Calculus

Given: $\dfrac{dy}{dx} + y = 2$, which is of the form $\dfrac{dy}{dx} + Py = Q$ with $P = 1,\ Q = 2$.

Integrating factor: $\text{I.F.} = e^{\int P\,dx} = e^{\int 1\,dx} = e^{x}$

The solution is $y\,(\text{I.F.}) = \int Q\,(\text{I.F.})\,dx + C$ $y\,e^{x} = \int 2e^{x}\,dx + C = 2e^{x} + C$

Dividing by $e^{x}$, $\boxed{y = 2 + C e^{-x}}$

Given: $Q_d=32-0.5P,\qquad Q_s=-8+0.3P$ and $\frac{dP}{dt}=0.25(Q_d-Q_s)$

First, $Q_d-Q_s=(32-0.5P)-(-8+0.3P)=40-0.8P$

So, $\frac{dP}{dt}=0.25(40-0.8P)$ $\frac{dP}{dt}=10-0.2P$ $\frac{dP}{dt}+\frac{1}{5}P=10$

This is the required differential equation.

Now solve it. Integrating factor: $IF=e^{\int \frac15 dt}=e^{t/5}$

So, $P e^{t/5}=\int 10e^{t/5}\,dt$ $P e^{t/5}=50e^{t/5}+C$ $P=50+Ce^{-t/5}$

Taking integration constant $C=15$, $P=50+15e^{-t/5}$

When $t=5$, $P=50+15e^{-1}$ $P=50+\frac{15}{e}$ $P\approx 50+5.518=55.518$

Hence, $\boxed{P=50+15e^{-t/5}}$ and at $t=5$, $\boxed{P\approx \text{Rs }55.52}$

$\int_{0}^{1}(x^2+2x+1)\,dx$

$=\left[\frac{x^3}{3}+x^2+x\right]_{0}^{1}$

$=\left(\frac{1^3}{3}+1^2+1\right)-0$

$=\frac13+1+1$

$=\frac{1+3+3}{3}=\frac73$

$\boxed{\int_{0}^{1}(x^2+2x+1)\,dx=\frac73}$

Given supply function:

$P = 60 + 0.75Q$

i) Elasticity of supply at $P = Rs. 100$

First find quantity supplied at $P = 100$:

$100 = 60 + 0.75Q$

$0.75Q = 40$

$Q = \frac{40}{0.75} = 53.33 \text{ units}$

Now,

$P = 60 + 0.75Q$

$\frac{dP}{dQ} = 0.75$

Elasticity of supply:

$E_s = \frac{P}{Q} \times \frac{dQ}{dP}$

Since $\frac{dQ}{dP} = \frac{1}{0.75} = 1.333$,

$E_s = \frac{100}{53.33} \times 1.333$

$E_s = 1.875 \times 1.333 = 2.5$

$\boxed{E_s = 2.5}$

ii) Interpretation

Since $E_s = 2.5 > 1$, supply is relatively elastic.

This means a 1% increase in price leads to a 2.5% increase in quantity supplied. The quantity supplied is highly responsive to a change in price.

iii) Percentage change in quantity supplied when price increases by 15%

Using elasticity of supply:

$E_s = \frac{\%\Delta Q_s}{\%\Delta P}$

$2.5 = \frac{\%\Delta Q_s}{15}$

$\%\Delta Q_s = 2.5 \times 15 = 37.5\%$

$\boxed{\text{Percentage change in quantity supplied} = 37.5\%}$

So, when price rises by 15%, quantity supplied increases by 37.5%.

The condition for maximization of a function $f(x)$ is:

$f''(x) < 0 \quad \text{(second derivative is negative)}$

along with the first-order condition $f'(x) = 0$.

$f'(x) = 0$

A stationary point is a point on the curve where the tangent is parallel to the x-axis, i.e. the slope of the tangent is zero. Hence the necessary condition is the first derivative equal to zero, $f'(x) = 0$; the values of $x$ satisfying this give the stationary points.

Profit function: $P = R - C = (Q - 3Q^2) - (Q^3 + 2Q)$ $P = -Q^3 - 3Q^2 - Q$

For quantity $Q \ge 0$, $P = -Q(Q^2 + 3Q + 1) \le 0$

So profit is maximum at $Q = 0$.

$P_{\max} = -(0)^3 - 3(0)^2 - 0 = 0$

Maximum profit = Rs 0 (i.e. there is no positive profit for any positive output).

Given:

At $Q=3$,

Elasticity of demand:

Since $P=30-4Q$,

So,

The magnitude of elasticity is $1.5$, so demand is elastic at $Q=3$.

Given, revenue function: $R(x)=2x^3+6x+5$

Marginal revenue is the derivative of revenue: $MR=R'(x)=\frac{d}{dx}(2x^3+6x+5)=6x^2+6$

At $x=10$, $MR=6(10)^2+6=600+6=606$

Marginal revenue at $x=10$ is Rs 606.

Given curve: $y = 4x$

Using the fundamental theorem of calculus,

$A = \int_{2}^{5} 4x\,dx = \left[2x^2\right]_{2}^{5}$

$= 2(5)^2 - 2(2)^2 = 50 - 8 = 42$

Therefore, the required area is 42 square units.

Example:
$\frac{dy}{dx}+2y=6$ Here the coefficient of $y$ is $2$, a constant, and the constant term is $6$.

Solution:
The equation is of the form
$\frac{dy}{dx}+Py=Q$ where $P=2$ and $Q=6$, both constants.

Integrating factor: $\text{I.F.}=e^{\int P\,dx}=e^{\int 2\,dx}=e^{2x}$

Multiplying both sides by $e^{2x}$: $e^{2x}\frac{dy}{dx}+2e^{2x}y=6e^{2x}$ $\frac{d}{dx}\left(y e^{2x}\right)=6e^{2x}$

Integrating both sides: $y e^{2x}=\int 6e^{2x}\,dx$ $y e^{2x}=3e^{2x}+C$

Therefore, $y=3+Ce^{-2x}$

Answer:
$\boxed{y=3+Ce^{-2x}}$ where $C$ is an arbitrary constant.

Given: $f(x) = \dfrac{1}{x}$, defined on $(-8, 0)$.

Step 1: Find the first derivative.

$f'(x) = -\frac{1}{x^2}$

Step 2: Find the second derivative.

$f''(x) = \frac{d}{dx}\left(-x^{-2}\right) = 2x^{-3} = \frac{2}{x^3}$

Step 3: Test the sign of $f''(x)$ on $(-8, 0)$.

For every $x \in (-8, 0)$, we have $x < 0$, so $x^3 < 0$. Hence

$f''(x) = \frac{2}{x^3} < 0 \quad \text{for all } x \in (-8, 0)$

Conclusion: Since $f''(x) < 0$ throughout the interval, $f'(x)$ is decreasing in $(-8, 0)$.

Verification (optional): Take $x_1 = -4$ and $x_2 = -2$ with $x_1 < x_2$. Then

$f'(-4) = -\frac{1}{16} = -0.0625, \qquad f'(-2) = -\frac{1}{4} = -0.25$

Here $f'(-4) > f'(-2)$, i.e. $f'(x)$ falls as $x$ increases. Hence $f'(x)$ is decreasing in $(-8, 0)$.

Derivative and anti-derivative are inverse of each other.

Differentiation turns a function into its rate of change, while anti-derivative (integration) recovers the original function from that rate. So one operation undoes the other.

Example: Let $f(x) = x^2$.

Derivative: $\dfrac{d}{dx}(x^2) = 2x$

Now take the anti-derivative of $2x$:

$\int 2x\,dx = x^2 + c$

Taking $c = 0$, we get back $x^2$, the original function. Hence differentiation and anti-derivative reverse each other, i.e. they are inverse operations.

Let $f(x)=\dfrac1x$. By the definition (first principles), $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}.$

Cancelling $h\ne0$, $f'(x)=\lim_{h\to0}\frac{-1}{x(x+h)}=-\frac{1}{x^2}.$

$\boxed{\frac{d}{dx}\left(\frac1x\right)=-\frac{1}{x^2}}$

Given $x=2at$ and $y=at^2$, where $t$ is the parameter.

Differentiating with respect to $t$: $\frac{dx}{dt}=2a,\qquad \frac{dy}{dt}=2at$

Therefore, $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2at}{2a}=t$

$\boxed{\frac{dy}{dx}=t}$

(Check: eliminating $t=\frac{x}{2a}$ gives $y=\frac{x^2}{4a}$, so $\frac{dy}{dx}=\frac{x}{2a}=t$.)

Total revenue is the integral of marginal revenue: $R=\int MR\,dx=\int (x^3+4x-2)\,dx=\frac{x^4}{4}+2x^2-2x+C$

When output is zero, revenue is zero, so $R(0)=0$, which gives $C=0$.

$\boxed{R=\frac{x^4}{4}+2x^2-2x}$

(b) Demand is elastic

Because $|E_d| = 2.25 > 1$, a percentage change in price causes a larger percentage change in quantity demanded.

(d) $\frac{1}{x}$

Here $P = -\frac{1}{x}$, so the integrating factor is $e^{\int P\,dx} = e^{-\int \frac{1}{x}dx} = e^{-\log x} = \frac{1}{x}$.

$AC = \dfrac{C}{Q}$

Average cost is total cost divided by output, so $AC = \dfrac{C}{Q}$.

The area bounded by the curve $y = f(x)$, the $x$-axis, and the vertical lines $x = a$ and $x = b$.

