Water is dripping out from a conical funnel, at the uniform rate of 2 cc/sec through a tiny hole at the vertex of the funnel. When the slant height of water is 5 cm, find the rate of decrease of the slant height of the water.


Let V be volume of the curve whose semi-vertical angle is α, radius = r, height = h and slant height = l.

therefore space space space space space space straight V space equals space 1 third πr squared straight h
Now space in space straight r. straight t. space angle straight d space increment OMP comma
space space space space space space space space space space space space space space straight r over straight l space equals space sin space straight alpha comma space space space space straight h over straight l space equals space cos space straight alpha
therefore space space space space space straight r space equals space straight l space sin space straight alpha comma space space space space straight h space equals space straight l space cos space straight alpha
therefore space space space space space space straight V space equals space 1 third straight pi left parenthesis straight l squared space sin squared straight alpha right parenthesis thin space left parenthesis straight l space cos space straight alpha right parenthesis space equals space straight pi over 3 straight l cubed space sin squared straight alpha space cos space straight alpha
therefore space space space space space space dV over dt space equals space straight pi over 3 space sin squared space straight alpha space cos space straight alpha. space open parentheses 3 straight l squared space dl over dt close parentheses
rightwards double arrow space space space space dV over dt space equals space πl squared sin squared straight alpha space cos space straight alpha. space space dl over dt
From space the space given space condition comma

               dV over dt space equals space minus 2
therefore space space space space space space space πl squared space sin squared straight alpha space cos space straight alpha. space dl over dt space equals space minus 2
therefore space space space space space space space dl over dt space equals space minus fraction numerator 2 over denominator πl squared space sin squared straight alpha space cos space straight alpha end fraction
When space straight l space equals space 4 space cm comma space we space have
space space space space space space space dl over dt space equals space minus fraction numerator 2 over denominator 16 straight pi space sin squared space straight alpha space cos space straight alpha end fraction space equals space minus fraction numerator 1 over denominator 8 straight pi space sin squared space straight alpha space cos space straight alpha end fraction
therefore space space space rate space of space decrease space of space slant space height space space equals space minus fraction numerator 1 over denominator 8 straight pi space sin squared space straight alpha space cos space straight alpha end fraction straight m divided by sec.
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A point source of light along a straight road is at a height of ‘a’ metres. A boy ‘b’ metres in height is walking along the road. How fast is his shadow increasing if he is walking away from the light at the rate of c metres per minute?

Let AB = a metres be the lamp-post and PQ = b metres the boy, CP = y be his shadow at time t. Let AP = x.
Now ∆CAB and ∆CPQ are equiangular and hence similar

therefore space space space space space space space PC over AC space equals space PQ over AB
rightwards double arrow space space space space space space space fraction numerator straight y over denominator straight x plus straight y end fraction space equals space straight b over straight a
therefore space space space space space space space space space space space space space space space space space space space space space space space space cy space equals space bx plus by
rightwards double arrow space space space space space space space space space space space space space space space space left parenthesis straight a minus straight b right parenthesis space straight y space equals space bx
therefore space space space space space space space space space space space space space space space space space space space space space straight y space equals space fraction numerator straight b over denominator straight a minus straight b end fraction straight x
therefore space space space space space space space space space space space space space space space space space space dy over dx space equals space fraction numerator straight b over denominator straight a minus straight b end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
Also comma space space space space space space space space space space space dx over dt space equals space straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
Now space space space space space dy over dt space equals space dy over dx space. dx over dt space space equals space fraction numerator straight b over denominator straight a minus straight b end fraction cross times straight c space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space space of space left parenthesis 1 right parenthesis comma space left parenthesis 2 right parenthesis close square brackets
therefore space space space space space dy over dt space space equals space fraction numerator bc over denominator straight a minus straight b end fraction comma space space space space space space which space gives space the space rate space at space which space boy apostrophe straight s space shadow space is space increasing. space space


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Water is running into a conical vessel, 15 cm deep and 5 cm in radius, at the rate of 0.1 cm2/sec. When the water is 6 cm deep, find at what rate is
(i) the water level rising?
(ii) the water surface area increasing?
(iii) the wetted surface of the vessel increasing?

