Derivative As A Rate Measurer

Total Surface Area of Cylinder = 2 π r² + 2 π r h

Given: Radius of the Cylinder varies.

Therefore, We need to find where S = Surface Area of Cylinder and r = radius of Cylinder.

Hence, Rate of change of total surface area of the cylinder when the radius is varying is given by (4 π r + 2 π h).

The volume of a Sphere =

Where D = diameter of the Sphere

We need to find, where V = Volume of the sphere and D = Diameter of the Sphere.

Hence, Rate of change of Volume of Sphere with respect to the diameter of the Sphere is given by .

Volume of Sphere =

Surface Area of Sphere = 4 π r²

We need to find, where V = Volume of the Sphere and S = Surface Area of the Sphere.

Area of a Circular disc = π r²

Circumference of a Circular disc = 2 π r

Where r = radius of Circular Disc.

Now we need to find where A = Area of Circular disc and C = Circumference of the Circular disk.

The volume of Cone =

Where r = radius of the cone

h = height of the cone

We need to find, where V = Volume of cone and r = radius of the cone.

Area of a Circle = π r²

Where, r = radius of a circle.

We need to find, , where A = Area of the Circle and r = radius of Circle.

At r = 5 cm we have,

Volume of a Ball = Volume of Sphere =

Where r = radius of the ball.

We need to find , where V = Volume of Ball and r = radius of the ball.

At r = 2 cm,

Given: Total Cost, C (x) = 0.007 x³ – 0.003x² + 15x + 4000

Where x = Number of units

Marginal Cost is given by,

Marginal Cost M(x) =

Therefore, M(x) = 0.007 × 3 x² – 0.003 × 2 x + 15

At Number of units = 17, x = 17.

So, M(17) = 0.021 (17)² – 0.006 (17) + 15

M(17) = 20.97

Hence, the Marginal Cost of 17 units is 20.97.

Given: Total Revenue, R(x) = 13 x² + 26 x + 15

Where x = number of units.

Marginal Revenue is given by,

M(x) =

Therefore,

M(x) = 26 (x + 1)

For x = 7,

M(x) = 26(7 + 1)

M(x) = 26 × 8 = Rs. 207

Given: Marginal Revenue, M(x) =

M(x) proportional to Money spent on the welfare of employees.

Thus, M(x) = k (Money spent on the welfare of employees)

M(x) = 6 x + 36

At x = 5, M(x) = 66

Now at x = 6, M(x) = 72

Thus Marginal Revenue for x = 5 is 66

And when x increased from x = 5 to x = 6,

Marginal revenue also increases and thus the money spent on the welfare of employees increases.

Given: the side of a square sheet is increasing at the rate of 4 cm per minute.

To find rate of area increasing when the side is 8 cm long

let the side of the given square sheet be x cm at any instant time.

Then according to the given criteria,

Rate of side of the sheet increasing is,

Then the area of the square sheet at any time t will be

A = x² cm².

Applying derivative with respect to time on both sides we get,

[from from equation(i)]

So when the side is 8cm long, the rate of area increasing will become

[from equation(ii)]

Hence the area is increasing at the rate of 64cm²/min when the side is 8 cm long

Given: the edge of a variable cube is increasing at the rate of 3 cm per second.

To find rate of volume of the cube increasing when the edge is 1 cm long

Let the edge of the given cube be x cm at any instant time.

Then according to the given criteria,

Rate of edge of the cube increasing is,

Then the volume of the cube at any time t will be

V = x³ cm³.

Applying derivative with respect to time on both sides we get,

[from equation(i)]

When the edge of the cube is 1cm long the rate of volume increasing becomes

Hence the volume of the cube increasing at the rate of 9cm³/sec when the edge of the cube is 1 cm long

Given: the side of a square is increasing at the rate of 0.2 cm/sec.

To find rate of increase of the perimeter of the square

Let the edge of the given cube be x cm at any instant time.

Then according to the given criteria,

Rate of side of the square increasing is,

Then the perimeter of the square at any time t will be

P = 4x cm.

