mathematics-class-12-unit-06-chapter-10-derivatives-l-10-12-4ny-cquzikk

### <Success style="font-size:25px"> Derivatives L-10 </Success>
### <primary style="font-size:24px"> Rate of change of quantities </primary>
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- Let $t$ denotes the time and let $x$ and $y$ be two quantities depending on the time $t$. So, $x = x(t)$ & $y= y(t)$.  
- Suppose we are given $y$ as a function of $x$. Now the rate of changes of $x$ and $y$ are the derivatives $\frac{d x}{d t}$ and $\frac{d y}{d t}$ of $x$ and $y$ w.r.t. time $t$.

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<footer style = "font-size: 18px"> $\rightarrow$ Derivatives lecture-10 $\rightarrow$ <span style = "color:blue"> Rate of change of quantities </span> $\rightarrow$ Rate of change of quantities $\rightarrow$ Problem-1 </footer>

### <Success style="font-size:25px"> Derivatives L-10 </Success>
### <primary style="font-size:24px"> Problem-1 </primary>
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- Examples:

- (1) The radius of a circle is increasing at the rate of $3 \mathrm{cm} / \mathrm{sec}$. Find the rate of change of the area of the circle when the radius is $10 \mathrm{cm}$.

- Solution: Area of tha circle, $A=\pi r^2, \quad r=$ radius.

- Given: $\frac{d r}{d t}=3 \mathrm{cm} / \mathrm{sec}$.

- To find: $\frac{d A}{d t}$ when $r=10 \mathrm{cm}$.

- $\frac{d A}{d t}=\frac{d A}{d r} \cdot \frac{d r}{d t} =(2 \pi r) \frac{d r}{d t}=(2 \pi r) \times 3 \mathrm{cm} / \mathrm{s}$  
- $\therefore \quad \frac{d A}{d t}|_{\text {when} , r=10 \mathrm{cm}}=(2 \pi \times 10 \mathrm{cm}) \times 3 \mathrm{cm} / \mathrm{sec}$  
- $=60 \pi \mathrm{cm}^2 / \mathrm{sec}$.

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<footer style = "font-size: 18px">Rate of change of quantities $\rightarrow$ Rate of change of quantities $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Problem-2 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Derivatives L-10 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- To find: $\frac{d A}{d t}$ when $x=12 \mathrm{cm}$.

- Since $V=x^3, \quad \frac{d V}{d t}=\frac{d V}{d x} \cdot \frac{d x}{d t}=3 x^2 \frac{d x}{d t}$

- $ \Rightarrow \quad \frac{d x}{d t}=\frac{1}{3 x^2} \frac{d v}{d t}=\frac{1}{3 x^2} \times 8 \mathrm{cm}^3 / \mathrm{s}$

- $A=6 x^2 \Rightarrow \frac{d A}{d t}=\frac{d A}{d x} \cdot \frac{d x}{d t}=12 x \cdot \frac{1}{3 x^2} 8 \mathrm{cm}^3 / \mathrm{s}$

- $=\frac{32}{x} \mathrm{cm}^3 / \mathrm{s} \text {. }$

- $\therefore \quad \frac{d A}{d t} \text { when } x=12 \mathrm{cm}=\frac{32}{12cm}\cdot \mathrm{cm}^3 / \mathrm{s}$

- $=\frac{8}{3} \mathrm{cm}^2 / \mathrm{s}$.

- Thus the surface area is increasing at the rate of $\frac{8}{3} \mathrm{cm}^2 / \mathrm{s}$ when $x=12 \mathrm{cm}$.

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<footer style = "font-size: 18px">Problem-1 $\rightarrow$ Problem-2  $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-3 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Derivatives L-10 </Success>
### <primary style="font-size:24px"> Problem-3 </primary>
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- The length $x$ of a rectangle is decreasing at $5 \mathrm{cm} / \mathrm{min}$ and the width $y$ is increasing at $4 \mathrm{cm} / \mathrm{min}$. When $x=8 \mathrm{cm}$ and $y=6 \mathrm{cm}$, find the rates of change of

- (a) the perimeter and (b) the area of the rectangle.