Let V be the volume of the water in the cone i.e. the volume of the water cone CA 'B' at any time t.
Let CO' = h, O' A' = r and CA' = I.
Let α be the semi-vertical angle of the cone. CAB where CO = 15 cm, OA = 5 cm
CO = 15 cm
OA = 5 cm
Then,      tan space straight alpha space equals space OA over CO space equals space 5 over 15 space equals space 1 third space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis

 Also,         tan space straight alpha space equals space fraction numerator straight O apostrophe straight A apostrophe over denominator CO apostrophe end fraction space equals space straight r over straight h                              ...(2)
From (1) and (2), we get,
       therefore space space space space space space space space space space space space space space space space space space space space space space space space 1 third space equals space straight r over straight h

rightwards double arrow space space space space space space space space space 3 straight r space equals space straight h                                                        ...(3)
Now,      straight V space equals space 1 third πr squared straight h space equals space space 1 third straight pi open parentheses straight h over 3 close parentheses squared straight h space equals space straight pi over 27 straight h cubed             
therefore space space space space space space space dV over dt space equals space fraction numerator 3 straight pi over denominator 27 end fraction straight h squared space dh over dt
rightwards double arrow space space space space space space 0.1 space equals space fraction numerator 3 straight pi over denominator 27 end fraction straight h squared dh over dt space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space space dV over dt space equals space 0.1 space cm cubed divided by sec space left parenthesis given right parenthesis close square brackets
rightwards double arrow space space space space space space space dh over dt space equals space fraction numerator 2.7 over denominator 3 πh squared end fraction
rightwards double arrow space space space space space space space space space open parentheses dh over dt close parentheses subscript straight h equals 6 end subscript space equals space fraction numerator 2.7 over denominator 3 space straight pi space left parenthesis 36 right parenthesis end fraction space equals space fraction numerator 1 over denominator 40 space straight pi end fraction
therefore space space space space space the space water space level space is space rising space at space the space rate space of space fraction numerator 1 over denominator 40 space straight pi end fraction cm divided by sec.
(ii) Let A be the water surface area at any time t. Then, A = πr squared 
therefore space space space space space space straight A space equals space straight pi straight h squared over 9 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space space of space left parenthesis 3 right parenthesis close square brackets
rightwards double arrow space space space space space dA over dt space equals space fraction numerator 2 πh over denominator 9 end fraction space dh over dt
When space straight h space equals space 6 comma space space space space dh over dt space equals space fraction numerator 1 over denominator 40 space straight pi end fraction comma space space we space have
space space space space space space space space dA over dt space equals space fraction numerator 2 straight pi cross times 6 over denominator 9 end fraction cross times space fraction numerator 1 over denominator 40 space straight pi end fraction space equals space 1 over 30 cm squared divided by sec
therefore space space space space the space water space surface space area space is space increasing space at space the space rate space of space 1 over 30 cm squared divided by sec.
(iii) Let S be the wetted surface area of the vessel at any time t. Then. S = πrl.
Now,       straight l squared space equals space CA to the power of apostrophe squared end exponent space equals space CO to the power of apostrophe squared end exponent space plus space straight O apostrophe straight A to the power of apostrophe 2 end exponent
space rightwards double arrow space space space space straight l squared space equals space straight h squared plus straight r squared
space rightwards double arrow space space space space straight l squared space equals space straight h squared plus straight h squared over 9 space equals space fraction numerator 10 space straight h squared over denominator 9 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space of space left parenthesis 3 right parenthesis close square brackets
space rightwards double arrow space space space space space straight l space equals space fraction numerator square root of 10 space straight h over denominator 3 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 4 right parenthesis
therefore space space space space space space straight S space equals space πrl
rightwards double arrow space space space space space straight S space equals space straight pi open parentheses straight h over 3 close parentheses space open parentheses fraction numerator square root of 10 space straight h over denominator 3 end fraction close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space of space left parenthesis 4 right parenthesis close square brackets
rightwards double arrow space space space space space space straight S space equals space straight pi over 9 square root of 10 space straight h squared
rightwards double arrow space space space space space space dS over dt space equals space fraction numerator 2 straight pi square root of 10 straight h over denominator 9 end fraction dh over dt
Since space straight h space equals 6 comma space space space dh over dt space equals space fraction numerator 1 over denominator 40 space straight pi end fraction
therefore space space space space space space space dS over dt space equals space fraction numerator 2 space straight pi space square root of 10 over denominator 9 end fraction cross times 6 cross times fraction numerator 1 over denominator 40 space straight pi end fraction space equals space fraction numerator square root of 10 over denominator 30 end fraction space cm squared divided by sec.
therefore space space space the space wetted space surface space area space of space the space vessel space is space increasing space at space the space rate space of space fraction numerator square root of 10 over denominator 30 end fraction cm squared divided by sec.