Applying derivative with respect to time on both sides we get,

[from equation(i)]

Hence the rate of increase of the perimeter of the square will be 0.8cm/sec

Given: the radius of a circle is increasing at the rate of 0.7 cm/sec.

To find rate of increase of its circumference

Let the radius of the given circle be r cm at any instant time.

Then according to the given criteria,

Rate of radius of a circle is increasing is,

Then the circumference of the circle at any time t will be

C = 2r cm.

Applying derivative with respect to time on both sides we get,

[from equation(i)]

Hence the rate of increase of the circle’s circumference will be 1.4 cm/sec

Given: the radius of a spherical soap bubble is increasing at the rate of 0.2 cm/sec.

To find rate of increase of its surface area, when the radius is 7 cm

Let the radius of the given spherical soap bubble be r cm at any instant time.

Then according to the given criteria,

Rate of radius of the spherical soap bubble is increasing is,

Then the surface area of the spherical soap bubble at any time t will be

S = 4r² cm².

Applying derivative with respect to time on both sides we get,

[from equation(i)]

So when the radius is 7cm, the rate of surface area will become,

Hence the rate of increase of its surface area, when the radius is 7 cm is 11.2 cm²/sec

Given: Spherical balloon inflated by pumping in 900 cubic centimetres of gas per second

To find the rate at which the radius of the balloon is increasing when the radius is 15 cm

Let the radius of the given spherical balloon be r cm and let V be the volume of the spherical balloon at any instant time

Then according to the given criteria,

As the balloon is inflated by pumping 900 cubic centimetres of gas per second hence the rate of volume of the spherical balloon increases by,

We know volume of the spherical balloon is .

Applying derivative with respect to time on both sides we get,

[from equation(i)]

So when the radius is 15cm, the above equation becomes,

Hence the rate at which the radius of the balloon is increasing when the radius is 15 cm will be cm/sec.

Given: radius of an air bubble is increasing at the rate of 0.5 cm/sec

To find the rate at which the volume of the bubble increasing when the radius is 1 cm

Let the radius of the given air bubble be r cm and let V be the volume of the air bubble at any instant time

Then according to the given criteria,

Rate of increase in the radius of the air bubble is,

We know volume of the air bubble is .

Applying derivative with respect to time on both sides we get,

[from equation(i)]

So when the radius is 1cm, the above equation becomes,

Hence the rate at which the volume of the air bubble is increasing when the radius is 1 cm will be 2 cm³/sec.

Given: man 2 metres high walks at a uniform speed of 5 km/hr away from a lamp - post 6 metres high

To find the rate at which the length of his shadow increases

Let AB be the lamp post and let MN be the man of height 2m.

Let AL = l meter and MS be the shadow of the man

Let length of the shadow MS = s (as shown in the below figure)

Given man walks at the speed of 5 km/hr

So the rate at which the length of the man’s shadow increases will be

Consider ΔASB

Now consider ΔMSN, we get

So from equation(ii) and (iii),

Applying derivative with respect to time on both sides we get,

[from equation(i)]

Hence the rate at which the length of his shadow increases by 2.5 km/hr, and it is independent to the current distance of the man from the base of the light.

Given: a stone is dropped into a quiet lake and waves move in circles at a speed of 4 cm/sec.

To find the instant when the radius of the circular wave is 10 cm, how fast is the enclosed area increasing

Let r be the radius of the circle and A be the area of the circle

When stone is dropped into the lake waves moves in circle at speed of 4cm/sec.

i.e., radius of the circle increases at a rate of 4cm/sec

i.e.,

We know that

Area of the circle is

Now,

So when the radius of the circular wave is 10 cm, the above equation becomes,

Hence the enclosed area is increasing at the rate of 80 cm²/sec

Given: man 160cm tall walks away from a source of light situated at the top of a pole 6 m high, at the rate of 1.1m/sec

To find the rate at which the length of his shadow increases when he is 1m away from the pole

Let AB be the lamp post and let MN be the man of height 160cm or 1.6m.

