Differentiation (Part-38) | S N De

 

$~16.~~$ In a certain culture the number of bacteria at any instant increases at a rate proportional to the cube root of the number present at that instant. If the number becomes $\,8\,$ times in $\,3\,$ hours, when the number will be $\,64\,$ times?

Sol. Let $\,x\,$ be the number of bacteria at a time $~t~$ so that by the given condition,

$~~~\frac{dx}{dt} \propto x^{1/3} \\ \therefore \frac{dx}{dt}=kx^{1/3} \\ \text{or,}~~\int{x^{-1/3}~dx}=k\int{dt} \\ \text{or,}~~ \frac{x^{-1/3+1}}{(-1/3)+1}=kt+c \\ \text{or,}~~\frac 32 x^{2/3}=kt+c \rightarrow(1)\\~~~[c\rightarrow \text{constant of integration.}]$ 

Let for $~t=0,~x=x_0~$ so that from $\,(1)\,$ we get,

$~~\frac 32(x_0)^{2/3}=k \times 0+c \\ \text{or,}~~c=\frac 32(x_0)^{2/3} \rightarrow(2)$

So, from $\,(1)\,$ and $\,(2)\,$ we get,

$~~\frac 32 \cdot x^{2/3}=kt+\frac 32\cdot (x_0)^{2/3} \rightarrow(3)$

Now, for $~~t=3,~~x=8x_0~$ we get from $\,(3)\,$

$~~\frac 32 (8x_0)^{2/3}=3k+\frac 32 \cdot x_0^{2/3} \\ \text{or,}~~ \frac 12\cdot 8^{2/3}x_0^{2/3}=k+\frac 12 \cdot x_0^{2/3} \\ \text{or,}~~ \frac 12 \cdot (2^3)^{2/3}x_0^{2/3}=k+\frac 12 \cdot x_0^{2/3}\\ \text{or,}~~2^2x_0^{2/3}=2k+x_0^{2/3}\\~~~~[\text{Multiplying both sides by 2}] \\ \text{or,}~~ 2k=4x_0^{2/3}-x_0^{2/3} \\ \text{or,}~~ k=\frac 32 x_0^{2/3} \rightarrow(4)$

Again, for $~~x=64x_0,~~t=T~\text{(say)}$ 

Hence from $\,(3)\,$ and $\,(4)\,$ we get,

$~~\frac 32 (64x_0)^{2/3}=\frac 32 x_0^{2/3}\cdot T+\frac 32 \cdot (x_0)^{2/3} \\ \text{or,}~~ (64)^{2/3} x_0^{2/3}=x_0^{2/3} \cdot T+x_0^{2/3} \\ \text{or,}~~ (2^6)^{2/3} \cdot x_0^{2/3}=x_0^{2/3} \cdot T+x_0^{2/3} \\ \text{or,}~~2^4=T+1\\~~~[\text{cancelling }~~x_0^{2/3}~~\text{from both sides}] \\ \text{or,}~~ T=2^4-1 \\ \text{or,}~~ T=16-1 \\ \text{or,}~~ T=15.$

Hence the number will be $\,64\,$ times in $\,15\,$ hours. 

$~17.~~$  A radioactive substance is subject to the law of natural decay $\,\frac{dv}{dt}=-kv~\,$ where $\,v\,$ is the volume of the substance at time $\,t\,$ and $\,k\,$ is a positive constant. If $\,40\%\,$ of the substance disappear in $\,25\,$ years, find the time it takes to disappear $\,60\%\,$ of the substance.

