Learn marginal cost function as prediction of actual cost, practice examples of related rates and particle motion in calculus, ladder problem calculus.

If \(\Delta Q\) is the net charge that passes through a surface during time \(\Delta t\) then current during this time interval is defined as

Average current = \(\dfrac {\Delta Q}{\Delta t}=\dfrac {Q_2-Q_1}{t_2-t_1}\)

If \(\Delta t\rightarrow0\) we get the equation

\(I=\dfrac {d Q}{d t}=\lim\limits_{\Delta t\rightarrow 0}\dfrac {\Delta Q}{\Delta t}\)

\(I\) is called the current flowing through surface.

Let C(x) be the total cost that a company incurs when it produces x units of any commodity, then C(x) is called the cost function in commodities.

- C(x) = Total cost of producing x units of commodity.
- If the number of commodity items increase from x
_{1 }to x_{2}then additional cost

\(\Delta C=C(x_2)-C(x_1)\)

- Average rate of change of cost = \(\dfrac {\Delta C}{\Delta x}=\dfrac {C(x_2)-C(x_1)}{x_2-x_1}\)

\(=\dfrac {C(x_1+\Delta x)-C(x_1)}{\Delta x}\)

- If \(\Delta x\rightarrow 0\), then

- \(\lim\limits_{\Delta x\rightarrow 0}\dfrac {\Delta C}{\Delta x}=\dfrac {dC}{dx}=\)Marginal cost = C'(x)

- \(\dfrac {dC}{dx}=\) approximately the cost of producing \((x+1)^{th}\) item.

- \(\dfrac {dC}{dx}\simeq C(x+1)-C(x)\)

A 3 + 0.2 x

B 7 + 0.1 x

C 45 + 8x

D 84 + 0.7 x

Suppose a factory manufactures certain item, there it is often important for the owner of the factory to know what will be the incremental cost of producing an extra item of a particular level of production.

Let C(x) = Total cost of producing 'x' units

C (x + 1) = Total cost of producing (x + 1) units

\(\therefore\) Actual cost of producing (x + 1 )^{th} item

= C(x + 1) – C(x)

- Marginal cost gives the rate at which cost are increasing w.r.t. production levels.

Marginal Cost = \(\dfrac {dC}{dx}=C'(x)\)

- It is a good approximation to cost of producing (x+1)
^{th}unit.

- In these problem we are given (or we know) the relation between two quantities. We differentiate both sides with respect to time to get the rate of change of one with respect to rate of change of other.
- Suppose x and y are connected by the equation x
^{2}+ y^{2}= 2, and we differentiate both sides with respect to another variable say 't', then

\(2x\dfrac {dx}{dt}+2y\dfrac {dy}{dt}=0\)

\(\Rightarrow \)\(x\dfrac {dx}{dt}+y\dfrac {dy}{dt}=0\)

Now, if out of 4 quantities x, y, \(\dfrac {dx}{dt}\) and \(\dfrac {dy}{dt}\), three are known then the fourth can be found.

A \(\dfrac {-3\sqrt 2}{2}\)

B \(\dfrac {7\sqrt 2}{5}\)

C \(\dfrac {-14}{\sqrt 2}\)

D \(5\sqrt 5\)

- Suppose a particle is moving on a fixed curve whose equation is given to us.
- Say \(f(x, y) = 0\). If we differentiate both sides w.r.t. time (t) then we obtain a relation between \(\dfrac {dx}{dt}\) and \(\dfrac {dy}{dt}\).
- So if rate of change of one of the coordinates (x or y) is given then the rate of change of other can be found.

e.g.

Let \(x^2y=2\) be a curve, then

\(x^2\;\dfrac {dy}{dt}+2x\;\dfrac {dx}{dt}\,y=0\)

Suppose \(\dfrac {dx}{dt}=5\,cm/sec.\) when x = 1 then,

\(1×\;\dfrac {dy}{dt}+2×1×\;\dfrac {dx}{dt}\,×2=0\) (when x = 1, y = 2)

\(\Rightarrow\,\dfrac {dy}{dt}+4\;\dfrac {dx}{dt}\,=0\)

\(\Rightarrow\,\dfrac {dy}{dt}=-4×5=-20\,cm/sec\)

A \(\dfrac {70}{71}\;cm/min\)

B \(\dfrac {-14}{3}\;cm/min\)

C \(5\;cm/min\)

D \(\dfrac {-15}{4}\;cm/min\)

- Suppose there are two particles which move in directions which are \(\perp\) to each other, then if the speed of both are given in the direction in which they are moving, we can find the rate at which the distance between them is increasing at a particular time.

- Suppose a particle x is moving westwards and y is traveling northwards as shown. Both move towards intersection of road (Point P).

Let 'C' be the distance between them at any time 't', then

\(c^2=a^2+b^2\) (By Pythagoras Theorem)

\(\therefore\) \(2c\, \dfrac {dc}{dt}=2a\, \dfrac {da}{dt}+2b\, \dfrac {db}{dt}\)

\(\Rightarrow\)\(c\, \dfrac {dc}{dt}=a\, \dfrac {da}{dt}+b\, \dfrac {db}{dt}\)

A \(9.3\, m/min\)

B \(3.6\, m/min\)

C \(27\, m/min\)

D \(12.80\, m/min\)

- Consider a ladder leaning against a wall, such that one of its end (bottom) is on the floor and the other (top) is on the wall. We can relate the rate at which the top slides down and the rate at which bottom moves on the floor.
- When top of a ladder slides down, the bottom slides away from the wall. At any instance let

x = height of ladder

y = distance of bottom of ladder from the wall

then

\(z^2=x^2+y^2\) (z = length of ladder)

\(\Rightarrow 0=2x\;\dfrac {dx}{dt}+2y\;\dfrac {dy}{dt}\) (Differentiating both sides w.r.t. t, z is a constant)

\(\Rightarrow 0=x\;\dfrac {dx}{dt}+y\;\dfrac {dy}{dt}\)

A \(3\sqrt 5\;m/sec\)

B \(-1\;m/sec\)

C \(15\;m/sec\)

D \(7\sqrt 6\;m/sec\)