Let AL = l meter and MS be the shadow of the man

Let length of the shadow MS = s (as shown in the below figure)

Given man walks at the speed of 1.1 m/sec

So the rate at which the length of the man’s shadow increases will be

Consider ΔASB

Now consider ΔMSN, we get

So from equation(ii) and (iii),

Applying derivative with respect to time on both sides we get,

[from equation(i)]

Hence the rate at which the length of his shadow increases by 0.4 m/sec, and it is independent to the current distance of the man from the base of the light.

Given: man 180cm tall walks at a rate of 2 m/sec away; from a source of light that is 9 m above the ground

To find the rate at which the length of his shadow increases when he is 3m away from the pole

Let AB be the lamp post and let MN be the man of height 180cm or 1.8m.

Let AL = l meter and MS be the shadow of the man

Let length of the shadow MS = s (as shown in the below figure)

Given man walks at the speed of 2m/sec

So the rate at which the length of the man’s shadow increases will be

Consider ΔASB

Now consider ΔMSN, we get

So from equation(ii) and (iii),

Applying derivative with respect to time on both sides we get,

[from equation(i)]

Hence the rate at which the length of his shadow increases by 0.6 m/sec, and it is independent to the current distance of the man from the base of the light.

Given: a ladder 13 m long leans against a wall. The foot of the ladder is pulled along the ground away from the wall, at the rate of 1.5m/sec

To find how fast is the angle θ between the ladder and the ground is changing when the foot of the ladder is 12 m away from the wall

Let AC be the position of the ladder initially, then AC = 13m.

DE be the position of the ladder after being pulled at the rate of 1.5m/sec, then DE = 13m as shown in the below figure.

So it is given that foot of the ladder is pulled along the ground away from the wall, at the rate of 1.5m/sec

……………(i)

Consider ΔABC, it is right angled triangle, so by applying the Pythagoras theorem, we get

AB² + BC² = AC²

⇒ y² + x² = h²……….(ii)

⇒ y² + (12)² = (13)²

⇒ y² = 169 - 144 = 25

⇒ y = 5

And in same triangle,

Now differentiate equation(ii) with respect to time, we get

Now substituting the values of x, y, h and (from equation(i)), we get

The value of h is always constant as the ladder is not increasing or decreasing in size, hence the above equation becomes,

And considering the same triangle,

Differentiating the above equation with respect to time we get

[applying quotient rule ]

Substituting the values of (from equation(iii)), x, y, (from equation(iv) and (from equation(i)) the above equation becomes,

Hence the angle θ between the ladder and the ground is changing at the rate of - 0.3 rad/sec when the foot of the ladder is 12 m away from the wall.

Given: a particle moves along the curve y = x² + 2x.

To find the points at which the curve are the x and y coordinates of the particle changing at the same rate

Equation of curve is y = x² + 2x

Differentiating the above equation with respect to x, we get

When x and y coordinates of the particle are changing at the same rate, we get

Now substitute the value from eqn(i), we get

2x + 2 = 1 ⇒ 2x = - 1

Substitute this value of x in the given equation of curve, we get

y = x² + 2x

Hence the points at which the curve are the x and y coordinates of the particle changing at the same rate is

Given: equation of curve y = 7x – x³ and x increases at the rate of 4 units per second.

To find how fast is the slope of the curve changing when x = 2

Equation of curve is y = 7x – x³

Differentiating the above equation with respect to x, we get slope of the curve

Let m be the slope of the given curve then the above equation becomes,

m = 7 - 3x²………(ii)

And it is given x increases at the rate of 4 units per second, so

Now differentiating the equation of slope i.e., equation(ii) we get

[by substituting the value of from equation (iii)]

When x = 2, equation(iv) becomes,

Hence the slope of the curve is changing at the rate of - 48 units/sec when x = 2

Given: a particle moves along the curve y = x³.

