# Miscellaneous(Exercise 6)

Question 1: Using differentials; find the approximate value of each of the following.

(a)

(b)

Answer : (a) Let

Let and

Then,

Now, dy is approximately equal to and given by,

Hence, the approximate value of is

(b) Let

Let and

Then,

Now, dy is approximately equal to and is given by,

Hence, the approximate value of is

Question 2: Show that the function given by has maximum at .

Answer : The given function is

Now,

Now,

Now,

Therefore, by second derivative test, is the maximum at .

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

Answer : Let be isosceles where is the base of fixed length b. Let the length of the two equal sides of be ‘a’, draw

Now, in , by applying the Pythagoras theorem,

Area of triangle

The rate of change of the area with respect to time is given by,

It is given that the two equal sides of the triangle are decreasing at the rate of per second.

Then, when we have:

Hence, if the two equal sides are equal to the base, then the area of the triangle is decreasing at the rate of .

Question 4: Find the equation of the normal to curve at the point .

Answer : The equation of the given curve is ,

Differentiating with respect to , we have

Now, the slope of the normal at point is

Equation of the normal at

is i.e. .

Question 5: Show that the normal at any point to the curve
is at a constant distance from the origin.

Answer : We have

Since,

Now,

Slope of the normal=

The equation of the normal at a given point is given by,

Now, the perpendicular distance of the normal from the origin is constant.

, which is independent of .

Hence, the perpendicular distance of the normal from the origin is constant.

Question 6: Find the intervals in which the function given by is (i) Increasing (ii) Decreasing.

Answer :

Now or

But

Now, and divides into three disjoint intervals, i.e.

In intervals and .

Thus, is increasing for and .

In the interval .

Thus, is decreasing for .

Question 7: Find the intervals in which the function f given by is (i) increasing (ii) decreasing.

Answer :

Then, i.e.

Now, the points and divide the real line into three disjoint intervals i.e., and

In intervals and i.e., when and

Thus, when and is increasing.

In interval i.e., when

Thus, when is decreasing.

Question 8: Find the maximum area of an isosceles triangle inscribed in the ellipse with its vertex at one end of the major axis.

Answer : The given ellipse is .

Let the major axis be along the -axis.

Let be the triangle inscribed in the ellipse where
vertex is at .

Let the point

and

Area of

Squaring both side

Differntiate with respect to x

For max and minima

or

or

Since, is not possible

Again differentiate of eq (ii)

At

Hence area of triangle is max when

Putting in eq (i)

Max area

Question 9: A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is and volume is . If building of tank costs ₹70 per square meters for the base and ₹45 per square metre for sides. What is the cost of least expensive tank?

Answer : Let , and represent the length, breadth, and height of the tank respectively.

Then, we have height

Volume of the

Volume of the tank

Now, area of the base

Total cost

Differentiate with respect to ‘l’

For max and minima

Again differentiate with respect to l

At

Therefore cost of tank is min when

Putting in (i)

minimum cost

Question 10: The sum of the perimeter of a circle and square is , where is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

Answer : Let be the radius of the circle and be the side of the square. Then, we have:

where is constant

The sum of the areas of the circle and the square is given by,

Now .

Now, .

When

.

The sum of the area is least when

Hence, it has been proved that the sum of their areas is least when the side of the square is double the radius of the circle.

Question 11: A window is in the form of rectangle surmounted by a semicircular opening. The total perimeter of the window is . Find the dimensions of the window to admit maximum light through the whole opening

Answer: Let and be the length and breadth of the rectangular window.

Radius of the semicircular opening

It is given that the perimeter of the window is .

Area of the window is given by,

Now,

Thus, when then .

Therefore, by second derivative test, the area is the maximum when length .

Now,

Hence, the required dimensions of the window to admit maximum light is given by

Length and breath .

Question 12: A point on the hypotenuse of a triangle is at distance and from the sides of the triangle. Show that the minimum length of the hypotenuse is

Answer: Let be right-angled at . Let and .

Let be a point on the hypotenuse of the triangle such that is at a distance of and from the sides and respectively. Let

We have,

Now, and

It can be clearly shown that when

Therefore, by second derivative test, the length of the hypotenuse is minimum when .

Now, if , we have:

Hence, the minimum length of the hypotenuses is .

Question 13: Find the points at which the function given by has

(i) local maxima

(ii) local minima

(iii) point of inflexion.