- Sol: Given: $\frac{d x}{d t}=-5 \mathrm{cm} / \mathrm{min}$.

- $\frac{d y}{d t}=4 \mathrm{cm} / \mathrm{min}$.

- Let $P$ and $A$ denote the perimeter and the area of the rectangle, respectively.

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<footer style = "font-size: 18px">Problem-2 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-3 </span> $\rightarrow$ Solution $\rightarrow$ Problem-4 </footer>

### <Success style="font-size:25px"> Derivatives L-10 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- To find: $\frac{d P}{d t}$ and $\frac{d A}{d t}$ when $x=8 \mathrm{cm} ~ $ & $ ~ y=6 \mathrm{cm}$.
- We know $P=2(x+y)~ $ & $ ~ A=x y$.

- $\therefore \quad \frac{d P}{d t}=2\left(\frac{d x}{d t}+\frac{d y}{d t}\right)$  
- $=2(-5+4) \mathrm{cm} / \mathrm{min}$    
- $=-2 \mathrm{cm} / \mathrm{min}$.

- Thus the perimeter is decreasing at the rate of $2 \mathrm{cm} / \mathrm{min}$.

- $ \quad \frac{d A}{d t}=\frac{d x}{d t} y+x \frac{d y}{d t}=-5 y+4 x $

- $\therefore  \frac{d A}{d t}|_{x=8 , y=6}$
- $=-5 \times 6+4 \times 8=2 \mathrm{cm}^2 / \mathrm{min}$

- So, the area is increasing at $2 \mathrm{cm}^2 / \mathrm{min}$.

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-3 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-4 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Derivatives L-10 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Solution.  
- $V=\frac{4}{3} \pi r^3 ; r=$ radius of sphere  
- Given: $\frac{d V}{d t}=900 \mathrm{cm}^3 / \mathrm{s}$.

- To find: $\frac{d r}{d t}$ when $r=15 \mathrm{cm}$.  
- $V=\frac{4}{3} \pi r^3 \Rightarrow \frac{d V}{d t}=\frac{4}{3} \pi\left(3 r^2\right) \frac{d r}{d t}=4 \pi r^2 \frac{d r}{d t}$
- $\therefore \frac{d r}{d t}=\frac{1}{4 \pi r^2} \cdot \frac{d V}{d t}=\frac{1}{4 \pi r^2} \times 900 \mathrm{cm}^3 / \mathrm{s}$.

- When $r=15 \mathrm{cm}$.

- $\frac{d r}{d t} =\frac{1}{4 \pi(15 \mathrm{cm})^2} \times 900 \mathrm{cm^3} / \mathrm{s}$

- $=\frac{900}{4 \pi \times 15 \times15} \mathrm{cm} / \mathrm{s}=\frac{1}{\pi} \mathrm{cm} / \mathrm{s}$ .

- So, the radius is increasing at $\frac{1}{\pi} \mathrm{cm} / \mathrm{s}$, when $r=15 \mathrm{cm}$.

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-4 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem 5 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Derivatives L-10 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Solution.   
- Given: $\frac{d x}{d t}=2 \mathrm{cm} / \mathrm{s}$.

- To find: $\frac{d y}{d t}$ when $x=4 \mathrm{m}$.

- We have, $x^2+y^2=5^2$

- Differentiate w.r.t. $'t'$, we get  
- $2 x \frac{d x}{d t}+2 y \frac{d y}{d t}=0$  
- $\Rightarrow \frac{d y}{d t}=-\frac{x}{y} \frac{d x}{d t}$  
- when $x=4 \mathrm{m}, y=\sqrt{5^2-4^2}=3 \mathrm{m}$.  
- $\therefore$ When $x=4 \mathrm{m}, \frac{d y}{d t}=-\frac{4 m}{3 m} \cdot 2 \mathrm{cm} / \mathrm{s}=\frac{-8}{3} \mathrm{cm} / \mathrm{s}$   
- So, the height in decreasing at $\frac{8}{3} \mathrm{cm} / \mathrm{s}$.