To find the points on the curve at which the y - coordinate changes three times more rapidly than the x - coordinate

Equation of curve is y = x³

Differentiating the above equation with respect to t, we get

When y - coordinate changes three times more rapidly than the x - coordinate, i.e.,

Equating equation (i) and equation (ii), we get

⇒ x² = 1 ⇒ x = ±1

When x = 1, y = x³ = (1)³⇒ y = 1

When x = - 1, y = x³ = ( - 1)³⇒ y = - 1

Hence the points on the curve at which the y - coordinate changes three times more rapidly than the x - coordinate are (1, 1) and ( - 1, - 1).

which increases twice as fast as its cosine.

As per the given condition,

θ = 2 cos θ

Now differentiating the above equation with respect to time we get

Hence,

So the value of angle θ which increases twice as fast as its cosine is

whose rate of increase twice is twice the rate of decrease of its consine

As per the given condition,

as the rate of increase is twice the rate of decrease, hence the minus sign.

Hence,

So the value of angle θ hose rate of increase twice is twice the rate of decrease of its consine is

Given: the top of a ladder 6 metre long is resting against a vertical wall on a level pavement, when the ladder begins to slide outwards, at the moment when the foot of the ladder is 4 metres from the wall, it is sliding away from the wall at the rate of 0.5 m/sec

To find how fast is the top - sliding downwards at this instance

Let AC be the position of the ladder initially, then AC = 6m.

DE be the position of the ladder after being pulled at the rate of 0.5m/sec, then DE = 6m as shown in the below figure.

So it is given that foot of the ladder is pulled along the ground away from the wall, at the rate of 0.5m/sec

……………(i)

Consider ΔABC, it is right angled triangle, so by applying the Pythagoras theorem, we get

AB² + BC² = AC²

⇒ y² + x² = h²……….(ii)

⇒ y² + (4)² = (6)²

⇒ y² = 36 - 16 = 20

⇒ y = 4.47

And in same triangle,

Now differentiate equation(ii) with respect to time, we get

Now substituting the values of x, y, h and (from equation(i)), we get

The value of h is always constant as the ladder is not increasing or decreasing in size, hence the above equation becomes,

And considering the same triangle,

Differentiating the above equation with respect to time we get

[applying quotient rule ]

Substituting the values of (from equation(iii)), x, y, (from equation(iv) and (from equation(i)) the above equation becomes,

Hence the angle θ between the ladder and the ground is changing at the rate of - 0.11 rad/sec when the foot of the ladder is 4 m away from the wall.

Given: a balloon in the form of a right circular cone surmounted by a hemisphere, having a diameter equal to the height of the cone, is being inflated

To find: how fast is its volume changing with respect to its total height h, when h = 9 cm

Solution:

Let height of the cone be h’

And the radius of the hemisphere be r

As per the given criteria,

Let the total height of the balloon be h

Then

…………(i)

So,

Volume of the balloon (V) = Volume of the cone + Volume of the hemisphere

This is the volume of the balloon

Now will substitute the value of h’ from equation (i), we get

Now differentiate the above equation with respect to h, we get

When h = 9cm, we get

Hence at the rate of 12 cm² the volume changes with respect to its total height.

Given: The water is running into an inverted cone at the rate of π cubic metres per minute. The height of the cone is 10 metres, and the radius of its base is 5 m.

To find how fast the water level is rising when the water stands 7.5 m below the base.

Let the height of the cone be H = 10m (given)

Let the radius of the base be R = 5m (given)

Let O’Y = r and CO’ = h

Now from the above figure,

ΔCOB~ΔCO’Q

So,

Let the volume of the water in the vessel at any time t be V

Then,

(from equation (i))

Now differentiate the above equation with respect to t, we get

But given the water is running at the rate of m³/min, i.e.,

So the above equation becomes

So when the water stands 7.5 m below the base

So h = 10 - 7.5 = 2.5m, the rate becomes

Hence the rate of water level rising when the water stands 7.5m below the base is 0.64 metres per min

Given: man 2m tall walks at a speed of 6km/h away; from a source of light that is 6 m above the ground

To find the rate at which the length of his shadow increases

Let AB be the lamp post and let MN be the man of height 2m.

Let AL = l meter and MS be the shadow of the man

Let length of the shadow MS = s (as shown in the below figure)

Given man walks at the speed of 6 km/h

So the rate at which the length of the man’s shadow increases will be

Consider ΔASB

Now consider ΔMSN, we get

So from equation(ii) and (iii),

Applying derivative with respect to time on both sides we get,

[from equation(i)]

Hence the rate at which the length of his shadow increases by 0.6 km/hr, and it is independent to the current distance of the man from the base of the light.

Given: the surface area of a spherical bubble is increasing at the rate of 0.2 cm²/sec

To find rate of increase of its volume, when the radius is 6cm

Let the radius of the given spherical bubble be r cm at any instant time.

It is given that the surface area of a spherical bubble is increasing at the rate of 0.2 cm²/sec

So the surface area of the bubble will be,

SA = 4r²

Now differentiating the above equation with respect to time we get

This is the rate of surface area increasing = 0.2cm²/sec, hence the rate at which the radius of the bubble is increasing becomes,

Then the volume of the spherical bubble at any time t will be

V = 4r³ cm³.

Applying derivative with respect to time on both sides we get,

[from equation(i)]

So when the radius is 6cm, the rate of volume will become,

Hence the rate of increase of its volume, when the radius is 6cm is 1.8 cm³/sec

Given: the radius of a cylinder is increasing at the rate 2 cm/sec and its altitude is decreasing at the rate of 3 cm/sec

To find the rate of change of volume when radius is 3 cm and altitude 5cm

Let V be the volume of the cylinder, r be its radius and h be its altitude at any instant of time ‘t’.

We know volume of the cylinder is

V = r²h

Differentiating this with respect to time we get

Now will apply the product rule of differentiation, i.e.,

, so the above equation becomes,

But given of a cylinder is increasing at the rate 2 cm/sec, i.e., and its altitude is decreasing at the rate of 3 cm/sec, i.e., , by subsitituting the above values in equation (i) we get

When radius of the cylinder, r = 3cm and its altitude, h = 5cm, the equation (ii) becomes,

Hence the rate of change of volume when radius is 3 cm and altitude 5cm is 33 cm³/sec

Let the inner radius be r, outer radius be R and volume be V of a hollow sphere at any instant of time

We know the volume of the hollow sphere is

Differentiating the volume with respect to time, we get

This is the rate of the volume of the hollow sphere and it is given this is constant hence

Given that the rate of increase in inner radius of the hollow sphere,, So the above equation becomes,

So when the radii are 4cm and 8 cm, the above equation becomes,

Therefore the rate of increase of the outer radius when the radii are 4 cm and 8cm respectively is 0.25 cm/sec

Let the volume be V, height be h and radius be r of the cone at any instant of time.

We know, volume of the cone is

And its given height of the cone is always one half of the radius of its base, i.e.,

So the new volume becomes

Differentiate the above equation with respect to time, we get

And it is also given that the sand is being poured onto a conical pile at the constant rate of 50 cm³/minute, so , so the above equation becomes,

Now when height of the pile is 5cm, the above equation becomes

Therefore, the rate of height of the pile increasing when the sand is 5 cm deep is cm/min.

Let P be the position of the kite and PR be the position of the string. Let QR = x and PR = y. Then from figure by applying Pythagoras theorem, we get

PR² = PQ² + QR²

⇒ y² = (120)² + x²……..(i)

Differentiating the above equation with respect to time, we get

Now the kite is moving away horizontally at the rate of 52m/sec, so , so the above equation becomes

…..(ii)

So when the string is 130m, y = 130m equation (i) become,

y² = x² + (120)²

⇒ (130)² = x² + 14400

⇒ x² = 16900 - 14400 = 2500

Therefore x = 50m

Substituting the value of x and y in equation (ii), we get

Hence the rate at which the string is being paid out is 20m/sec

Given: a particle moves along the curve .

To find the points on the curve at which the y - coordinate is changing twice as fast as the x - coordinate.

