Version 005 Eam Review Practice Problems NOT FOR A GRADE aleander 5575 This print-out should have 47 questions Multiple-choice questions may continue on the net column or page find all choices before answering 00 00 points Find an equation for the tangent to the graph of f at the point P, f when f y + 5 + 5 0 5 y 5 5 correct y 5 9 5 4 y 7 5 y + 5 + 5 0 If,thenf, sowehavetofind an equation for the tangent line to the graph of 5 f at the point, Now the Newtonian quotient for f at a general point, f is given by f+h f h First let s compute the numerator of the Newtonian Quotient: f+h f Thus 5 +h 5 5 5{ +h} { +h} f+h f h 5h h +h 5 +h Hence f lim h 0 5 +h At, therefore, f 5 5 6 5, so by the point slope formula an equation for the tangent line at, is y + 5 which after simplification becomes y 5 5 00 00 points Find when + y 5 4 5 6 / y y / correct y/ y/ y / / y
Version 005 Eam Review Practice Problems NOT FOR A GRADE aleander 5575 Differentiating implicitly with respect to, we see that + y 0 y y / 00 00 points Determinethevalueof/at when and / 4 5 y 5 9 correct 7 Differentiating implicitly with respect to t we see that At, therefore, 9 004 00 points Determine f when f sin sin+4 f cos sin+4 f 5sincos sin+4 f 5cos sin+4 4 f sincos sin+4 5 f 6 f cos sin+4 By the Quotient Rule, 5cos sin+4 correct f sin+4cos sin cos sin+4 But sin+4 cos sin cos 5 cos Thus f 5cos sin+4 keywords: derivative of trig functions, derivative, quotient rule 005 00 points The points P and Q on the graph of y y 5y +0 0 have the same -coordinate Find the point of intersection of the tangent lines to the graph at P and Q 5 intersect at 7, 0 correct 7 intersect at intersect at 0, 0 7 5 7, 0 7
Version 005 Eam Review Practice Problems NOT FOR A GRADE aleander 5575 4 intersect at 5 intersect at 0 7, 5 5, 0 7 The respective y-coordinates at P, Q are the solutions of y y 5y +0 0 at ; ie, the solutions of Thus y 7y +0 y 5y 0 P, 5, Q, To determine the tangent lines we need also the value of the derivative at P and Q But by implicit differentiation, so Thus y +5 y 0 y y 5 5 P, Q and y + 0 respectively These two tangent lines 5 intersect at 7, 0 7 006 00 points At noon, ship A is 50 miles due west of ship B Ship A is sailing south at 0 mph while ship B is sailing north at 0 mph At what speed is the distance between the ships changing at 5:00 pm? speed 4 mph speed 8 mph speed mph correct 4 speed 0 mph 5 speed 6 mph Let a at be the distance travelled by ship A up to time t and b bt the distance travelled by ship B Then, if s st is the distance between the ships, the relative positions and directions of movement of the ships are shown in By the point-slope formula, therefore, the equation of the tangent line at P is 50 B while that at Q is y 5 5, s b y the tangent lines at P and Q are y 5 + 5 A a
Version 005 Eam Review Practice Problems NOT FOR A GRADE aleander 5575 4 By Pythagoras, therefore, s a+b +50 Thus, after implicit differentiation, in which case, s ds da a+b + db, ds a+b a+b 0+0 40 s s when ships A and B are travelling at respective speeds 0 mph and 0 mph But at 5:00 pm, a 50, b 50, s 50 at 5:00 pm, the distance between the ships is increasing at a speed ds mph t5 007 00 points Determine / when 4 5 6 y cos 5 y cos y tan correct y tan y cot y cot y sin After implicit differentiation with respect to we see that y sin +y cos 0 y sin cos 008 00 points y tan A 5 foot ladder is leaning against a wall If the foot of the ladder is sliding away from the wall at a rate of ft/sec, at what speed is the top of the ladder falling when the foot of the ladder is 4 feet away from the base of the wall? speed 6 ft/sec correct speed 5 ft/sec speed 49 ft/sec 4 speed 46 ft/sec 5 speed 47 ft/sec Let y be the height of the ladder when the foot of the ladder is feet from the base of the wall as shown in figure ft 5 ft We have to epress / in terms of, y and / But by Pythagoras theorem, +y 5,
