4x=0 2yy'-4=0 y'= 4/2y = 2/y y"= (y(0)-2y') / y y"y = -2 (2/y) substitute t=4 y'= 2/4 = 1/2 y"=-4/(4) = -1/16 substitute y' and y"to solve for R: R=22.3624Find the minimun distance from the point (4,2) to the parabola y=8xa.43 b.22 c.3 d.23b. 22d=(x-4)+(y-2) --->Eq.1 y=8x x=y/8 --->Eq.2 substitute Eq. 2 in Eq.1 d=(y/8-4)^2+(y-2)^2 d'=[2(y^2 / 8-4)(2y / 8+2(y-2)] / 2d 0=[2(y / 8-4)(2y / 8+2(y-2)] / 2d 0=(y / 8-4)(2y / 8)+(y-2) 0=y / 32 -y+y-2 2=y / 32 y=4 substitute y=4 in Eq.2 y=(4) / 8=2 substitute x=2 and y=4 in Eq.1 d=(2-4)+(4-2) d=2225A farmer has enough money to build only 100 meters of fence. What ara the dimensions of the field he can enclose the maximum area?a.25mx25m b.15mx35m c. 20mx30m d.22.5mx27.5ma. 25mx25m2x+2y=100 y=50-x ---> Eq.1 A=xy ---> Eq.2 substitute Eq.1 in Eq.2: A= x(50-x) differentiate: dA/dx=50-2x 0=50-2x x=25 substitute x=25 in Eq.1: y=50-25 y=2526Find the minimum amount of tin sheet that can be made into a closed cylinder having a volume of 108 cu. Inches in square inches.a. 125.50b. 127.50c. 129.50d. 123.50a. 125.50V=rh 108=rh h=108/r --->Eq.1 A=2rh+2(r) --->Eq.2 substitute Eq.1 in Eq.2: A=216/r+2r --->Eq.3 differentiate: dA/dr=-216/r +4r 0=-216/r +4r 216/r=4r r=216/4 r=2.8in substitute r in Eq.3: A=216/2.58+2(2.58) A=125.544 in27A poster is to contain 300(cm Aquare) of printed matter with margins of 10 cm at the top and bottom and 5 cm at each side. Find the overall dimensions if the total area of the poster is minimum.a. 27.76cm, 47.8cmb. 20.45cm, 35.6cmc. 22.24cm, 44.5cmd. 25.55cm, 46.7cmc. 22.24cm, 44.5cmLet A=total area Ap=area with the picture A=(10+x)(20+y) A=200+10y+20x+xy ---> Eq.1 Ap=xy 300=xy y=300/x --->Eq.2 substitute Eq.2 in Eq.1: A=200+10(300 / x)+20x+300 A=500+3000 / x+20x --->Eq.3 differentiate: dA/dx=0-3000 / x+20 0=-3000 / x+20 3000 / x=20 x=150 x=12.24cm substitute x=12.24 in Eq.2: y=300 / 12.24=24.5cm Thus the dimensions of the poster is, (12.24)(20+24.5) or 22.24cm by 44.5cm28A normal window is in the shape of a rectangle surmounted by a semi circle. What is the ratio of the width of the rectangle to the total height so that it will yield a window admitting the most ligth for a given perimeter?a. 1b. 1/2c. 2d. 2/3a. 1let P= perimeter P=2x+2y+1/2(2x)y=P-2x-(x) / 2y=.5P-2.57x --->Eq.1A=y(2x)+1/2x --->Eq.2subtitute Eq.1 in Eq.2A=2x(.5P-2.57x)+x/2A=Px-5.14x+x/2 A=PX-3.57xdifferentiate:dA/dx = P-7.14x0=P7.14xx=.14Py=.5P-2.57(.14p)y=.14Psolving for h:Ratio= 2x/h =2x/x+y =2(.14P)/.14P+.14PRatio= 129Type of derivative obtained from a finction whiich depends upon two or more independent variable.a. Ordinary derivativeb. Partial derivativec.derivative of a function d. none of the aboveb.Partial derivativeThere are two types of Derivative:

Ordinary Derivative - the derivative obtained from a function which depends upon a single independent variable.