Equation of curve is

Differentiating the above equation with respect to t, we get

When y - coordinate is changing twice as fast as the x - coordinate, i.e.,

Equating equation (i) and equation (ii), we get

⇒ x² = 1 ⇒ x = ±1

When x = 1,

When x = - 1,

Hence the points on the curve at which the y - coordinate changes twice as fast as the x - coordinate are and

Given: a point on the curve y² = 8x.

To find the point on the curve at which the abscissa and ordinate change at the same rate

Equation of curve is y² = 8x

Differentiating the above equation with respect to t, we get

When abscissa and ordinate change at the same rate, i.e.,

Equating equation (i) and equation (ii), we get

⇒ y = 4

When y, y² = 8x ⇒ (4)² = 8x ⇒ x = 2

Hence the point on the curve at which the abscissa and ordinate change at the same rate is (2,4)

Let x be the edge of the cube, V be the volume of the cube at any instant of the time.

We know,

V = x³

Differentiating the above equation with respect to time, we get

But is is given that the volume of the cube is increasing at the rate of 9cm³/sec, so the above equation becomes,

We also know that the surface area of the cube is

S = 6x²

Again differentiating the above equation with respect to time is

Substitute equation (ii) in above equation, we get

When the edge is of length 10 cm, we get

Hence the rate at which the surface area increasing when the length of an edge is 10 cm is 3.6 cm²/sec

Given: the volume of a spherical balloon is increasing at the rate of 25 cm³/sec.

To find the rate of change of its surface area at the instant when the radius is 5 cm

Let the radius of the given spherical balloon be r cm, and V be its volume at any instant time.

Then according to the given criteria,

The rate of the volume of the spherical balloon is increasing is,

But volume of the spherical balloon is,

Applying derivative with respect to time on both sides we get,

Substituting the value from equation (i) in above equation, we get

Now the surface area of the spherical balloon at any time t will be

S = 4r² cm².

Applying derivative with respect to time on both sides we get,

Substituting the value from equation (ii), we get

So when the radius is 5cm, the rate of surface area will become,

Hence the rate of change of its surface area at the instant when the radius is 5 cm is 10cm²/sec

Given the length of the rectangle is x cm and width of the rectangle is y cm.

As per given criteria, length is decreasing at the rate of 5cm/min

And width is increasing at the rate of 4cm/min

(i) Let P be the perimeter of the rectangle

And we know,

P = 2(x + y)

Differentiating both sides with respect to t, we get

Substituting the values from equation (i) and (ii),we get

When x = 8 cm and y = 6 cm, the rates of change of the perimeter is - 2cm/min (it is decreasing in nature) and is independent on length and width of the rectangle.

(ii) Let A be the area of the rectangle

And we know,

A = xy

Differentiating both sides with respect to t, we get

Now will apply the product rule of differentiation, i.e.,

, so the above equation becomes

Substituting the values from equation (i) and (ii),we get

When x = 8 cm and y = 6 cm, the above equation becomes,

When x = 8 cm and y = 6 cm, the rates of change of the area is 2cm/min (it is increasing in nature) and is dependent on length and width of the rectangle

Let r be the radius of the circular disc and A be the area of the circular disc at any instant of time.

We know that, the area of the circle

Differentiating both sides with respect to t, we get

As per the given criteria, the circular disc expands on heating with the rate of change of radius is 0.05cm/s, i.e.,

Substituting this value in equation (i), we get

When the radius is 3.2 cm, the above equation becomes

So the rate at which its area is increasing when radius is 3.2 cm is 0.32 cm²/sec

CORRECTION: taking instead of

⇒

Substituting values of r=10 and , we get

The area of an equilateral triangle, with side a, is defined as

– (1)

Given that and a=10cm, we have to calculate

Differentiating (1) with respect to t, we get

Substituting the values, we get

=10√3 cm² /sec

The surface area of a sphere, of radius r, is defined by

A(r)=4πr² – (1)

Given that and r=200cm, we have to calculate

Differentiating (1) with respect to t, we get

Substituting values, we get

The volume of a cone, with height h and radius r, is defined by

- (1)