Version 005 Eam Review Practice Problems NOT FOR A GRADE aleander 5575 5 so by implicit differentiation, In this case +y y 0 ButagainbyPythagoras,if 4,theny Thus, if the foot of the ladder is moving away from the wall at a speed of ft/sec, and 4, then the velocity of the top of the ladder is given by 4 the speed at which the top of the ladder is falling is speed 6 ft/sec The slope, m, of the tangent line at the point P,f on the graph of f is the value of the derivative f + at, ie, m 7 On the other hand, f Thus by the point-slope formula, an equation for the tangent line at P,f is y 7, which after simplification becomes y 7 the tangent line at P has y-intercept 00 00 points Determine the derivative of f arcsin f 6 9 keywords: speed, ladder, related rates 009 00 points Find the y-intercept of the tangent line at the point P,f on the graph of y-intercept f ++ y-intercept correct f f 4 f 5 f 6 f 9 6 9 correct y-intercept 6 4 y-intercept 6 5 y-intercept 6 y-intercept Use of d arcsin, together with the Chain Rule shows that f /
Version 005 Eam Review Practice Problems NOT FOR A GRADE aleander 5575 6 f 9 0 00 points Find the derivative of f f f sin e e e 6 correct e 6 f e +e 6 4 f +e 6 5 f e +e 6 6 f 7 f 8 f Since d sin +e 6 e 6 e e 6, the Chain Rule ensures that f d ea ae a, e e 6 0 00 points The height of a triangle is increasing at a rate of 4 cm/min while its area is increasing at a rate of sq cms/min At what speed is the base of the triangle changing when the height of the triangle is cms and its area is 5 sq cms? speed cms/min speed cms/min speed cms/min 4 speed 5 cms/min 5 speed 4 cms/min correct Let b be the length of the base and h the height of the triangle Then the triangle has area A bh Thus by the Product Rule, and so da db da h h b dh +hdb, dh b da A h since b A/h Thus, when dh 4, and da we see that db 4A h h dh,, cms/min at the moment when h and A 5, the base length is changing at a speed 4 cms/min
Version 005 Eam Review Practice Problems NOT FOR A GRADE aleander 5575 7 recall: speed is non-negative 0 00 points Use linear approimation with a 6 to estimate the number 6 as a fraction 4 5 6 4 40 correct 6 4 80 6 4 80 6 4 0 6 4 6 For a general function f, its linear approimation at a is defined by L fa+f a a and for values of near a f L fa+f a a provides a reasonable approimation for f Now set f, f Then, if we can calculate a easily, the linear approimation a+h a+ h a provides a very simple method via calculus for computing a good estimate of the value of a+h for small values of h In the given eample we can thus set For then a 6, h 0 6 4 40 Find y when y y +5+ 04 00 points y +5+ 6 y y +5+ y y +5+4 4 y 5+ y 5 y y +5+4 6 y y +5+4 correct Differentiating implicitly with respect to we see that Thus and so d y +5+ d 6 y +y+5+4 0, y y 5 4 y y +5+4 05 00 points Let f, g be differentiable functions Consider the following statements: A If F {f /f}, then { F f f+ } f
Version 005 Eam Review Practice Problems NOT FOR A GRADE aleander 5575 8 B If F fg, then F f g+fg C If F f/g, then F f g+fg g Which of these statements are true? C only B and C only A and B only 4 A only 5 all of them 6 B only correct 7 none of them 8 A and C only By the respective Product and Quotient Rules, while d fg f g+fg d f g f g g f g Applying these to F f f we see that F f f + f, { f } f A Not True B True C Not True 06 00 points Find the derivative of f when f 4+7 f +7 f 6+7 correct f 6 7 4 f 4 7 5 f 6 7 6 f 4+7 By the Product Rule f +7 + After simplification this becomes f +7+4 07 00 points Find the derivative of f 6+7 f sin 9 sin
Version 005 Eam Review Practice Problems NOT FOR A GRADE aleander 5575 9 f cos sin f 6 9 sin 4 f 6cos sin 5 f 6 f 9 sin 6 9 sin correct The Chain Rule together with d shows that sin a f a a 6 9 sin 08 00 points Find anequationfor thetangent linetothe π π graph of f at the point P 4 4, f when f 9tan y 0+ π 4 y 8+9 π correct y +8 π 4 4 y 4+ π 4 5 y 7+4 π When π, then f 9, so P 4 π 4, 9 Now f 9sec Thus the slope of the tangent line at P is f π 9sec π 8 4 4 By the point-slope formula, therefore, an equation for the tangent line at P is given by y 9 8 π, 4 which after simplification becomes Find f when y 8+9 π 09 00 points f 4 5 f 5 4 5 f 5 correct f 0 4 5 4 f 5 5 f 5 Using the Quotient Rule for differentiation we see that f 45 54 5 f 5 00 00 points
