+91-8076753736 support@wiredfaculty.com
logo
logo

Sponsor Area

Application Of Derivatives

Question
CBSEENMA12035048
Wired Faculty App

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.

Solution
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
                       

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation