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**Q.1. A spherical ball of salt is dissolving in water in such a manner that the rate of decrease of the volume at any instant is proportional to the surface. Prove that the radius is decreasing at a constant rate.****Ans.**

Ball of salt is spherical

âˆ´ Volume of ball, where r = radius of the ball

As per the question, where S = surface area of the ball

â‡’[âˆµ S = 4pr^{2}]

â‡’

â‡’(K = Constant of proportionality)

â‡’

âˆ´

Hence, the radius of the ball is decreasing at constant rate.**Q.2. If the area of a circle increases at a uniform rate, then prove that perimeter varies inversely as the radius.****Ans.**

We know that:

Area of circle, A = Ï€r^{2}, where r = radius of the circle.

and perimeter = 2Ï€r

As per the question,

= K, where K = constant

â‡’

âˆ´...(1)

Now Perimeter c = 2Ï€r

Differentiating both sides w.r.t., t, we get

â‡’

â‡’[From (1)]

â‡’

Hence, the perimeter of the circle varies inversely as the radius of the circle.**Q.3. A kite is moving horizontally at a height of 151.5 meters. If the speed of kite is 10 m/s, how fast is the string being let out; when the kite is 250 m away from the boy who is flying the kite? The height of boy is 1.5 m.****Ans.**

Given that height of the kite (h) = 151.5 m

Speed of the kite(V) = 10 m/s

Let FD be the height of the kite and AB be the height of the boy.

Let AF = x m

âˆ´ BG = AF = x

m and

From the figure, we get that

GD = DF â€“ GF â‡’ DF - AB

= (151.5 â€“ 1.5) m = 150 m [âˆµ AB = GF]

Now in Î”BGD,

BG^{2} + GD^{2} = BD^{2} (By Pythagoras Theorem)

â‡’ x^{2} + (150)^{2} = (250)^{2}

â‡’ x^{2} + 22500 = 62500 â‡’ x^{2} = 62500 â€“ 22500

â‡’ x^{2} = 40000 â‡’ x = 200 m

Let initially the length of the string be y m

âˆ´ In Î”BGD

BG^{2} + GD^{2} = BD^{2} â‡’ x^{2} + (150)^{2} = y^{2}

Differentiating both sides w.r.t., t, we get

â‡’

â‡’ 2 x 200 x 10 = 2 x 250

âˆ´

Hence, the rate of change of the length of the string is 8 m/s.**Q.4. Two men A and B start with velocities v at the same time from the junction of two roads inclined at 45Â° to each other. If they travel by different roads, find the rate at which they are being seperated.****Ans.**

Let P be any point at which the two roads are inclined at an angle of 45Â°.

Two men A and B are moving along the roads PA and PB respectively with the same speed â€˜Vâ€™.

Let A and B be their final positions such that AB = y

âˆ APB = 45Â° and they move with the same speed.

âˆ´ âˆ APB is an isosceles triangle. Draw

AB = y âˆ´and PA = PB = x (let)

âˆ APQ = âˆ BPQ =

[âˆµ In an isosceles D, the altitude drawn from the vertex, bisects the base]

Now in right Î”APQ ,

â‡’

Differentiating both sides w.r.t, t, we get

Hence, the rate of their separation isunit/s.**Q.5. Find an angle Î¸, 0 < Î¸ <which increases twice as fast as its sine.****Ans.**

As per the given condition,

â‡’= 2 cosâ‡’ 1 = 2 cos Î¸

âˆ´ cos Î¸ =â‡’ cos Î¸ =

Hence, the required angle is**Q.6. Find the approximate value of (1.999) ^{5}.**

(1.999)

Let x = 2 and Î”x = â€“ 0.001

Let y = x

Differentiating both sides w.r.t, x, we get

= 5x

Now Î”y == 80 Ã— (- 0.001) = - 0.080

âˆ´ (1.999)^{5} = y + Dy

= x^{5} â€“ 0.080 = (2)^{5} â€“ 0.080 = 32 â€“ 0.080 = 31.92

Hence, approximate value of (1.999)^{5} is 31.92.**Q.7. Find the approximate volume of metal in a hollow spherical shell whose internal and external radii are 3 cm and 3.0005 cm, respectively.****Ans.**

Internal radius r = 3 cm

and external radius R = r + Î”r = 3.0005 cm

âˆ´ Î”r = 3.0005 â€“ 3 = 0.0005 cm

Let y = r^{3} â‡’ y + Î”y = (r + Î”r)^{3} = R^{3} = (3.0005)^{3} ...(i)

Differentiating both sides w.r.t., r, we get

= 3r^{2}

âˆ´ Î”y == 3r^{2 }x 0.0005

= 3 x (3)^{2} x 0.0005 = 27 x 0.0005 = 0.0135

âˆ´ (3.0005)^{3} = y + Î”y [From eq. (i)]

= (3)^{3} + 0.0135 = 27 + 0.0135 = 27.0135

Volume of the shell =

= 4Ï€ Ã— 0.005 = 4 x 3.14 x 0.0045 = 0.018 Ï€ cm^{3}

Hence, the approximate volume of the metal in the shell is 0.018 Ï€ cm^{3}.**Q.8. A man, 2m tall, walks at the rate oftowards a street light which is ****above the ground. At what rate is the tip of his shadow moving? At what ****rate is the length of the shadow changing when he is****from the base of ****the light?****Ans.**

Let AB is the height of street light post and CD is the height of the man such that

and CD 2 m

Let BC = x length (the distance of the man from the lamp post) and CE = y is the length of the shadow of the man at any instant.

From the figure, we see that

Î”ABE ~ Î”DCE [by AAA Similarity]

âˆ´ Taking ratio of their corresponding sides, we get

â‡’

â‡’ 8y = 3x + 3y â‡’ 8y - 3y = 3x â‡’ 5y = 3x

Differentiating both sides w.r.t, t, we get

â‡’

[âˆµ man is moving in opposite direction]

= - 1 m/s

Hence, the length of shadow is decreasing at the rate of 1 m/s.

Now let u = x + y

(u = distance of the tip of shadow from the light post)

Differentiating both sides w.r.t. t, we get

Hence, the tip of the shadow is moving at the rate of

towards the light post and the length of shadow decreasing at the rate of 1 m/s.**Q.9. A swimming pool is to be drained for cleaning. If L represents the number of litres of water in the pool t seconds after the pool has been plugged off to drain and L = 200 (10 â€“ t) ^{2}. How fast is the water running out at the end of 5 seconds? What is the average rate at which the water flows out during the first 5 seconds?**

Given that L = 200(10 â€“ t)

where L represents the number of litres of water in the pool.

Differentiating both sides w.r.t, t, we getâˆ´

= 200 x 2(10 - t) (-1) = - 400(10 - t)

But the rate at which the water is running out

= 400(10 - t)...(1)

Rate at which the water is running after 5 seconds

= 400 x (10 - 5) = 2000 L/s (final rate)

For initial rate put t = 0 = 400(10 - 0) = 4000 L/s

The average rate at which the water is running out

Hence, the required rate = 3000 L/s.

Let x be the length of the cube

âˆ´ Volume of the cube V = x

Given that

Differentiating Eq. (1) w.r.t. t, we get

âˆ´

Now surface area of the cube, S = 6x

Differentiating both sides w.r.t. t, we get

â‡’(4K = constant)

Hence, the surface area of the cube varies inversely as the length of the side.

Let area of the first square A

and area of the second square A

Now A

Differentiating both A

âˆ´

Hence, the rate of change of area of the second square with respect to first is

2x

The two circles intersect orthogonally if the angle between the tangents drawn to the two circles at the point of their intersection is 90Â°.