Version 005 Eam Review Practice Problems NOT FOR A GRADE aleander 5575 0 A circle of radius r has area A and circumference C are given respectively by A πr, C πr If r varies with time t, for what value of r is the rate of change of A withrespect to t twice the rate of change of C with respect to t? r r π r correct 4 r 5 r π 6 r π Differentiating A πr, C πr implicitly with respect to t we see that da πr dr, dc π dr Thus the rate of change, da/, of area is twice the rate of change, dc/, of circumference when da dc, ie, when πr dr This happens when π dr r 0 00 points Find the derivative of f cos sin f cos sin f +sin cos f +cos sin 4 f sin cos 5 f +cos sin 6 f sin cos By the Quotient Rule, But correct f sinsin cos cos sin cos +sin cos sin cos +sin f cos sin keywords: DerivTrig, DerivTrigEam, quotient rule 0 00 points Determine f when f tan Hint : first simplify f f + f
Version 005 Eam Review Practice Problems NOT FOR A GRADE aleander 5575 f 4 f 5 f correct + If tanθ, then by Pythagoras theorem applied to the right triangle f sec 8+sectan f 8sec+ 8+sec 4 f 8cos+ 8cos+ 5 f 8+cos 8cos+ correct 6 f 8sin+cos 8cos+ Now, θ tan 8+sec Thus sin cos 8+ cos sin 8cos+ we see that Thus sinθ f sin f cos 8cos+ + 8sin 8cos+ cos8cos++8sin 8cos+ f f 8+cos 8cos+, Alternatively, we can differentiate f using the Chain Rule and the fact that d tan + 0 00 points since cos +sin 04 00 points If y y is defined implicitly by 6e 5y y +6, find the value of / at, y, 0 Find the derivative of f when f f tan 8+sec sec 8+sectan 9 9 correct 5 7
Version 005 Eam Review Practice Problems NOT FOR A GRADE aleander 5575 4 5 6 7 7 Differentiating 6e 5y y +6 implicitly with respect to we see that 0e 5y In this case, 0e 5y y + +6 y +6 at, y, 0 we see that 9 5 rate 4 ccs/min After differentiation of PV C with respect to t using the Product Rule we see that dpv P dv +V dp which after rearrangement becomes When therefore, dv V P V 400, P 80, dv 400 8 80 dp dp 40 0, 8, the volume is decreasing at a rate of 40 ccs/min keywords: implicit differentiation, eponential function, 05 00 points Boyle s Law states that when a sample of gas is compressed at a constant temperature, the pressure and volume satisfy the equation PV C,whereC isaconstant Supposethat at a certain instant the volume is 400 ccs, the pressure is 80 kpa, and the pressure is increasing at a rate of 8 kpa/min At what rate is the volume decreasing at this instant? rate 6 ccs/min rate 40 ccs/min correct 06 00 points Determine f when f sin + Hint : first simplify f f f f 4 f + + correct + + rate 4 ccs/min 5 f + 4 rate 8 ccs/min
Version 005 Eam Review Practice Problems NOT FOR A GRADE aleander 5575 If sinθ +, 4 L then by Pythagoras theorem applied to the right triangle 5 L + 6 L + The linearization of f is the function θ But for the function L f0+f 0 we see that f + + /, Thus tanθ f θ tan f + Alternatively, we can differentiate f using the Chain Rule and the fact that d sin 07 00 points Find the linearization of at 0 f + L 4 correct L + 4 L + 4 the Chain Rule ensures that and so f + / f0, f 0 4, L 4 08 00 points Find the value of f a when f a a+ f a f a 4 a+ ft t+ t+ a+ correct 4 f 4 a a+
Version 005 Eam Review Practice Problems NOT FOR A GRADE aleander 5575 4 5 f a a+ 6 f a By definition, 4 a+ f fa+h fa a lim h 0 h Now, for the given f, while Thus fa+h a+h+ a+h+, fa a+ a+ fa+h fa a+h+ a+h+ a+ a+ But and a+h+a+ a+h+a+ a+h+a+ {a+h+}a+ ha++a+a+, a+h+a+ ha++a+a+ fa+h fa h in which case h{a+ a+} ha+h+a+ f a a+h+a+, a+ 09 00 points Find the derivative of f sin+cos f cos f sin f + sin 4 f cos 5 f +cos correct 6 f sin By the Product Rule while d sin sin+ cos, d cos cos sin Consquently, f +cos keywords: DerivTrig, DerivTrigEam, 00 part of 00 points A point is moving on the graph of When the point is at 4 +5y y P 9, 9, its -coordinate is decreasing at a speed of 4 units per second
