Application of Derivatives - Exemplar Solutions

 CBSE Class–12 Mathematics

NCERT Exemplar
Chapter - 6
APPLICATION OF DERIVATIVES - Short Answer Questions

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1. For the curve if increases at the rate of 2 units/sec, then how fast is the slope of curve changing when ?

Sol. To find slope of tangent to the given curve at different point, differentiate it w.r.t x 

        dydx=5−6x2$\frac{dy}{dx}=5-6x^2$

To find the rate of change of slope, differentiate the slope w.r.t. time(t)

⇒ddt(dydx)=−12x.dxdt⇒ddt(dydx)x=3=−12.(3).(2)=−72units/sec.$\begin{array}{rl}& \Rightarrow \frac{d}{dt}\left(\frac{dy}{dx}\right)=-12x.\frac{dx}{dt}\\ & \Rightarrow \frac{d}{dt}{\left(\frac{dy}{dx}\right)}_{x=3}=-12.(3).(2)\\ & =-72units/\sec .\end{array}$

Thus, slope of curve is decreasing at the rate of when is increasing at the rate of 

2. Water is dripping out from a conical funnel of semi-vertical angle at the uniform rate of in the surface area, through a tiny hole at the vertex of the bottom. When the slant height of cone is 4 cm, find the rate of decrease of the slant height of water.

Sol.If s represents the surface area, then

Also, on using trigonometric ratios, radius of cone can be taken as

r=lsinπ4$r=l\sin \frac{\pi }{4}$

Therefore, 

when 

Thus, rate of decrease of slant height of water is 2√4πcm/sec.$\frac{\sqrt{2}}{4\pi }cm/\sec .$

3. Find the angle of intersection of the curves 

Sol.Solving the given equations, we have 

Therefore, 

i.e. points of intersection are 

Further 

and 

At the slope of the tangent to the curve is parallel to y-axis and the tangent to the curve is parallel to x-axis.

angle of intersection 

At (1, 1) slope of the tangent to the curve is equal to and that of is 2.

4. Prove that the function is strictly decreasing on .

Sol.

When−π3<x<π3,1<secx<2$\frac{-\pi }{3}<x<\frac{\pi }{3},1<\sec x<2$

Therefore, 

Thus for 

Hence is strictly decreasing on .

5. Determine for which values of x, the function is increasing and for which values, it is decreasing.

Sol.

Now, 

Since and is continuous in and [0, 1].

Therefore, is decreasing in and is increasing in 

Note: Here is strictly decreasing in and is strictly increasing in .

6. Show that the function has neither maxima nor minima.

Sol.

Since for all and for all 

Hence is a point of inflexion i.e., neither a point of maxima nor a point of minima.

is the only critical point, and has neither maxima nor minima.

7. Using differentials, find the approximate value of 

Sol.Let 

Using f(x+Δx)=f(x)+Δx.f′(x),taking,x=0.09,andΔx=−0.008$f\left(x+\mathrm{\Delta }x\right)=f(x)+\mathrm{\Delta }x.f^{\prime }(x),taking,x=0.09,and\mathrm{\Delta }x=-0.008$

we get f(0.09−0.008)=f(0.09)+(−0.008).f′(0.09)$f\left(0.09-0.008\right)=f(0.09)+(-0.008).f^{\prime }(0.09)$

= 0.3 – 0.0133 = 0.2867.

8. Find the condition for the curves to intersect orthogonally.

Sol.Let the curves intersect at . Therefore,

slope of tangent at the point of intersection 

Again .

For orthoganality, 

9. Find all the point of local maxima and local minima of the function 

Sol.f′(x)=−3x3−24x2−45x$f^{\prime }(x)=-3x^3-24x^2-45x$

. Therefore, is point of local maxima

. Therefore, is point of local minima

. Therefore, is point of local maxima.

10. Show that the local maximum value of is less than local minimum value.

Sol.Let 

Hence local maximum value of is at and the local maximum value .

Local minimum value of is at and local minimum value .

Therefore, local maximum value is less than local minimum value .

 

Long Answer Questions

 

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11. Water is dripping out at a steady rate of through a tiny hole at the vertex of the conical vessel, whose axis is vertical. When the slant height of water in the vessel is 4 cm, find the rate of decrease of slant height, where the vertical angle of the conical vessel is 

Sol.Given that where is the volume of water in the conical vessel.

From the Fig.6.2, 

Therefore, 

 

Therefore, Rate of decrease of slant height, when slant height is 4 units. 

Therefore, the rate of decrease of slant height .