Partial Derivative - the derivative obtained from a function which depends upon two or more independent variable.30The cost a fuel in runing a locomotive is proportional to the square of the speed and is $25per hour for a speed of 25 miles/hr.other costs amount to $100/hr,regardless of the speed. What is the speed which will make the cost per mile a minimum?a. 40b. 55c. 50d. 45c. 50let C=total cost per hour N=speed in miles per hourC=fuel cost+others C=kN+100--->Eq.1Fuel cost=kN25=k(25)k=1/25substitute k in Eq.1C=1/25N+100let x=total cost per milex=total cost per hour/speed in miles per hourx=1/25N+100/Nx=N/25+100/Ndx/dN=1/25+(-100)/NN=2500N=50mph31The cost C of a product is a function of the quantity x of the product: C(x)=x-4000x+50.Find the quantity for which the cost is minimum.a. 1000b. 1500c. 2000d. 3000c. 2000C=x-4000x+50dC/dx=2x-40000=2x-40002x=4000x=200032An open top rectangular tank with square bases is to have a volume of 10cu.m. The materials for its bottom are to cost P15 per square meter and that for the sides,P6 per square meter.Find the most economical dimensions for the tank.a. 15m x 1.5m x 4.4mb. 2m x 2m x 2.5mc. 4m x 4m x .6md. 3m x 3m x 1.1mb. 2m x 2m x 2.5mlet C=total costC=x(15)+4(xh)(6)C=15x+24xh--->Eq.1V=xh10=xhh=10/x ---Eq.2substitute Eq.2 in Eq.1C=15x+24x(10/x)C=15x+240/xdifferentiate:dC/dx=30x-240/x0=30x-240/x240/x=30xx=240/30x=2msubstitute x=2 in Eq.2h=10/(2)=2.5mtherefore dimensions are 2m x 2m x 2.5m33A boatman ai at A which is 4.5km from the nearest point B on a straight shore BM. He wishes to reach in minimum time a point C situated on the shore 9km from B. How far from C should he land if he can row at the rate of 6 kph and can walk at the rate of 7.5 kph?a. 4.15 kmb.3 km c. 3.25 kmd. 4 kmb. 3 kmlet T=total time needed T=time to row + time to walkT=d/6+x/7.5T=(4.5)+(9-x)^2/6 + x/7.5differentiate:dT/dx=(1/6)2(9-x)(-1)/2(45)+(9-x) +1/7.5-(9-x)/6(45)+(9-x) + 1/7.5 = 09-x/(4.5)+(9-x) = 6/7.5(4.5)+(9-x)=1.25(9-x)(1.25)(9-x) = (4.5) + (9-x).5625(9-x) = 20.25(9-x) = 369-x = 6x= 3km34A function y=f(x) is said to be continous at x=a are follows except:a.f(a) is defined b. lim xa f(x) exists c. lim xa f(x) non existsd. lim xa f(x)=f(a)c.lim xa f(x) non existsA function y=f(x) is said to be continous at x=a if (1) f(a) is defined (2) lim xa f(x) exists (3) lim xa f(x)=f(a)35A fencing is limited to 20 ft. length. What is the maximum rectangular area that can be fenced in using two perpendicular corner sides of an existing wall?a. 120b. 100c. 140d. 190b. 100x+y = 20y=20-x --->Eq.1A=xy --->Eq.2substitute Eq.1 in Eq.2A=x(20-x)differentiate:dA/dx = 20 - 2x0 = 20 - 2xx =10substitute x= 10 in Eq.1y=20-10= 10ft.Substitute x and y in Eq.2A=(10)(10)= 100ft36Given a cone of diameter x and altitude of h. What percent is the volume of the largest cylinder which can be inscribed in the cone to the volume of the cone?a. 44%b. 46%c. 56%d. 65%a. 44%let R= radius of the cone H= heigth of the cone r= radius of the inscribed cylinder h= heigth of the inscribed cylinder V= volume of the cylinderV= rh ---> Eq.1by ratio and proportion:R/H=r/H-hHr=RH-Rhh=(RH-Hr)/R --->Eq.2substitute Eq.2 in Eq.1V=R[(RH-Hr)/R]V=rH- (pi rH/R)differentiate: dV/dr=2rH - (3rH/R)0=2rH- (3rH/R)r=2/3Rsubstitute r in Eq.2h=RH-H(2/3R)/R = 1/3Hsubstitute r and h in Eq.1:V=(2/3R) (1/3H)V=(4/27) RHlet Vc = volume of the coneVc=1/8 RHratio = V/Vcratio = .4437In FDT the function y=f(x) has a maximun point at x=a excepta.y'=0 for x=0b. y'>0 for x0 for x049A triangle has variable sides x,y,z subject to the constrait such that the perimeters is fixed t o 18cm. What is the maximum possible area for the triangle?a. 15.59 cmb. 18.71 cm c. 17.15 cmd. 14.03 cma. 15.59 cmnote: for maximum area the triangle must be a n equilateral.3x = 18x=6A=1/2 x sin60A= 1/2 (6) sin60A=15.59 cm50What is the maximum profit when the profit-versus-production function is as given below? P is profit and x is unit of production.P=200,000 - x -(1.1/x+1)^8a. 285,000b. 200,000c. 250,000d. 305,000b. 200,000P=200,000 - x -(1.1/x+1)^8 ---> Eq.1differentiate:dP/dx = -1.8(1.1/x+1)[-1.1/(x+1)]0 = -1 -8(1.1/x+1)^7 [-1.1/(x+1)]0 = -1 +8(1.1)^8/(x+1)^9(x+1)^9 = 8(1.1)^8x = 0.371substitute x in Eq.1:P=200,000 - 0.371 - (1.1/0.371+1)^8P = 199,999.457 = 200,000

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