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Short Answer Type

31.

The length x of a rectangle is decreasing at the rate of 3 cm/minute and the width y is increasing at the rate of 2 cm/minute. When x = 10 cm and y = 6 cm, find the rate of change of (a) the perimeter and (b) the area of rectangle

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32.

The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minutc. When x = 8 cm and y = 6 cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.

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Long Answer Type

33.

The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base?

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Short Answer Type

34.

A particle moves along the curve 6y = x³ + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.

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Long Answer Type

35.

At what points of the ellipse 16x² + 9y² = 400, does the ordinate decrease at the same rate at which the abscissa increases?

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36.

A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4m away from the wall?

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Short Answer Type

37.

Sand is pouring from a pipe at the rate of 12 cm³/s. The falling sand forms a cone on the ground in such a way what the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand-cone increasing when the height is 4 cm?

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Long Answer Type

38. A water tank has the shape of an inverted right circular cone with its axis vertical and vertex lowermost. Its semi-vertical angle is tan^–1 (0.5). Water is poured into it at a constant rate of 5 cubic metre per hour. Find the rate at which the level of the water is rising at the instant when the depth of water in the tank is 4 m.

Let r be the radius, h be the height and a be semi-vertical angle of right circular cone.
$therefore space space space space tan space straight alpha space equals space straight r over straight h space space space or space space space straight alpha space equals space tan to the power of negative 1 end exponent open parentheses straight r over straight h close parentheses$
But $straight alpha space equals space tan to the power of negative 1 end exponent left parenthesis 0.5 right parenthesis$ (given)
$therefore space space space space space straight r over straight h space equals space 0.5 space space space space space rightwards double arrow space space space space space straight r space equals space straight h over 2 space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis$

Let V be the volume of the cone.
$therefore space space space space space space space space space space space space straight V space equals space 1 third πr squared straight h space equals space 1 third straight pi open parentheses straight h over 2 close parentheses squared space straight h space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space of space left parenthesis 1 right parenthesis close square brackets therefore space space space space space space space space space space space space straight V space equals space πh cubed over 12 therefore space space space space space space space space space space space dV over dt space equals space straight d over dh open parentheses fraction numerator straight pi space space straight h cubed over denominator 12 end fraction close parentheses space. space dh over dt space equals space fraction numerator 3 πh squared over denominator 12 end fraction space equals space dh over dt therefore space space space space space space space space dV over dt space equals space straight pi over 4 straight h squared space dh over dt space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis Now space space space dV over dt space equals space 5 space straight m cubed divided by straight h space space space space space space and space space space space space straight h space equals space 4 space metres$
$therefore space space space space space from space left parenthesis 2 right parenthesis comma space we space get comma space space space space space space space space space space space space space space space space space space space space space space 5 space equals space straight pi over 4 left parenthesis 4 right parenthesis squared space dh over dt space space space space space or space space space space 5 space equals space 4 straight pi dh over dt therefore space space space space space space space space space space space dh over dt space equals space fraction numerator 5 over denominator 4 straight pi end fraction space equals space fraction numerator 5 cross times 7 over denominator 4 cross times 22 end fraction space space straight m divided by straight h space equals space 35 over 88 space straight m divided by straight h therefore space space space space space rate space of space change space of space water space level space space equals space 35 over 88 straight m divided by straight h.$

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Short Answer Type

39.

A man 2 metres high walks at a uniform speed of 6 metre/sec away from a lamp-post 6 metres high. Find the rate at which the length of his shadow increases.

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40.

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.

170 Views