RD Sharma Class 12 Exercise 12.1 Derivative as a rate measure Solutions Maths - Download PDF Free Online

Derivative as a rate measure Exercise: 12.1

Derivative as a rate measure exercise 12.1 question 1

Answer:
$4\pi r+2\pi h$
Hint:
To find the rate of change of the total surface area of the cylinder, we need to differentiate it with respect to the radius
Total surface area of cylinder,$A=2\pi r(r+h)$
Given:
$radius(r)andheight(h)ofcylinder.$
Solution:
Here we have to find rate of change of the total surface area of cylinder. here radius=$r$, height=$h$
so, total surface area of cylinder.
$A=2\pi r(r+h)$
$\therefore A=2\pi r^2+2\pi rh....(i)$
Let’s differentiate equation (i) with respect to radius
$\therefore \frac{dA}{dr}=\frac{d}{dr}(2\pi r^2+2\pi rh)$
$\frac{dA}{dr}=\frac{d}{dr}\left(2\pi r^2\right)+\frac{d}{dr}(2\pi rh)$
$\therefore \frac{dA}{dr}=4\pi r+2\pi h$ $[\because \frac{d(x^n)}{dx}=n{x}^{n-1}]$
Which is the rate of change of total surface area with respect to radius.
Note:
Here we have taken rate with respect to radius because radius was varying.

Derivative as a rate measure exercise 12.1 question 2

Let’s differentiate equation (i) with respect to diameter,

$\therefore \frac{dV}{dD}=\frac{d}{dD}\left(\frac{\pi D^3}{6}\right)$

$\therefore \frac{dV}{dD}=\frac{\pi D^2}{2}$ $[\because \frac{d\left(x^n\right)}{dx}=n{x}^{n-1}]$
This is the rate of change of the volume of sphere with respect to diameter.
Note:
Here answer is with respect to diameter but with respect to radius, rate of change of volume of sphere is
$\therefore \frac{dV}{dR}=4\pi R^2$

Derivative as a rate measure exercise 12.1 question 3

Let’s differentiate with respect to Area (A)

$\Rightarrow \frac{dV}{dA}=4\pi r^2\frac{dr}{dA}$

$\Rightarrow \frac{dV}{dA}=\frac{4\pi r^2}{8\pi r}$ {From the Equation (i)}

$\therefore \frac{dV}{dA}=\frac{r}{2}$
To putting value of $r=2cm$
$\therefore \frac{dV}{dA}=1cm$
Note:
We cannot put Area directly in formula of volume,
$V=\frac{4}{3}\pi r^2$
$So,wecannotwrite,V=\frac{A.r}{3}$

Derivative as a rate measure exercise 12.1 question 4

Answer: 3cm
Hint:
We have to differentiate area of circular disc ($A$) with respect to its circumference ($L$)
Area of circular disc, $A=\pi r^2$
Circumference of circular disc, $L=2\pi r$
Given:
Radius of a circular disc, $r=3cm$
Solution:
Here we have,
Radius of disc, $r=3cm$
Circumference of disc, $L=2\pi r$
Let’s differentiate with the circumference with respect to radius $r$
$\frac{dL}{dr}=2\pi$
$\therefore \frac{dr}{dL}=\frac{1}{2\pi }.....(i)$ $[\because \frac{d\left(x^n\right)}{dx}=n{x}^{n-1}]$
Area of circular disc, $A=\pi r^2$.
Let’s differentiate with the area with respect to $L$
$\therefore \frac{dA}{dL}=2\pi r\frac{dr}{dL}$
$\therefore \frac{dA}{dL}=\frac{2\pi r}{2\pi }$ {From eq.(i)}
$\therefore \frac{dA}{dL}=r$
Putting value of $r=3cm$$r=3cm$
Note:
Here we cannot write directly,
$A=\frac{L}{2}r$ It's wrong

Derivative as a rate measure exercise 12.1 question 5

$\frac{2}{3}\pi rh$
Hint:
Here we have to differentiate volume of cone ($V$) with respect to radius of base ($r$)
$Volumeofcone,V=\frac{1}{3}\pi r^{2}h$
Where $h$ is height of cone.
Given:
Volume of cone.
Solution:
Here we have,
$r$ = Radius of base of cone
$h$ = Height of cone
$Volumeofcone,V=\frac{1}{3}\pi r^{2}h$
Let’s differentiate $V$with respect to $r$,
$\therefore \frac{dV}{dr}=\frac{2}{3}\pi rh$ $[\because \frac{d\left(x^n\right)}{dx}=n{x}^{n-1}]$
Note:
Here height ($h$) is constant so we don’t have to differentiate it.

