Informative line

Rate Of Change In Physics And Economics

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

Relation between Current and Charge in Physics

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.

 

Illustration Questions

The amount of charge Q in Coulomb that passed through a point in wire up to time 't' (in seconds) is given by Q(t) = 2t3 – 4t2 + 7t + 5 Find the current when t = 3 sec.

A 25 A

B 5 A

C 1 A

D 37 A

×

\(I=\dfrac {dQ}{dt}=\dfrac {d}{dt}\) (2t3 – 4t2 + 7t + 5)

= 6t2 – 8t + 7 

\(\Rightarrow\)\( I \)at (t = 3) = 6 × 9 – 8 × 3 + 7

= 54 – 24 + 7 = 37 C/s = 37 Ampere

The amount of charge Q in Coulomb that passed through a point in wire up to time 't' (in seconds) is given by Q(t) = 2t3 – 4t2 + 7t + 5 Find the current when t = 3 sec.

A

25 A

.

B

5 A

C

1 A

D

37 A

Option D is Correct

Marginal Cost Function

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 x1 to x2 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.

 

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

Illustration Questions

The cost in dollars of producing x units of certain commodity is given by the cost function. C(x) = 5000 + 3x + 0.1x2 Find the marginal cost function.

A 3 + 0.2 x 

B 7 + 0.1 x

C 45 + 8x

D 84 + 0.7 x

×

Marginal cost function = C'(x) where C(x) is the cost function.

In this case

\(C'(x) =\)\(\dfrac {d}{dx}C(x)=\dfrac {d}{dx}\) (5000 + 3x + 0.1x2) = 3 + 0.2 x

\(\therefore\)  \(C'(x) =\)3 + 0.2 x

The cost in dollars of producing x units of certain commodity is given by the cost function. C(x) = 5000 + 3x + 0.1x2 Find the marginal cost function.

A

3 + 0.2 x 

.

B

7 + 0.1 x

C

45 + 8x

D

84 + 0.7 x

Option A is Correct

Actual Cost of Producing a Particular Numbered Item

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)

Illustration Questions

The cost in dollars of producing x units of certain commodity is given by the cost function. C(x) = 5000 + 3x + 0.1x2 Find the actual cost of producing 101st item.

A $ 23.1 

B $ 57.8

C $ 64

D $ 11.3

×

Actual cost of producing (x + 1 )th item

 = C(x + 1) – C(x)

In this case,

C (x + 1) = 5000 + 3 (x + 1) + 0.1 (x+1)2

C(x) = 5000 + 3x + 0.1x2

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

 = C(x + 1) – C(x)

= (5000 + 3(x+1)+ 0.1(x+1)2) –  (5000 + 3x + 0.1x2)

= 3 + 0.1(2x + 1)

= 3 + 0.2 x + 0.1

\(\therefore\) Actual cost of producing 101st item

 = 3 + 0.2 ×100 + 0.1

= $ 23.1    (Putting x = 100 )

The cost in dollars of producing x units of certain commodity is given by the cost function. C(x) = 5000 + 3x + 0.1x2 Find the actual cost of producing 101st item.

A

$ 23.1 

.

B

$ 57.8

C

$ 64

D

$ 11.3

Option A is Correct

Marginal Cost as Prediction of Actual Cost

  • 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.

Illustration Questions

The cost of producing x units of commodity in a factory is given by C(x) = 1500 + 10x – 0.2x2 + 0.005x3 where C(x) is in $. Find the marginal cost of producing 201st unit of commodity.

A $ 408

B $ 73

C $ 8

D $ 530

×

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

In this case,

\(C'(x) = \dfrac {d}{dx}\) (1500 + 10 x – 0.2x2 + 0.005x3  )

= 10 – 0.4x + 0.015 x2

\(\therefore\)  C'(200) = Marginal cost of producing 201st unit

= 10 – 0.4 × 200 + 0.015 × 200 × 200      (Putting x = 200 in previous formula)

= 10 – 80 + 600

= $ 530 

The cost of producing x units of commodity in a factory is given by C(x) = 1500 + 10x – 0.2x2 + 0.005x3 where C(x) is in $. Find the marginal cost of producing 201st unit of commodity.

A

$ 408

.

B

$ 73

C

$ 8

D

$ 530

Option D is Correct

Related Rates

  • 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 x2 + y2 = 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.

Illustration Questions

If \(9y^2 + 4x^2 = 36\) where \(x\) and \(y\) depend on 't' and \(\dfrac {dy}{dt}=\dfrac {1}{2}\), find \(\dfrac {dx}{dt}\) where \(x=1\) and \(y = \dfrac {4\sqrt 2}{3}\).

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

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

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

D \(5\sqrt 5\)

×

Given relation between \(x\) and \(y\) is \(9y^2 + 4x^2 = 36 \)

Differentiate both sides with respect to 't'

\(\Rightarrow 18y\dfrac {dy}{dt}+8x\dfrac {dx}{dt}=0\)

\(\Rightarrow 18×\dfrac {4\sqrt2}{3} ×\dfrac {1}{2}+8×1×\dfrac {dx}{dt}=0\)

(Put \(x = 1\)\(y=\dfrac {4\sqrt2}{3}\) and \(\dfrac {dy}{dt}=\dfrac {1}{2}\))

\(\Rightarrow \dfrac {dx}{dt}=\dfrac {-12\sqrt2}{8}\)

\(\Rightarrow \dfrac {dx}{dt}=\dfrac {-3\sqrt2}{2}\)

– sign indicates that \(x\) decreases with respect to 't'.

If \(9y^2 + 4x^2 = 36\) where \(x\) and \(y\) depend on 't' and \(\dfrac {dy}{dt}=\dfrac {1}{2}\), find \(\dfrac {dx}{dt}\) where \(x=1\) and \(y = \dfrac {4\sqrt 2}{3}\).

A

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

.

B

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

C

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

D

\(5\sqrt 5\)

Option A is Correct

Relation between Rate of change of x and y co-ordinates

  • 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)

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

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

Illustration Questions

A particle is moving along the curve \(x^3y^2=8\), when it is at the point (2, 1) the x co-ordinate is increasing at the rate of \(5\; cm/min\). How fast is the y-coordinate changing then?

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

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

C \(5\;cm/min\)

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

×

The equation of the curve is \(x^3\,y^2 = 8\)

Differentiate both sides w.r.t. 't'

\(\Rightarrow x^3 × 2y\,\dfrac {dy}{dt}+3x^2\,\dfrac {dx}{dt}×y^2=0\) ( Product Rule)

\(\Rightarrow 2x^3 y\,\dfrac {dy}{dt}+3x^2\,y^2\,\dfrac {dx}{dt}=0\)

Put \(x = 2,\, y = 1\) and \(\dfrac {dx}{dt}= 5\, cm/min\)

\(\Rightarrow 2×8×1×\dfrac {dy}{dt}+3×4×1×5\;=0\)

\(\Rightarrow 16\dfrac {dy}{dt}+60\;=0\)

\(\Rightarrow \dfrac {dy}{dt}=\dfrac {-15}{4}\;cm/min\) (Negative sign means y is decreasing with time)

A particle is moving along the curve \(x^3y^2=8\), when it is at the point (2, 1) the x co-ordinate is increasing at the rate of \(5\; cm/min\). How fast is the y-coordinate changing then?

A

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

.

B

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

C

\(5\;cm/min\)

D

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

Option D is Correct

Particles Moving in Directions Perpendicular to Each Other

  • 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)

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

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

Illustration Questions

Two persons start from the some point. One travels north at the speed of \(10\, m/min\). and other travels east at \(8\, m/min\). At what rate is the distance between them increasing after 3 hours?

A \(9.3\, m/min\)

B \(3.6\, m/min\)

C \(27\, m/min\)

D \(12.80\, m/min\)

×

Let x be the distance  of first person A from the starting point after t min. Let y be the distance of second person B from starting point after 't' min. Let z be the distance between them.

image

By Pythagoras Theorem

\(z^2 = x^2+y^2\)

image

Differentiate both sides w.r.t. 't'

\(\Rightarrow\) \(2z\, \dfrac {dz}{dt}=2x\, \dfrac {dx}{dt}+2y\, \dfrac {dy}{dt}\)

\(\Rightarrow\)\(z\, \dfrac {dz}{dt}=x\, \dfrac {dx}{dt}+y\, \dfrac {dy}{dt}\)

image

After 3 hours i.e. \(180\, min.\)

\(x=180×10=1800\, min.\\ y=180×8=1440\, min.\)

\(z=\sqrt {(1800)^2+(1440)^2}\) \(=\sqrt {5313600}=2305.12\)

image

\(\therefore\) \(2035.12\,\dfrac {dz}{dt}=1800×10+1440×8\)

\(\dfrac {dz}{dt}=\dfrac {1800×10+11520}{2305.12}=12.8\,m/min\)

image

Two persons start from the some point. One travels north at the speed of \(10\, m/min\). and other travels east at \(8\, m/min\). At what rate is the distance between them increasing after 3 hours?

image
A

\(9.3\, m/min\)

.

B

\(3.6\, m/min\)

C

\(27\, m/min\)

D

\(12.80\, m/min\)

Option D is Correct

The Ladder Problem

  • 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)

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

Illustration Questions

The top of a ladder 6 m long is resting against a vertical wall on a level pavement when the ladder begins to slide outwards. At the moment when the foot of ladder is 4 m from the wall, it is sliding away from the wall at rate of 5 m/sec. How fast is the top sliding down at this instance?

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

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

C \(15\;m/sec\)

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

×

Let A = Top of ladder

B = Bottom of ladder

At any instance

\(x^2+y^2=6^2\)

\(\Rightarrow\;x^2+y^2=36\)

image

Differentiate both sides w.r.t. 't'

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

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

Put y = 4, \(x=\sqrt {36-16}=2\sqrt 5,\;\dfrac {dy}{dt}=5\,m/sec.\)

\(\therefore\;2\sqrt 5\;\dfrac {dx}{dt}+4×5=0\)

\(\Rightarrow \dfrac {dx}{dt}=\dfrac {-20}{2\sqrt 5}=-1\;m/sec\)

\(\therefore\;-\dfrac {dx}{dt}=\) rate of top sliding down = \(-1\,m/sec.\) (x decreases as 't' increases)

The top of a ladder 6 m long is resting against a vertical wall on a level pavement when the ladder begins to slide outwards. At the moment when the foot of ladder is 4 m from the wall, it is sliding away from the wall at rate of 5 m/sec. How fast is the top sliding down at this instance?

A

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

.

B

\(-1\;m/sec\)

C

\(15\;m/sec\)

D

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

Option B is Correct

Practice Now