Version 005 Eam Review Practice Problems NOT FOR A GRADE aleander 5575 5 What is the speed of the y-coordinate at that time? speed y-coord units/sec speed y-coord units/sec speed y-coord units/sec 4 speed y-coord units/sec 5 speed y-coord units/sec correct Differentiating 4 +5y y implicitly with respect to t we see that Thus Now at P while +5y 9, 9 y 5y Hence, at P, y 5y, 9 9 9 5 9 y + 9, 9 6 When the -coordinate at P is decreasing at a rate of 4 units per second, therefore, 4, so the speed y-coord units/sec 0 part of 00 points In which direction is the y-coordinate moving at that time? direction increasing y correct direction decreasing y Since, at P, the y-coordinate of the point is moving in the at that time Find when 4 5 direction increasing y 0 00 points +y 9y 0 y y +y y y y + y y correct y + y + We use implicit differentiation For then 6 +y 9y 9 0,
Version 005 Eam Review Practice Problems NOT FOR A GRADE aleander 5575 6 which after solving for / and taking out the common factor gives y+ y 0 y y keywords: implicit differentiation, Folium of Descartes, derivative, 0 00 points Use linear appproimation to estimate the value of 7 /4 Hint: 6 /4 7 /4 6 7 /4 6 7 /4 6 4 7 /4 5 7 /4 65 correct Set f /4, so that f6 as the hint indicates Then df 4 /4 By differentials, therefore, we see that fa+ fa df a 4a /4 Thus, with a 6 and, 7 /4 04 00 points Find the derivative of g when g 4 cos g 4sin cos g 4cos sincorrect g 4cos+sin 4 g 4sin+cos 5 g 4 cos sin 6 g 4 sin cos By the Product rule, g 4 sin+cos 4 g 4cos sin Find when 05 00 points tan y +y sec y+ sec y+ sec y sec y 7 /4 65 +sec y sec y+
Version 005 Eam Review Practice Problems NOT FOR A GRADE aleander 5575 7 4 5 6 sec y sec y sec y sec y+ correct sec y sec y+ Differentiating implicitly with respect to, we see that sec y After rearranging, this becomes + sec y+ sec y keywords: sec y sec y+ 06 00 points If the radius of a melting snowball decreases at a rate of ins/min, find the rate at which the volume is decreasing when the snowball has diameter 4 inches rate 6π cuins/min correct Thus by implicit differentiation, dv 4πr dr 4πr, since dr/ ins/min When the snowball has diameter 4 inches, therefore, its radius r and dv 44π when the snowball has diameter 4 inches, the volume of the snowball is decreasing at a rate 6π cuins/min 07 00 points A point is moving on the graph of When the point is at 6 +4y y P 0, 0, its y-coordinate is increasing at a speed of 7 units per second What is the speed of the -coordinate at that time and in which direction is the - coordinate moving? speed 9 units/sec, increasing 4 speed 7 8 units/sec, decreasing rate 8π cuins/min speed 9 4 units/sec, decreasing rate 7π cuins/min 4 rate 5π cuins/min 5 rate 4π cuins/min The volume, V, of a sphere of radius r is given by V 4 πr 4 speed units/sec, increasing 5 speed 7 units/sec, increasing 4 6 speed units/sec, decreasing 7 speed 7 units/sec, decreasing 4 correct
Version 005 Eam Review Practice Problems NOT FOR A GRADE aleander 5575 8 8 speed 7 8 Differentiating units/sec, increasing 6 +4y y implicitly with respect to t we see that Thus Now at P, while 8 +y Hence, at P, y 8 y y + y 50, 8 y 5 4 When the y-coordinate at P is increasing at a rate of 7 units per second, therefore, 7 4 the -coordinate is moving at speed 7 4 units/sec, speed 5π sq ft/sec speed 9 sq ft/sec 4 speed 6π sq ft/sec correct 5 speed 6 sq ft/sec 6 speed 5 sq ft/sec 7 speed 8 sq ft/sec 8 speed 7π sq ft/sec The area, A, of a circle having radius r is given by A πr Differentiating implicitly with respect to t we thus see that When da r, πr dr dr 4, therefore, the speed at which the area of the ripple is increasing is given by Find when speed 6π sq ft/sec 09 00 points tany y and the negativesign indicates that it is moving in the direction of decreasing 08 00 points Arockisthrownintoastillpondandcauses a circular ripple If the radius of the ripple is increasing at a rate of 4 ft/sec, at what speed is the area of the ripple increasing when its radius is feet? speed 8π sq ft/sec 4 +ysec y sec y+ ysec y sec y+ correct +sec y ysec y sec y ysec y