Equation of the two circles are given as

2x = y

and 2 = k ...(ii)

Differentiating eq. (i) and (ii) w.r.t. x, we get

(m

â‡’ 2xy = k

âˆ´

[m

If the two tangents are perpendicular to each other,

then m

â‡’

Now solving 2x = y

and 2xy = k [From (ii)]

From eq. (ii)

Putting the value of y in eq. (i)

â‡’ 8x

Hence, the required condition is k

Given circles are xy = 4 ...(i)

and x

Differentiating eq. (i) w.r.t., x

â‡’...(iii)

where, m

Differentiating eq. (ii) w.r.t. x

where, m

To find the point of contact of the two circles

Putting the value of y

x

âˆ´ x = Â± 2

âˆµ x

âˆ´ The point of contact of the two circles are (2, 2) and ( - 2, 2).

Q.14. Find the co-ordinates of the point on the curve at which tangent is equally inclined to the axes.

Equation of curve is given by

Let (x

âˆ´

Differentiating both sides w.r.t. x

Since the tangent to the given curve at (x

âˆ´ Slope of the tangent

So, from eq. (i) we get

Squaring both sides, we get

Putting the value of y

â‡’

Since y

âˆ´ y

Hence, the required point is (4, 4).

We know that the angle of intersection of two curves is equal to the angle between the tangents drawn to the curves at their point of intersection.

The given curves are y = 4 â€“ x

Differentiating eq. (i) and (ii) with respect to x, we have

m

and

m

So, m

Now solving eq. (i) and (ii) we get

â‡’ 4 â€“ x

So, m

Let Î¸ be the angle of intersection of two curves

âˆ´

âˆ´

Hence, the required angle is tan

Given that the equation of the two curves are y

and x

Differentiating (i) w.r.t. x, we get

Slope of the tangent at (1, 2), = 1

Differentiating (ii) w.r.t. x â‡’ 2x + 2y

â‡’

âˆ´ Slope of the tangent at the same point (1, 2)

â‡’

We see that m

Hence, the given circles touch each other at the same point (1, 2).

We have equation of the curve 3x

Differentiating both sides w.r.t. x, we get

â‡’

Slope of the tangent to the given curve =

âˆ´ Slope of the normal to the curve =

Now differentiating both sides the given line x + 3y = 4

â‡’

Since the normal to the curve is parallel to the given line x + 3y = 4.

âˆ´

Putting the value of y in 3x

3x

âˆ´ y = Â± 2

âˆ´ The points on the curve are (2, 2) and (â€“ 2, â€“ 2).

Now equation of the normal to the curve at (2, 2) is

â‡’ 3y â€“ 6 = â€“ x + 2 â‡’ x + 3y = 8

â‡’ 3y + 6 = â€“ x â€“ 2 â‡’ x + 3y = â€“ 8

Hence, the required equations are x + 3y = 8 and x + 3y = â€“ 8 or x + 3y = Â± 8.

Given that the equation of the curve is

x

Differentiating both sides w.r.t. x, we have

â‡’

Since the tangent to the curve is parallel to the y-axis.

âˆ´ Slope

So, from eq. (ii) we get

Now putting the value of y in eq. (i), we get

â‡’ x

â‡’ x

â‡’ x

â‡’ x(x â€“ 3) + 1(x â€“ 3) = 0 â‡’ (x - 3) (x + 1) = 0

â‡’ x = â€“ 1 or 3

Hence, the required points are (â€“ 1, 2) and (3, 2).

**at the point where the curve intersects the axis of y.****Ans.**

Given that y = b Ã— e^{â€“ x/a}, the equation of curve

andthe equation of line.