Given that h=2r, we get - (1)

and πr²=1m²=10000cm², we have to find out

Differentiating (1) with respect to t, we get

Substituting the values, we get

⇒

The volume of a cylinder, with radius r and height h, is defined by

V(r,h)=πr²h

Substituting r=0.5m, get

- (1)

Given that , we have to calculate

Differentiating (1) with respect to t, we get

Substituting values, we get

⇒

If we are given the distance travelled(x) as a function of time, we can calculate velocity(v) and acceleration(a) by

and a

x(t)= t³ – 12t² + 6t + 8

differentiating w.r.t. time, we get

Differentiating again w.r.t. time, we get

a

Given that a=0 ⇒ 6t-24=0 or t=4 units

v(4)=3×4² -24×4+6 =-42

The relation between height h, radius r and semi-vertical angle α is defined by

Given that h=20cm

-(1)

Given that α=30° and , we have to find

Differentiating (1) with respect to t, we get

Substituting values, we get

⇒

⇒

Let P(x)=x³-5x²+5x+8 – (1)

Given that , We have to calculate x

Differentiating (1) with respect to t, we get

3x²-10x+5=2

⇒ 3x²-10x+3=0

Factorizing the above quadratic equation, we get

(3x-1)(x-3)=0

⇒ or x=3

Taking ellipse to be 16x²+9y²=400 instead of 46x²+9y²=400

Let E(x,y):16x²+9y²=400

Solving for y, we get

- (1)

Given that , we have to calculate x,y

Differentiating (1) with respect to t, we get

Substituting values, we get

Simplifying the equation, we get

Squaring both sides

256x²=9(400-16x²)

Solving the equation, we get

x²=9

x=±3

Substituting in (1), we get

By seeing the graph of E(x,y), we can conclude that the point has to lie in I^st or III^rd quadrant, as only in these quadrants, the increase in abscissa leads to decrease in ordinate.

Hence the points are

The lateral surface area of a cone, with radius r and height h, is defined as

-(1)

Given that r=7cm, h=24cm, and , we have to calculate

Differentiating (1) with respect to t, we get

Substituting values, we get

Simplifying above equation, we get

The volume of a sphere, of radius r, is defined by

– (1)

Given that r=15cm, , we have to calculate

Differentiating (1) with respect to t, we get

Substituting values, we get

The volume of a sphere, of radius r, is defined by

– (1)

Given that r=2cm, , we have to calculate

Differentiating (1) with respect to t, we get

Substituting values, we get

⇒

If we are given the distance travelled(s) as a function of time, we can calculate velocity(v) by

s = 45t + 11t² – t³

Differentiating with respect to time, we get

If the particle is at rest, then its velocity will be 0

v(t)=45+22t-3t²=0

Solving the quadratic equation, we get

or t=9 sec

Since time can’t be negative therefore t=9secs.

The volume of a sphere, of radius r, is defined by

– (1)

Given that V=288π cm³,

Solving for r, we get r=6cm

Given that , we have to calculate

Differentiating (1) with respect to t, we get

Substituting values, we get

⇒

The volume of a sphere, of radius r, is defined by

Given that

We get

⇒ 4πr²=1

⇒ units

The area of a circle, of radius r, is defined by

A(r)=πr²

Given that

We get

⇒ πr=1

⇒ units

The area of an equilateral triangle, with side a, is defined as

– (1)

Given that and a=2cm, we have to calculate

Differentiating (1) with respect to t, we get

Substituting the values, we get

If we are given the distance travelled(s) as a function of time, we can calculate velocity(v) and acceleration(a) by

and a

s(t)= t³ – 4t² + 5

differentiating w.r.t. time, we get

Differentiating again w.r.t. time, we get

a

Given that a=0 ⇒ 6t-8=0 or units

If we are given the distance travelled(s) as a function of time, we can calculate velocity(v) and acceleration(a) by

and a

s(t)= 2t² + sin2t

differentiating w.r.t. time, we get

– (1)

Differentiating again w.r.t. time, we get

a

Given that a=2 ⇒ 4-4sin2t=2 or

⇒

The circumference of a circle, with radius r, is defined as

A(r)=πr² – (1)

Given that r=12cm, , we have to calculate

Differentiating (1) with respect to t, we get

The circumference of a circle, with radius r, is defined as

A(r)=πr² – (1)

Given that r=π cm, , we have to calculate

Differentiating (1) with respect to t, we get

Given that h_m=2m, h_l=5m, v_m=4.8km/hr=4800m/hr. We have to calculate v_s(speed of shadow).