214 Views

A man of height 2 metres walks at a uniform speed of 5 km/h away from a lamp post which is 6 metres high. Find the rate at which the length of his shadow increases.


Let AB be the lamp-post and PQ the man, CP be his shadow at time t.
Let AP PC = y. Also AB = 6 m, PQ = 2 m. Now ∆CAB and ∆CPQ are equiangular and hence similar.

therefore space space space space space PC over AC space equals space PQ over AB space space space space space space space rightwards double arrow space space space space space fraction numerator straight y over denominator straight x plus straight y end fraction space equals space 2 over 6 space space space space space or space space space fraction numerator straight y over denominator straight x plus straight y end fraction space equals space 1 third
therefore space space space space space space 3 straight y space equals space straight x plus straight y comma space space space space space space space or space space space space 2 straight y space equals space straight x
therefore space space space space space space space space space space space space space space space space space straight x space equals space 2 straight y
therefore space space space space space space space space space dx over dt space equals space 2 dy over dt
But space dx over dt space equals space 5 space km divided by straight h
therefore space space space space space 5 space equals space 2 space dy over dt space space space rightwards double arrow space space space space space dy over dt space equals space 5 over 2 km divided by straight h
therefore space space space space length space of space the space shadow space increases space at space the space rate space of space 5 over 2 km divided by straight h.

170 Views

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An inverted cone has a depth of 10 cm and a base of radius 5 cm. Water is poured into it at the rate of 3 over 2 straight c. straight c. space per space minute. Find the rate at which the level of water in the cone is rising when the depth is 4 cm.


Let α be the semi-vertical angle of the cone CAB whose height CO is 10 cm and radius OB = 5 cm. Then
tan space straight alpha space equals space 5 over 10 space equals space 1 half
tan space straight alpha space equals space fraction numerator straight O apostrophe straight B apostrophe over denominator CO apostrophe end fraction space equals space fraction numerator straight O apostrophe straight B apostrophe over denominator straight h end fraction
rightwards double arrow space space space space straight O apostrophe straight B apostrophe space space equals space straight h space tan space straight alpha



Let V be the volume of the water in the cone i.e. the volume of the cone CA 'B' after time t minutes and h be the height of water. Then,
                            straight V space equals space 1 third straight pi left parenthesis OB apostrophe right parenthesis squared space left parenthesis CO apostrophe right parenthesis
rightwards double arrow space space space space space space space space space space straight V space equals space 1 third πh cubed tan squared straight alpha
rightwards double arrow space space space space space space space space space space space space space space space space space straight V space equals space straight pi over 12 straight h cubed space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space space space tan space straight alpha space equals space 1 half close square brackets
therefore space space space space space space dV over dt space space equals space straight pi over 12 space 3 space straight h squared space dh over dt space equals straight pi over 4 straight h squared dh over dt space
rightwards double arrow space space space space space space 3 over 2 space space equals space fraction numerator straight pi space straight h squared over denominator 4 end fraction space space dh over dt space space space space space space space space space space space space space space space space space space space space space space open square brackets because space space dV over dt space equals 3 over 2 cm cubed divided by minute space left parenthesis given right parenthesis close square brackets space
rightwards double arrow space space space space dh over dt space equals space 6 over πh squared
rightwards double arrow space space space space space open parentheses dh over dt close parentheses subscript straight h space equals space 4 end subscript space space equals space fraction numerator 6 over denominator straight pi left parenthesis 4 right parenthesis squared end fraction space equals space fraction numerator 3 over denominator 8 space straight pi end fraction space space cm divided by min. space
therefore space space space space required space rate space space equals space fraction numerator 3 over denominator 8 space straight pi end fraction cm divided by sec. space space space space space space space space space space space space space space
                       
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