Let the coordinates of the point where the curve intersects the y-axis be (0, y_{1}) Now differentiating y = b Ã— e^{â€“ x/a} both sides w.r.t. x, we get

So, the slope of the tangent, m_{1 }=

Differentiating both sides w.r.t. x, we get

So, the slope of the line, m_{2 }=

If the line touches the curve, then m_{1} = m_{2}

â‡’

â‡’(Taking log on both sides)

â‡’â‡’ x = 0

Putting x = 0 in equation y = b Ã— e ^{â€“ x/a}

â‡’ y = b Ã— e^{0} = b

Hence, the given equation of curve intersect at (0, b) i.e. on y-axis.**Q.20. Show that f (x) = 2x + cot ^{â€“1}x + logis increasing in R.**

Given that f (x) = 2x + cot

Differentiating both sides w.r.t. x, we get

For increasing function, f â€²(x) â‰ 0

â‡’

Squaring both sides, we get 4x

â‡’ 4x

which is true for any value of x âˆˆ R.

Hence, the given function is an increasing function over R.

Given that: f (x) = âˆš3 sin x - cos x - 2ax + b, aâ‰¥ 1

Differentiating both sides w.r.t. x, we get

f â€²(x) = âˆš3 cos x + sin x- 2a

For decreasing function, f â€²(x) < 0

âˆ´ âˆš3 cos x + sin x- 2a < 0

â‡’

â‡’

â‡’

Since cos x âˆˆ [ - 1, 1] and a â‰¥ 1

âˆ´ f â€²(x) < 0

Hence, the given function is decreasing in R.

Given that: f (x) = tan

Differentiating both sides w.r.t. x, we get

For an increasing function f â€²(x) â‰¥ 0

âˆ´

â‡’ cos x â€“ sin x â‰¥ 0

â‡’ cos x â‰¥ sin x, which is true for

Hence, the given function f(x) is an increasing function in

Given that: y = â€“ x

Differentiating both sides w.r.t. x, we get= â€“ 3x

Let slope of the curve

âˆ´ z = -3x

Differentiating both sides w.r.t. x, we get= â€“ 6x + 6

For local maxima and local minima,= 0

âˆ´ â€“ 6x + 6 = 0 â‡’ x = 1

â‡’

Put x = 1 in equation of the curve y = (â€“ 1)

= â€“ 1 + 3 + 9 â€“ 27 = â€“ 16

Maximum slope = â€“ 3(1)

Hence, (1, â€“ 16) is the point at which the

slope of the given curve is maximum and maximum slope = 12.

We have: f (x) = sin x + âˆš3 cosx =

(Maxima)

Maximum value of the function at

Hence, the given function has maximum value atand the maximum value is 2.

Let Î”ABC be the right angled triangle in which âˆ B = 90Â°

Let AC = x, BC = y

âˆ´

âˆ ACB = Î¸

Let Z = x + y (given)

Now area of Î”ABC,A =

Squaring both sides, we get

Differentiating both sides w.r.t. y we get

...(i)

For local maxima and local minima,

â‡’ yZ â‰ 0 ( âˆµ y â‰ 0 and Z â‰ 0)

âˆ´ Z - 3y = 0

â‡’( âˆµ Z = x + y)

â‡’ 3y = x + y â‡’ 3y - y = x â‡’ 2y = x

â‡’

âˆ´

Differentiating eq. (i) w.r.t. y, we have

Hence, the area of the given triangle is maximum when the angle

between its hypotenuse and a side is

We have f (x) = x

â‡’ f â€²(x) = 5x

For local maxima and local minima, f â€²(x) = 0

â‡’ 5x

â‡’ 5x

âˆ´ x = 0, x = 1 and x = 3

Now f â€³(x) = 20x

â‡’ f â€³(x)

âˆ´ f (x) has the point of inflection at x = 0

f â€³(x)at x = 1 = 20(1)

= 20 - 60 + 30 = - 10 < 0 Maxima

f â€³(x)at x = 3 = 20(3)

= 540 - 540 + 90 = 90 > 0 Minima

The maximum value of the function at x = 1

f(x) = (1)

The minimum value at x = 3 is

f (x) = (3)

= 243 â€“ 405 + 135 â€“ 1 = 378 â€“ 406 = â€“ 28

Hence, the function has its maxima at x = 1 and the maximum value = 0 and it has minimum value at x = 3 and its minimum value is â€“ 28.

x = 0 is the point of inflection.