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<footer style = "font-size: 18px"> Solution $\rightarrow$ Problem-5 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-6 $\rightarrow$ Application of rate of change in economics </footer>

### <Success style="font-size:25px"> Derivatives L-10 </Success>
### <primary style="font-size:24px"> Problem-6 </primary>
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- A particle moves along the curve $6 y=x^3+2$. Find the points on the curve at which the $y$-coordinate is changing 8 times as fast as the $x$-coordinate.

- Solution: To find: $(x, y)$ such that, $\frac{d y}{d t}=8 \frac{d x}{d t}$.         
- $6 y=x^3+2 \Rightarrow 6 \frac{d y}{d t}=3 x^2 \frac{d x}{d t}$           
- $\Rightarrow \frac{d y}{d t}=\frac{x^2}{2} \frac{d x}{d t}$           
- $\frac{d y}{d t}=8 \frac{d x}{d t}$  
- $\Rightarrow \frac{x^2}{2}=8$  
- $\Rightarrow x^2=16 \Rightarrow x= \pm 4$

- When $x=4, y=\frac{4^3+2}{6}=\frac{64+2}{6}=11$
- When $x=-4, y=\frac{(-4)^3+2}{6}=\frac{-62}{6}=\frac{-31}{3}$
- So, the required points are $(4,11)$ and $\left(-4,-\frac{31}{3}\right)$.

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<footer style = "font-size: 18px"> Problem-5 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-6 </span> $\rightarrow$ Application of rate of change in economics $\rightarrow$ Problem-7 </footer>

### <Success style="font-size:25px"> Derivatives L-10 </Success>
### <primary style="font-size:24px"> Application of rate of change in economics </primary>
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- $x$ : number of units of an item produced.             
- $C(x)$ : cost of producing $x$ units.             
- $R(x)$ : revenue obtained by selling $x$ units.             
- $\text { Profit }=R(x)-C(x) \text {. }$             
- Marginal cost, $\operatorname{MC}(x)$ is defined to be the rate of change of $C(x)$ w.r.t. $x$, i.e.             
- $M C(x)=\frac{d C}{d x}$

- Similarly, marginal revenue,   
- $M R(x)=\frac{d R}{d x}$.

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-6 $\rightarrow$ <span style = "color:blue"> Application of rate of change in economics </span> $\rightarrow$ Problem-7 $\rightarrow$ Problem-8 </footer>

### <Success style="font-size:25px"> Derivatives L-10 </Success>
### <primary style="font-size:24px"> Problem-7 </primary>
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- The total cost $C(x)$ in rupees for producing $x$ units of an item is given by

- $C(x)=0.007 x^3-0.003 x^2+15 x+4000$

- Find the marginal cost when 17 units are produced.

- Solution:  
- $M C(x)=\frac{d C}{d x}=0.021 x^2-0.006 x+15$

- $\therefore M C(17) =(0.021)(17)^2-0.006 \times 17+15$

- $=0.021 \times 289-0.102+15$

- $=20.967$.

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<footer style = "font-size: 18px">Problem-6 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-7 </span> $\rightarrow$ Problem-8 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Derivatives L-10 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Suppose we want to maximize the profit.

- Profit, $P(x)=R(x)-C(x)$

- $\therefore \quad P^{\prime}(x)=R^{\prime}(x)-C^{\prime}(x)=M R(x)-M C(x)$

- So, in order to maximize the profit the number of unit $x$ should be such that

- $ P^{\prime}(x)=0 \quad\text { i.e. } ~ M R(x)=M C(x)$

- So, by equating the marginal revenue and the marginal we get the values of $x$ for which the profit is maximum.

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<footer style = "font-size: 18px">Problem-7 $\rightarrow$ Problem-8 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Thank you $\rightarrow$  </footer>