After time t hrs, the man would have moved distance 4800t m away from the lamp. Let the shadow move a distance of x m in time t hrs.

Consider ∆AEC and ∆BED

∠AED=∠BED=θ (common angle)

∠EAC=∠EBD=90°

Therefore, ∆AEC~∆BED by AA criteria

Substituting values, we get

Simplifying the equation, we get

x=3200t – (1)

Differentiating (1) with respect to t, we get

v_s=3.2km/hr

Given that h_m=6ft, h_l=15ft, v_m=9ft/sec. We have to calculate v_s (speed of shadow).

After time t secs, the man would have moved distance 9t ft away from the lamp. Let the shadow move a distance of x ft in time t secs.

Consider ∆AEC and ∆BED

∠AED=∠BED=θ (common angle)

∠EAC=∠EBD=90°

Therefore, ∆AEC~∆BED by AA criteria

Substituting values, we get

Simplifying the equation, we get

x=6t – (1)

Differentiating (1) with respect to t, we get

The volume and surface area of a sphere, with radius r, is defined as

- (1) and A(r)=4πr²

Differentiating (1) with respect to t, we get

=(Surface area of a sphere)×(rate of change of radius)

The surface area of a sphere, with radius r, is defined as

A(r)=4πr² – (1)

Differentiating (1) with respect to t, we get

=8π×(current radius)×(rate of change of radius)

The volume of a cylinder, of radius r and height h, is defined as

V(r,h)=πr²h

Given that r=10m, we get

V(h)=π×10²×h=100πh – (1)

Given that , we have to calculate

Differentiating (1) with respect to t, we get

Substituting values and using π=3.14, we get

⇒

The Velocity(v) of the particle, if the distance travelled(s) is given as a function of time, is defined as .

Since s(t)= t³ –6t² + 9t + 8

Therefore

Substituting t=0 in the equation, we get

v_t=0 =9units/unit time

The volume of a sphere, with radius r, is defined as

- (1)

Given that and r=2cm, we have to calculate .

Differentiating (1) with respect to t, we get

Substituting the values, we get

The area of an equilateral triangle, with side a, is defined as

– (1)

Given that and a=10cm, we have to calculate

Differentiating (1) with respect to t, we get

Substituting the values, we get

The perimeter of a square, with side a, is defined as

P(a)=4a – (1)

Given that , we have to calculate

Differentiating (1) with respect to t, we get

Substituting the values, we get

The circumference of a circle, with radius r, is defined as

C(r)=2πr – (1)

Given that , we have to calculate

Differentiating (1) with respect to t, we get

The perimeter of an equilateral triangle, with side a, is defined as

P(a)=3a – (1)

Given that , we have to calculate

Differentiating (1) with respect to t, we get

Substituting the values, we get

The volume of a sphere, of radius r, and its surface area are defined by

and A(r)=4πr²

Given that

We get

⇒ 4πr²=A(r)=1unit²

The volume of a sphere, of radius r, is defined by

Given that

We get

⇒ 4πr²=1

⇒ units

Given that P(x) = 0.0005x³ + 0.02x² + 30x

The marginal increase in the pollution level is

On differentiating both sides, we get

P’(x)=0.0005×3x² +0.02×2x +30

= 0.0015x² + 0.04x +30

Substituting x=3, we get

P’(3) = 0.0015×9 + 0.04×3 +30 = 30.255 units

Given that . We have to find when h=2m

Also – (1)

At b=2m,

Differentiating (1) with respect to t, we get

Substituting values, we get

⇒