12. Find the equation of all the tangents to the curve y = cos (x + y), that are parallel to the line 

Sol.Given that 

Since tangent is parallel to therefore slope of tangent 

Therefore, 

Since and 

Therefore, 

Therefore, 

Thus, 

Hence, the points are 

Therefore, equation of tangent at and equation of tangent at or .

13. Find the angle of intersection of the curves and .

Sol.Given that and . Solving (i) and (ii), we get

Or 

Therefore, the points of intersection are (0, 0) and 

Again, 

Therefore, at (0, 0) the tangent to the curve is parallel to y-axis and tangent to the curve is parallel to x-axis.

Angle between curves 

At (slope of the tangent to the curve (i)) =12(ab)13$=\frac{1}{2}{\left(\frac{a}{b}\right)}^{\frac{1}{3}}$

(slope of the tangent to the curve (ii)) 

Therefore, 

Hence, 

14. Show that the equation of normal at any point on the curve is .

Sol.We have 

Therefore, dxdθ=−3sinθ+3cos2θsinθ=−3sinθ(1−cos2θ)=−3sin3θ$\frac{dx}{d\theta }=-3\sin \theta +3{\cos }^2\theta \sin \theta =-3\sin \theta (1-{\cos }^2\theta )=-3{\sin }^3\theta$

Hence the equation of normal is

Or 

15. Find the maximum and minimum values of

Sol.

Therefore, 

Therefore, possible values of are 

or 

Again, 

. We not that

Therefore, is a point of maxima.

. Therefore, is a point of maxima.

. Therefore, is a point of minima.

. Therefore, is a point of minima.

Maximum Value of y at x = 0 is 1 + 0 = 1

Maximum Value of y at x = is -1 + 0 = -1

Minimum Value of y at is 

Minimum Value of y at is 

16. Find the area of greatest rectangle that can be inscribed in an ellipse 

Sol.Let ABCD be the rectangle of maximum area with sides where is a point on the ellipse as shown in the Fig. 6.3.

The area A of the rectangle is which gives 

Therefore,

Again, 

Now, 

At 

Thus, at is maximum and hence the area A is maximum.

Maximum area 

17. Find the difference between the greatest and least values of the function 

Sol.

Therefore, 

Clearly, is the greatest value and is the least.
Therefore, difference 

18. An isosceles triangle of vertical angle 2is inscribed in a circle of radius a. Show
that the area of triangle is maximum when 

Sol.Let ABC be an isosceles triangle inscribed in the circle with radius a such that AB = AC.
and (see fig. 16.4)

Therefore, area of the triangle ABC i.e. 

Therefore, 

Therefore, 

Therefore, Area of triangle is maximum when 

 

Objective Questions

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Choose the correct answer from the given four options in each of the following Examples 19 to 23.

19. The abscissa of the point on the curve the normal at which passes through origin is:

(A) 1

(B) 

(C) 2

(D) 

Sol.Let be the point on the given curve at which the normal passes through the origin. Then we have Again the equation of the normal at passing through the origin gives 

Since satisfies the equation, therefore, Correct answer is (A).

20. The two curves 

(A) touch each other

(B) cut at right angle

(C) cut at an angle 

(D) cut at an angle

Sol.From first equation of the curve, we have 

say and second equation of the curve gives

say

Since, Therefore, correct answer is (B).

21. The tangent to the curve given by make with x-axis an angle:

(A) 0

(B) 

(C) 

(D) 

Sol.

Therefore, and hence the correct answer is (D).

22. The equation of the normal to the curve y = sinx at (0, 0) is:

(A) 

(B) 

(C) 

(D) 

Sol.Therefore, slope of normal Hence the equation of normal is 

Therefore, correct answer is (C).

23. The point on the curve , where the tangent makes an angle of with x-axis is

(A) 

(B) 

(C) (4, 2)

(D) (1, 1)

Sol.

Therefore, correct answer is B.

Fill in the blanks in each of the following Examples 24 to 29.

24. The values of for which touches the axis of are _______________ .

Sol.

Therefore, 

Hence, the values of are 

25. If , then its maximum value is _____________.

Sol.For to be maximum, should be minimum i.e. giving the minimum value of .

Hence, maximum value of .

26. Let have second derivative at such that then is a point of __________.

Sol.Local minima.

27. Minimum value of if ____________.

Sol.-1

28. The maximum value of is ____________.

Sol.

29. The rate of change of volume of a sphere with respect to its surface area, when the radius is 2 cm, is ____________.

Sol.