Derivative as a rate measure exercise 12.1 question 6

Answer:
$10\pi c{m}^2/cm$
Hint:
Here we have to differentiate Area of circle ($A$) with respect to its radius ($r$)
Area of circle, $A=\pi r^2$
Given:
Radius of circle, $r=5cm$
Solution:
Here we have,
$r$ = Radius of circle = $5cm$
Area of circle, $A=\pi r^2$
Let’s differentiate $A$ with respect to radius
$\therefore \frac{dA}{dr}=\frac{d}{dr}\left(\pi r^2\right)$
$=2\pi r$ $[\because \frac{d\left(x^n\right)}{dx}=n{x}^{n-1}]$
Put, $r=5cm$
$\therefore \frac{dA}{dr}=10\pi c{m}^2/cm$
Note:
Here unit of rate is cm²/cm we can’t write unit cm because it shows rate means
Area ($A$) varying with respect to radius ($r$)

Derivative as a rate measure exercise 12.1 question 7

Answer:
$4\pi r^2,16\pi c{m}^3/cm$
Hint:
Here the volume of ball is the volume of sphere ($V$)
$Thevolumeoftheball=thevolumeofsphereV=\frac{4}{3}\pi r^3$
And we have to differentiate $V$ with respect to radius ($r$)
Given:
Radius of ball $=r=2cm$
Solution:
Here we have,
Radius of ball $=r=2cm$
$Volumeofball=V=\frac{4}{3}\pi r^3$
Let’s differentiate $V$ with respect to $r$
$\therefore \frac{dV}{dr}=\frac{d}{dr}\left(\frac{4}{3}\pi r^3\right)$ $[\because \frac{d\left(x^n\right)}{dx}=n{x}^{n-1}]$
$\therefore \frac{dV}{dr}=4\pi r^2.....(i)$
Equation ($i$) is rate of change of volume of ball with respect to radius.
Let's put. $r=2cm$
$\therefore \frac{dV}{dr}=4\pi 2^2$
$\therefore \frac{dV}{dr}=16\pi c{m}^3/cm$
Note:
It we take $r=2cm$
The correct answer would be
$\frac{dV}{dr}=16\pi c{m}^3/cm$
And we can’t write the unit of given rate cm² because it represents volume ($V$) varying with respect to radius ($r$)

Derivative as a rate measure exercise 12.1 question 8

So $20.967Rs$ is the marginal cost when 17 units are produced.

Derivative as a rate measure exercise 12.1 question 9

Answer: Rs 208
Hint:
To find marginal revenue we have to differentiate total revenue R($x$) with respect to units $x$
Given:
$R(x)=13x^2+26x+15$
Solution:
Here we have,
Units, $x=7$
$TotalrevenueR(x)=13x^2+26x+15$
Let’s differentiate R($x$) with respect to $x$
$\therefore \frac{dR(x)}{dx}=R^{\prime }(x)=\frac{d}{dx}(13x^2+26x+15)$
$\therefore R^{\prime }(x)=26x+26$ $[\because \frac{d\left(x^n\right)}{dx}=n{x}^{n-1}]$
Let's put $x=7$
$\therefore R^{\prime }(7)=26(7)+26$
$\therefore R^{\prime }(7)=208$
So, the marginal revenue for $x=7$ is 208.

Derivative as a rate measure exercise 12.1 question 10

Answer:
$MR=Rs66$. It indicates the extra money spent when number of employees increase from 5 to 6
Given:
$R(x)=3x^2+36x+5$
Solution:
Here we have,
Total revenue
$R(x)=3x^2+36x+5$
To find marginal revenue $MR$ let’s differentiate R($x$) with respect to $x$
$\therefore \frac{d}{dx}R(x)=MR=6x+36$ $[\because \frac{d\left(x^n\right)}{dx}=n{x}^{n-1}]$
To find $MR$
Put $x=5$ in $MR$
$\therefore MR=6(5)+36$
$\therefore MR=66Rs$
Here we can write
$MR=\underset{h\to 1}{lim}\frac{R(x+h)-R(x)}{h}=66$
So, we can say thet in first there were 5 employees but after they become 6 employees.
Note:
Here we have taken
$MR=\underset{h\to 1}{lim}\frac{R(x+h)-R(x)}{h}$
$Not\underset{h\to 0}{lim}becausehrepresents$