Version 005 Eam Review Practice Problems NOT FOR A GRADE aleander 5575 9 5 6 ysec y sec y sec y ysec y+ Differentiating implicitly with respect to, we see that sec y y + After rearranging, this becomes sec y+ ysec y keywords: ysec y sec y+ 040 00 points Find the derivative of + g + g 9 + g ++9 + g +9 + 4 g + 9 + 5 g ++9 + 6 g 9 + correct By the Quotient and Product Rules we see that + g + + + + + + + + + But +++ + +++ Consequently ++9 g ++9 + Find / when 4 5 6 y y 04 00 points +y 5 y correct y y y Diferentiating +y 5
Version 005 Eam Review Practice Problems NOT FOR A GRADE aleander 5575 0 implicitly with respect to we see that 6+4y 0 6 4y y 04 00 points A balloon is released feet away from an observer The balloon is rising vertically at a rate of ft/sec and at the same time the wind is carrying it horizontally away from the observeratarateofft/sec Atwhatspeedis the angle of inclination of the observer s line of sight changing seconds after the balloon is released? speed 7 rads/sec speed 7 rads/sec speed 5 54 rads/sec 4 speed rads/sec correct 8 5 speed 9 rads/sec Let y yt be the height of the balloon t seconds after release and let t be its horizontal distance from the point of release as shown in the figure Then by right triangle trigonometry, the angle of inclination θ satisfies the equation tanθ y + Differentiating this equation implicitly with respect to t we see that sec θ dθ + y + But sec θ +tan θ Thus dθ + y + +y + y + +y rads/sec Now after seconds, while y 6 ft/sec, 9 ft/sec Hence after seconds the speed at which the angle of inclination is changing is given by speed 8 rads/sec 04 00 points θ y + The dimensions of a cylinder are changing, but theheight is alwaysequal to the diameter of the base of the cylinder If the height is increasing at a speed of 5 inches per second, determine the speed at which the volume, V, is increasingin cubic inches per second when the height is inches dv 4π cub ins/sec
Version 005 Eam Review Practice Problems NOT FOR A GRADE aleander 5575 4 5 dv dv dv dv 6π cub ins/sec π cub ins/sec 5π cub ins/sec correct π cub ins/sec Since the height h of the cylinder is equal toitsdiameter D, the radiusofthecylinder is r h Thus, as a function of h, the volume of the cylinder is given by Vh πr h π 4 h Therateofchangeofthevolume, therefore, is dv π h dh 4 Now dh 5 ins/sec when h ins, dv Find / when 5π cub ins/sec 044 00 points y + 6 y y 6+ y y + y 4 5 6 y correct y y + y y + 6 y y + 6+ y Differentiating implicitly with respect to, we see that so + y y + y, y y y 045 00 points Find anequationfor thetangent linetothe graph of f sec+cos at the point π, f π y π y + π y π
Version 005 Eam Review Practice Problems NOT FOR A GRADE aleander 5575 4 y π correct 5 y π 6 y + π Since f π, we have to find an equation for the tangent line to the graph of f at the point P π, Now f sectan sin, so the slope at P is f π By the point-slope formula, therefore, an equation for the tangent line at P is given by y π, which after rearrangement and simplification becomes y π 046 00 points Find the derivative of f arctan f f f 4 f +9 arctan arctan correct +9 9+ arctan 9+ arctan 5 f sec tan 6 f sec tan Since d arctan +, the Chain Rule gives d arctan +9 Using the Chain Rule yet again, therefore, we see that f +9 arctan keywords: derivative, inverse tan, Chain Rule, 047 00 points A street light is on top of a foot pole A person who is 5 feet tall walks away from the pole at a rate of feet per second At what speed is the length of the person s shadow growing? speed ft/sec speed 5 7 speed 6 7 ft/sec 4 speed 7 7 ft/sec 5 speed 8 7 ft/sec ft/sec correct If denotes the length of the person s shadow and y denotes the distance of the person from the pole, then shadow and the lightpole are related in the following diagram
Version 005 Eam Review Practice Problems NOT FOR A GRADE aleander 5575 5 y +y By similar triangles, 5 +y, so 5y 5 Thus, after implicit differentiation with respect to t, 5 5 When /, therefore, the length of the person s shadow is growing with speed 5 7 ft/sec