Answer: The given function is

Now,

and or

Now, for values of close to and to the left of Also, for values of close to and to the right of .

Thus, is the point of local maxima.

Now, for values of close to 2 and to the left of Also, for values of close to 2 and to the right of .

Thus, is the point of local minima.

Now, as the value of varies through does not changes its sign.

Thus, is the point of inflexion.

Question 14: Find the absolute maximum and minimum values of the function given by

Answer:

Now, evaluating the value of at critical points and and at the end points of the interval (i.e., at and ),

we have:

Hence, the absolute maximum value of is occurring at and the absolute minimum value of is 1 occurring at and .

Question 15: Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius is .

Answer : Let be the centre and be the radius of the given sphere,

In ΔOBM, using pythagoras theorem

BM is the radius of cone

Volume of cone

Differentiating with respect to x

—(ii)

For max and minima

Hence or

or

Not possible

Agian differentiate with respect to x of eq (ii)

At

Hence volume of cone is max when

Height of cone

Question 16: Let be a function defined on such that , for all . Then prove that is an increasing function on .

Answer : We have to prove that function is always increasing, i.e.

for all Where

Let and be two numbers in the interval and .

Let the interval .

Function is continuous as well as differential in as it is continuous and differential in .

Using Mean value theorem, there exists , such that

Given that for all .

Therefore, for all .

Now, for the two point , where , we have .

Hence, the functions is increasing in the interval .

Question 17: Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius is . Also find the maximum volume.

Answer : Consider be the centre and be the radius of the given sphere,

In ΔOBM, using pythagoras theorem

BM is the radius of cone

Volume of cone

Differentiating with respect to x

For max and minima

Again differentiate of eq (ii)

At

Therefore volume of cylinde is max when

Height of cylinder

Question 18: Show that the height of the cylinder of greatest volume which can be inscribed in a right circular cone of height and having semi vertical angle is one third that of the cone and the greatest volume of the cylinder is

Solution: Let radius of cone = r and Radius of cylinder= x

Semivertical angle of cone and height of cone

In

Therefore, Height of cylinder

Volume of cylinder

Differentiating with respect to ‘x’

For max and minima

Again differentiate with respect to x

At

Therefore the volume of cylinder is max when

Now from eq (i)

Height of cylinder

Volume of cylinder

Volume of cylinder

Question 19: A cylindrical tank of radius is being filled with wheat at the rate of 314 cubic meter per hour. Then the depth of wheat is increasing at the rate of:

(A)

(B)

(C)

(D)

Solution: Option (A) is correct.

Consider be the depth of the wheat in the cylindrical tank whose radius is at time t.

Volume of wheat in cylindrical tank at time ,

So, cubic meter

We are given that cubic meter

So,

Therefore,

Question 20: The slope of the tangent to the curve at the point (2, is:

(A)

(B)

(C)

(D)

Solution: Option (B) is correct.

Solution: Equation of the curves are and and Slope of the tangent to the given curve at point is

At the given point and

At , from equation

At , From eq(2)

Here, common value of in the two sets of values is 2 .
Again, from equation (3),

Slope of the tangent to the given curve at point

Question 21: The line is a tangent to the curve if the value of is:

(A) 1

(B) 2

(C) 3

(D)

Solution: Option (A) is correct.

Equation of the curve is

Slope of the tangent to the given curve at point

Now

Putting the values of and in equation (1),

Question 22: The normal at the point on the curve is:

(A)

(B)

(C)

(D)

Solution: Option (B) is correct.

Equation of the given curve is

Slope of the tangent to the given curve at point is

Equation of the normal at is

Question 23: The normal to the curve passing through is:

(A)

(B)

(C)

(D)

Solution: Option is correct.

Equation of the curve is

Slope of the normal at is

Again slope of normal at given point

From equation (2) and (3), we have

From equation (1),

Now, at point , slope of the normal from equation

Equation of the normal is

or

Question 24: The points on the curve where the normal to the curve make equal intercepts with axes are:

(A)

(B)

(C)

(D)

Solution: Option (A) is correct.

Equation of the curve is

Slope of the tangent to curve (1) at any point is

Slope of the normal .

Slopes of lines with equal intercepts on the axis are

So

if

From (i) and (ii) we have

and

If

we have, and

Required points are