Sol. $~~~\frac{dv}{dt}=-kv~~~\text{(given)} \\ \text{or,}~~\frac{dv}{v}=-k~dt \\ \text{or,}~~ \displaystyle \int_{100}^{v'}{\frac{dv}{v}}=-k\int_0^t{dt}~~[*] \rightarrow(1)$

Note[*] : Initially when $~t=0,~~v=100~~$ and after a certain time $\,t,~~~v=v'~~\text{(say)}.$ 

Then from $\,(1)\,$ we get,

$~~\displaystyle\left[\log v\right]_{100}^{v'}=-kt \rightarrow(2)$

Since $\,40\%\,$ of the substance disappear in $\,25\,$ years, from $\,(2)\,$ we notice that for $\,t=25,~~v'=(100-40)\%=60\%.$

Hence from $\,(2)\,$ we get,

$~~\displaystyle \left[\log v\right]_{100}^{60}=-k\times 25 \\ \text{or,}~~\log( 60)-\log (100)=-25k \\ \text{or,}~~ -k=\frac{1}{25}[\log(60)-\log(100)]\\ \text{or,}~~ -k=\frac{1}{25}\left[\log \left(\frac{60}{100}\right)\right] \\ \text{or,}~~ -k=\frac{1}{25} \left[\log\left(\frac{6}{10}\right)\right]\rightarrow(3)$

Hence, from $\,(2)\,$ and $\,(3)\,$ we get,

$~~[\log v]_{100}^{v'}=\frac{1}{25}\left[\log\left(\frac{6}{10}\right)\right] \times t \rightarrow(4)$

 Let $\,T\,$ be the time it takes to disappear $\,60\%\,$ of the substance.

Then from $\,(4)\,$ we can say that when $~~t=T~\text{(say),}~~v'=(100-40)\%=60\%.$

So, from $\,(4)\,$ we can say,

$~~[\log(v)]_{100}^{40}=\frac{1}{25}\log\left(\frac{6}{10}\right) \times T \\ \text{or,}~~ [\log(40)-\log(100)]=\frac{T}{25} \times  \log\left(\frac{6}{10}\right)\\ \text{or,}~~\log\left(\frac{40}{100}\right)=\log\left(\frac{3}{5}\right) \times \frac{T}{25} \\ \text{or,}~~ \log\left(\frac{2}{5}\right)=\frac{T}{25} \times \log\left(\frac 35\right)\\ \text{or,}~~(\log 2 -\log 5)=\frac{T}{25} \times (\log 3-\log 5) \\ \text{or,}~~T=\frac{25(\log 2-\log 5)}{\log 3-\log 5} \\ \text{or,}~~T=\frac{25(\log 5-\log 2)}{\log 5-\log 3}~~\text{(ans.)}$

$\,18.~~$ Suppose the marginal cost of a product is given by rs. $~(10+24x-3x^2)$, where $\,x\,$ is the number of units produced. If the fixed cost is known to be rs. $\,40\,$, find the total cost function and the average cost function.

Sol. Marginal cost $~~\frac{dC}{dx}=10+24x-3x^2\\~~\text{where}~~C\rightarrow\text{Total cost function.} \\ \therefore dC=(10+24x-3x^2)~dx \\ \text{or,}~~\displaystyle\int{dC}=10\int{dx}+24\int{x~dx}\\~~~~~~-3\displaystyle\int{x^2~dx} \\ \text{or,}~~C=10x+24 \cdot\frac{x^2}{2}-3 \cdot\frac{x^3}{3}+k\\~~~[k\rightarrow\text{constant}] \\ \text{or,}~~C=10x+12x^2-x^3+k \rightarrow(1)$

Now, by the given problem, fixed cost $~C=40,~$ so that $~x=0.$

Then from $\,(1)\,$ we get, $~~40=k\rightarrow(2)$

Hence from $\,(1)\,$ and $\,(2)\,$ we get the total cost function $\,(C)\,$ which is given by : 

$~~~~C=10x+12x^2-x^3+40 \\ \text{or,}~~ C=40+10x+12x^2-x^3~~\text{(ans.)}$

Also, average cost function is : 

$~\frac{40+10x+12x^2-x^3}{x} \\=\frac{40}{x}+10+12x-x^2\\=10+12x-x^2+\frac{40}{x}~~\text{(ans.)}$

$~19.~~$ If the marginal revenue function is $\,(15-2x-x^2)\,$, find total revenue function and the demand function ($\,x\,$ being the number of units sold).

Sol. Let total revenue function be denoted by $~R$ so that marginal revenue function is :

$~~\frac{dR}{dx}=15-2x-x^2 \\ \therefore dR=(15-2x-x^2)~dx \\ \text{or,}~~\displaystyle\int{dR}=15 \int{dx}-2\int{x~dx}-\int{x^2~dx} \\ \text{or,}~~R=15x-2 \cdot\frac{x^2}{2}-\frac{x^3}{3}+k \rightarrow(1)\\~~~[k\rightarrow \text{constant of integration}]$

Clearly , for $~x=0,~~R=0.$

Hence, from $\,(1)\,$ we get, $~~k=0.$

Putting the value of $\,k\,$ in $\,(1)\,$ we get,

$~~~~R=15x-x^2-\frac{x^3}{3} \\ \therefore R=15x-x^2-\frac{x^3}{3}~~\text{(ans.)}$

If $~P(x)~$ be the demand function then according to the problem,

$~~x \cdot P(x)= 15x-x^2-\frac{x^3}{3} \\ \text{or,}~~ P(x)=\frac 1x\left(15x-x^2-\frac{x^3}{3} \right) \\ \text{or,}~~P(x)=15-x-\frac{x^2}{3}~~\text{(ans.)}$

$~20.~~$ The marginal cost function of manufacturing $\,x\,$ pairs of shoes is rs.$\,(6+10x-6x^2)\,$. The total cost of producing a pair of shoes is rs. $\,12~$. Find the total and average cost functions.

Sol. Let the total cost function be denoted by : $~C.$

So, according to the problem,

$~~\frac{dC}{dx}=6+10x-6x^2 \\ \therefore dC=(6+10x-6x^2)~dx \\ \text{or,}~~\displaystyle \int{dC}=6\int{dx}+10\int{x~dx}-6\int{x^2~dx} \\ \text{or,}~~ C=6x+10 \cdot\frac{x^2}{2}-6\cdot\frac{x^3}{3}+k~~[*] \\ \text{or,}~~ C=6x+5x^2-2x^3+k\rightarrow(1)$

For $~~x=1,~C=12~$ and so from $\,(1)\,$ we get,

$~~12=6\times 1+5\times 1^2-2\times 1^3+k \\ \text{or,}~~12=6+5-2+k \\ \text{or,}~~12=9+k \\ \text{or,}~~k=12-9 \\ \text{or,}~~k=3.$

So, putting the value of $~k~$, we get from $\,(1)\,$,

$~~~~C=6x+5x^2-2x^3+3 \\ \text{or,}~~C=3+6x+5x^2-2x^3~~\text{(ans.)}$

Again, the average cost function is :

$~~\frac Cx=\frac{3+6x+5x^2-2x^3}{x} \\~~~~~~=\frac 3x+6+5x-2x^2~~\text{(ans.)}$

Note[*] : $~k \rightarrow \text{constant of integration.}$

$~21.~~$ If $\,y\,$ is the total cost of $\,x\,$ units of output and it is given that marginal cost equals average cost, show that the average cost function is constant.

Sol. If $\,y\,$ is the total cost of $\,x\,$ units of output, then marginal cost is given by $~\frac{dy}{dx}.$

Again, average cost is : $~~\frac yx.$

Since it is given that marginal cost equals average cost,

$~~~\frac{dy}{dx}=\frac yx \\ \therefore \frac{dy}{y}=\frac{dx}{x} \\ \text{or,}~~ \displaystyle \int{\frac{dy}{y}}=\int{\frac{dx}{x}} \\ \text{or,}~~ \log y=\log x+\log c\\~~~~[\log c\rightarrow \text{constant of integration.}]\\ \text{or,}~~ \log y=\log(xc)\\ \text{or,}~~ y=cx \\ \text{or,}~~  \frac yx=c\rightarrow(1) $

From $\,(1)\,$, we can conclude that the average cost function is constant.