Let us consider that the company increases the annual subscription by â‚¹ x.

So, x is the number of subscribers who discontinue the services.

âˆ´ Total revenue, R(x) = (500 â€“ x) (300 + x)

= 150000 + 500x â€“ 300x â€“ x

= â€“ x

Differentiating both sides w.r.t. x, we get R â€²(x) = - 2x + 200

For local maxima and local minima, R â€²(x) = 0

- 2x + 200 = 0 â‡’ x = 100

R â€³(x) = - 2 < 0 Maxima

So, R(x) is maximum at x = 100

Hence, in order to get maximum profit, the company should increase its annual subscription by â‚¹ 100.

The given curve is ...(i)

and the straight line x cos Î± + y sin Î± = p ...(ii)

Differentiating eq. (i) w.r.t. x, we get

So the slope of the curve =

Now differentiating eq. (ii) w.r.t. x, we have

âˆ´

So, the slope of the straight line = â€“ cot Î±

If the line is the tangent to the curve, then

Now from eq. (ii) we have x cos Î± + y sin Î± = p

â‡’ a

â‡’

â‡’ a^{2} cos^{2} Î± + b^{2} sin^{2} Î± = p Ã— p

Hence, a^{2} cos^{2} Î± + b^{2 }sin^{2} Î± = p^{2}**Alternate method: **

We know that y = mx + c will touch the ellipse

Here equation of straight line is x cos Î± + y sin Î± = p and

that of ellipse is

x cos Î± + y sin Î± = p

â‡’ y sin Î± = â€“ x cos Î± + p

â‡’

Comparing with y = mx + c, we get

So, according to the condition, we get c^{2} = a^{2}m^{2 }+ b^{2}

â‡’

Hence, a^{2} cos^{2} Î± + b^{2} sin^{2} Î± = p^{2}

Hence proved.**Q.29. An open box with square base is to be made of a given quantity of card board of area c ^{2}. Show that the maximum volume of the box iscubic units.**

Let x be the length of the side of the square base of the cubical open box and y be its height.

âˆ´ Surface area of the open box

c

Now volume of the box, V = x x x x y

â‡’ V = x

Differentiating both sides w.r.t. x, we get

...(ii)

For local maxima and local minima,

â‡’

âˆ´

Now again differentiating eq. (ii) w.r.t. x, we get

Volume of the cubical box (V) = x

Hence, the maximum volume of the open box iscubic units.

Let x and y be the length and breadth of a given rectangle ABCD as per question, the rectangle be revolved about side AD which will make a cylinder with radius x and height y.

âˆ´ Volume of the cylinder V = Ï€r

â‡’ V = Ï€x

Now perimeter of rectangle P = 2(x + y) â‡’ 36 = 2(x + y)

â‡’ x + y = 18 â‡’ y = 18 â€“ x ...(ii)

Putting the value of y in eq. (i) we get

V = Ï€x

â‡’ V = Ï€(18x

Differentiating both sides w.r.t. x, we get

...(iii)

For local maxima and local minima

âˆ´ p(36x â€“ 3x

â‡’ 3x(12 â€“ x) = 0

â‡’ x â‰ 0 âˆ´ 12 â€“ x = 0 â‡’ x = 12

From eq. (ii) y = 18 â€“ 12 = 6

Differentiating eq. (iii) w.r.t. x, we get

at x = 12 = Ï€(36 â€“ 6 x 12)

= Ï€(36 - 72) = - 36Ï€ < 0 maxima

Now volume of the cylinder so formed = Ï€x

= Ï€ x (12)

Hence, the required dimensions are 12 cm and 6 cm and the maximum volume is 864Ï€ cm

Let x be the edge of the cube and r be the radius of the sphere.

Surface area of cube = 6x

and surface area of the sphere = 4Ï€r

âˆ´ 6x

Volume of the cube = x

âˆ´ Sum of their volumes (V) = Volume of cube + Volume of sphere

â‡’

â‡’

Differentiating both sides w.r.t. x, we get

âˆ´...(ii)

For local maxima and local minima,

âˆ´

â‡’

x â‰ 0 âˆ´ x -

Squaring both sides, we get

â‡’

âˆ´

Now putting the value of K in eq. (i), we get

6x

â‡’ 6x

âˆ´2r = x

âˆ´ x : 2r = 1 : 1

Now differentiating eq. (ii) w.r.t x, we have

Put

So it is minima.

Hence, the required ratio is 1 : 1 when the combined volume is minimum.**Q.32. AB is a diameter of a circle and C is any point on the circle. Show that the area of âˆ† ABC is maximum, when it is isosceles.****Ans.**

Let AB be the diameter and C be any point on the circle with radius r.

âˆ ACB = 90Â° [angle in the semi circle is 90Â°]

Let AC = x

âˆ´

â‡’...(i)

Now area of

â‡’

Squaring both sides, we get

Let A^{2} = Z

âˆ´

Differentiating both sides w.r.t. x, we get

...(ii)

For local maxima and local minima

âˆ´

x â‰ 0 âˆ´ 2r^{2} - x^{2} = 0

â‡’ x^{2} = 2r^{2} â‡’ x = âˆš2r = AC

Now from eq. (i) we have

So AC = BC

Hence, Î”ABC is an isosceles triangle.

Differentiating eq. (ii) w.r.t. x, we get

Put x = âˆš2r

âˆ´

Hence, the area of Î”ABC is maximum when it is an isosceles triangle.**Q.33. A metal box with a square base and vertical sides is to contain 1024 cm ^{3}. The material for the top and bottom costs Rs 5/cm^{2} and the material for the sides costs Rs 2.50/cm^{2} . Find the least cost of the box.**

Let x be the side of the square base and y be the length of the vertical sides.

Area of the base and bottom = 2x

âˆ´Cost of the material required = â‚¹ 5 x 2x

Area of the 4 sides = 4xy cm

âˆ´ Cost of the material for the four sides

= â‚¹ 2.50 x 4xy = â‚¹ 10xy

Total cost C = 10x

New volume of the box = x

â‡’ 1024 = x

âˆ´...(ii)

Putting the value of y in eq. (i) we get

Differentiating both sides w.r.t. x, we get

...(iii)

For local maxima and local minima

â‡’ 20x

Now from eq. (ii)

âˆ´ Cost of material used C = 10x

= 10 x 8 x 8 + 10 x 8 x 16 = 640 + 1280 = 1920

Now differentiating eq. (iii) we get

Put x = 8

Hence, the required cost is â‚¹1920 which is the minimum.

Let â€˜r â€™ be the radius of the sphere.

âˆ´ Surface area of the sphere = 4Ï€r

Volume of the sphere =

The sides of the parallelopiped are x, 2x and

âˆ´ Its surface area =

Volume of the parallelopiped =

As per the conditions of the question,

Surface area of the parallelopiped + Surface area of the sphere = constant

â‡’ 6x

âˆ´...(i)

Now let V = Volume of parallelopiped + Volume of the sphere

â‡’

â‡’[from eq. (i)]

â‡’

â‡’

â‡’

Differentiating both sides w.r.t. x, we have

For local maxima and local minima, we have

âˆ´

â‡’

â‡’

Here x â‰ 0 and

â‡’

Squaring both sides, we get

4Ï€x

â‡’ 4Ï€x

â‡’ ...(ii)

â‡’ 2x^{2}(2Ï€ + 27) = 9K

âˆ´

Now from eq. (i) we have

â‡’

â‡’

â‡’

Now we have

Differentiating both sides w.r.t. x, we get

Put

[ âˆµ 27 - 2Ï€ > 0]

so, it is minima.

Hence, the sum of volume is minimum for

âˆ´ Minimum volume,

Hence, the required minimum volume isand x = 3r.

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