Geometrically, the definite integral $\int_a^b f(x)\,dx$ gives the area under the curve $y = f(x)$ between $x = a$ and $x = b$ (areas below the $x$-axis being taken as negative).

The highest-order derivative present is $\dfrac{dy}{dx}$, so the order is 1. Its power is 1, so the degree is 1.

Revenue function $R = P \times Q$

If $P$ is the demand function (price expressed in terms of output $Q$), then total revenue is obtained by multiplying price by output, so the revenue function is $R = P \cdot Q$.

The formula for elasticity of demand is:

$E_d = \frac{\text{Percentage change in quantity demanded}}{\text{Percentage change in price}} = \frac{\%\Delta Q}{\%\Delta P}$

Equivalently, using the ratio form:

$E_d = \frac{\Delta Q}{\Delta P} \times \frac{P}{Q}$

where $P$ and $Q$ are the original price and quantity, and $\Delta P$, $\Delta Q$ are their respective changes.

At market equilibrium, demand price = supply price.

$196-x^2=x^2+4x+126$

$70=2x^2+4x$

$x^2+2x-35=0$

$(x+7)(x-5)=0$

Since quantity cannot be negative,
$x_0=5$

Equilibrium price: $P_0=196-(5)^2=196-25=171$

So equilibrium is at $(x_0,P_0)=(5,171)$.

Consumer surplus $CS=\int_0^{x_0} P_d\,dx-P_0x_0$

$=\int_0^5 (196-x^2)\,dx-171(5)$

$=\left[196x-\frac{x^3}{3}\right]_0^5-855$

$=\left(980-\frac{125}{3}\right)-855$

$=\frac{2815}{3}-855=\frac{250}{3}=83.33$

Producer surplus $PS=P_0x_0-\int_0^{x_0} P_s\,dx$

$=171(5)-\int_0^5 (x^2+4x+126)\,dx$

$=855-\left[\frac{x^3}{3}+2x^2+126x\right]_0^5$

$=855-\left(\frac{125}{3}+50+630\right)$

$=855-\frac{2165}{3}=\frac{400}{3}=133.33$

Therefore,

$\boxed{CS=\text{Rs }83.33,\qquad PS=\text{Rs }133.33}$

Given:
$\frac{dP}{dt}=0.15(6000-P),\qquad P=0 \text{ when } t=0$

a. Solve the differential equation

Separate the variables: $\frac{dP}{6000-P}=0.15\,dt$

Integrating both sides: $\int \frac{dP}{6000-P}=\int 0.15\,dt$

$-\ln(6000-P)=0.15t+C$

$\ln(6000-P)=-0.15t-C$

$6000-P=e^{-0.15t-C}=Ae^{-0.15t}$

where $A=e^{-C}$.

So, $P=6000-Ae^{-0.15t}$

Using $P=0$ when $t=0$: $0=6000-A$

$A=6000$

Hence the required solution is: $\boxed{P=6000\left(1-e^{-0.15t}\right)}$

b. Population after 10 years

Put $t=10$: $P=6000\left(1-e^{-0.15\times 10}\right)$

$P=6000\left(1-e^{-1.5}\right)$

Using $e^{-1.5}\approx 0.22313$, $P=6000(1-0.22313)$

$P=6000\times 0.77687$

$P\approx 4661.22$

$\boxed{P\approx 4661 \text{ people}}$

So, the population after 10 years is approximately 4661.

Given: Total length of fencing wire = perimeter = $144$ m.

Let the length of the rectangular garden be $x$ m and the breadth be $y$ m.

Step 1: Form the constraint (perimeter).

$2(x + y) = 144$ $x + y = 72 \quad \Rightarrow \quad y = 72 - x$

Step 2: Express the area as a function of one variable.

$A = xy = x(72 - x) = 72x - x^2$

Step 3: Apply the condition for maximum.

$\frac{dA}{dx} = 72 - 2x$

Put $\dfrac{dA}{dx} = 0$:

$72 - 2x = 0 \quad \Rightarrow \quad x = 36$

Step 4: Test for maximum (second derivative).

$\frac{d^2A}{dx^2} = -2 < 0$

Since the second derivative is negative, the area is maximum at $x = 36$.

Step 5: Find the other side.

$y = 72 - x = 72 - 36 = 36 \text{ m}$

Result: Length = breadth = $36$ m. So the garden must be a square of side $36$ m.

Justification: For a fixed perimeter, the rectangle of maximum area is always a square. Here the two adjacent sides come out equal ($36$ m each), which confirms the square shape. The maximum area is $36 \times 36 = 1296$ m².

The curve $y=x^3$ crosses the x-axis at $x=0$.
Given $y=8$, we have $x^3=8$, so $x=2$.
Thus the required limits are $x=-2$ to $x=2$.

Since $y=x^3$ is below the x-axis for $x<0$, take absolute value: