The Derivative as a Rate of Change |

Since the **first derivative** of a continuous function *y = f(x)* represents the **slope of the tangent to the curve**, and, since a slope defines a **change in the dependent variable (***y***)** corresponding to a **change in the independent variable ( ***x***), the derivative is a RATE OF CHANGE**. That's why the best notation for those of us not too lazy to use it is *dy/dx** *or* ds/dt *etc*.* since this notation indicates **a ratio of change** in one variable to change in another.

Now we consider 2 specific rates of change:

Change in **position**to change in**time**--**velocity**Change in **velocity**to change in**time**--**acceleration**.If *s*(*t*)*position*of a point**P**with respect to time*t*, then

the**first derivative***ds/dt*

**change in position with respect to a change in time**or**VELOCITY****.**The

**second derivative**(*d*²*s / dt*² ) represents the

**change in VELOCITY****with respect****to a change in time**or**ACCELERATION**Let's investigate the

**units**involved in both derivatives. If we're measuring**position in meters**and**time in seconds**,The

**1st derivative**units will be**meters / second**which is a speed or**velocity**.

The**2nd derivative**units will be**meters / second**which is^{ 2}**acceleration**..

Because the derivative is found by taking the limit as either

*h*or*delta*t approach 0, when we use the**first derivative**of a position function to find velocity, we find the**instantaneous velocity****at**the specified**point in time**. When we need**average velocity over an****interval of time**, we must find the actual change in position rather than an approximation of it -- (see Increments and Differentials).**average velocity**over interval (*t , t +**D**t*) =.

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**Example 1**The position function

*s*of a point**P**on a coordinate line is:*s*(*t*)*= t*^{ 3}– 12*t*² + 36*t*– 20 ,*t*in*seconds*,*s(t)*in*cm.*Describe the motion of

**P**during the time interval [–1, 9].Since

*s'*(*t*)*= v*(*t*)*= the velocity function*, and*s''*(*t*)*= a*(*t*)*= the acceleration function*,The signs of the 2 derivatives will tell us if the point

a) is moving to the left (

*v*(*t*)*< 0*)*,*or to the right, (*v*(*t*)*> 0*)b) where velocity is decreasing (

*a*(*t*)*< 0*)*,*and where it is increasing(*a*(*t*)*> 0*).*s'*(*t*)*= v*(*t*)*= 3t*^{ 2}*–**24t*^{ }+ 36 = 3(t*–**2)(t**–**6)**s''*(*t*)*= a*(*t*)*= 6t*^{ }*–**24 = 6(t**–**4)**t =*–1 2 4 6 9 *s*(*t*)*=*–69 12 –4 –20 61 *s'*(*t*)*= v*(*t*)*=*63 0 –12 0 63 *s''*(*t*)*= a*(*t*)*=*–30 –12 0 12 30 So,

**P**is at (–1, –69) to begin, it moves right until (2, 12) where*v*(*t*) = 0.From (2, 12),

**P**moves left to (4, –4) where the acceleration*a*(*t*) = 0.It continues left to (6, –20) where

*v*(*t*) = 0Then it begins to move right until (9, 61).

The velocity and acceleration at time

*t*are as shown in the table.**Temperature Change over Time****Example 2**We've created a new synthetic material called

*Goodstuff*, so we're testing it's reactions to temperature changes. The mad scientists on our team tell us that if we heat it for*t minutes*, the temperature in degrees Celsius will beThey warn us that we can heat it for up to 5 minutes, past that -- no one knows what happens.

We need to know:

1) the

**average rate of temperature change**over the interval**between 4 and 4.41 minutes**.and

2) the

**temperature change**at exactly**4 minutes**.1) We want the

**average rate of change**, so we'll set*t = 4 min*and*D**t = 0.41 min.*Then,

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2) We want the

**instantaneous****rate of change**, so we'll find*g'(t)*at*t = 4*..

Derivative = slope velocity and acceleration Commerce applications practice solutions .

The commerce applications of this theory involve 4 functions:

-- the*C(x)***Cost**function -- usually**a polynomial in**the number of items produced -- represents the*x,***cost of materials and labour**needed to make the product.-- the*p***demand**function -- is the**unit selling price**(ususally in $) of the product.=*R(x)*-- the*px***Revenue**function --**income**from sellingunits at*x*each.*$p*-- the*P(x)***Profit**function =*R(x) - C(x)**--*(money in) - (money out)*C(x)*is the cost of producing*x*units of product, so is the**average Cost function**or the production cost per unit.The

**average Cost function**is denoted*c(x)*or said*C bar*, or*C hat*.Generally,

*C(x)*will have a constant term which represents the fixed costs of running a business such as heat, electricity, rent, etc. Since this fixed cost doesn't affect the cost of producing*n*more units, we consider the change in cost per unit produced called**marginal cost = C'(x)**.Just as we consider

**average and marginal Cost function**, we also consider the**average****and****marginal****Revenue and Profit functions**denoted R'(x) and P '(x) respectively.And there's also the

**average marginal****Cost, Revenue and Profit functions**respectively.

denotedJust as we use

*dy/dx*or*f '(x) to approximate changes to the function when the change in x is small*, we use*C '(x) -- the marginal cost function*to approximate the cost of producing

the (*x + 1)th**unit*when x is large to begin with. The logic being that a change of 1 unit when we started at 550 or 1000 units is small enough to allow the approximation to be within acceptable norms. So, if asked to find the cost of producing the 501st*thingy*we'll evaluate*C '(500)*.(see example 4 part b).

To **minimize**the average cost,set

**marginal cost = average cost**.*C'(x) = c(x)*To

**maximize revenue,**set

**marginal revenue = marginal cost.***R'(x) = C'(x)*.

**Example 3**A widget manufacturer has a monthly

**fixed cost of $10,000**, a production**cost of $12 per**widget and he**sells**the widgets at**$20 each**.Find:

a)

**C(**cost function,*x*)**c(***x***)**avg. cost function,**R(***x***)**revenue function , and**P(***x***)**profit function.b) Find values for C(

), c(*x*), R(*x*), and P(*x*) if*x***x = 1000**.c) How many widgets must be sold to break even? (no profit, no loss)

**Solution**a) C( ) = 12*x*+ 10,000*x*c( ) = 12 + 10,000/*x**x*R( ) = 20*x**x*P( ) = 8*x*– 10,000*x*b) C(1000) = 22,000 c(1000) = 22 R(1000) = 20,000 P(1000) = –2000 c) set profit , so 8*P*(*x*) = 0– 10,000 = 0 which means*x*to break even.*x*= 1250.

**Example 4**A paint manufacturer determines that the total cost in dollars of producing

*x*gallons of paint per day is:*C*(*x*) = 5000 +*x*+ 0.001*x*²We want to:

a) Find the marginal cost of producing 500 gallons per day.

b) Use this marginal cost to

**approximate**the**cost**of producing the**501st gallon**of paint.c) Find the

**exact cost**of producing the 501st gallon of paint.**Solution**a) The marginal cost is

*C '*(*x*)*x*, so,*C '*(500 ) = 1 + 0.002(500) =**2**.b) Since

*C '*(500 ) = 2, the approximate cost of producing the**501st**gallon of paint is $2.00.c) The

**exact cost**of producing the 501st gallon of paint isC(501 ) – C(500 ) =

{5000 + 501 + 0.001(501)²} - {5000 + 500 + 0.001(500)²

*}*=**2.001**.

Derivative = slope velocity and acceleration Commerce applications practice solutions .

1) Our company estimates that the cost in dollars of producing

*x*boblets per day is:*C*(*x*) = 2600 + 2*x*+ 0.001*x*²a) Find the

**cost**,**average cost**and**marginal cost**of producing**1000**boblets, and**2000**boblets.b) What production level (

*x*) will**minimize the average cost**?c) What is the

**minimum average cost**per boblet?.

2) The

**average cost**of producing*x*units of a commodity is:*c*(*x*) = 21.4 – 0.002*x*Find the

**marginal cost function**and evaluate it at a production level of 1000 units..

3) The cost in dollars of producing

*x*meters of fine silk fabric is:*C*(*x*) = 1200 + 12*x –*0.1*x*² + 0.0005*x*³The demand function

*p*-- the selling price in dollars per meter -- is:*p*(*x*)*=*29 – 0.00021*x*What production level will

**maximize the profits**?.

4) The position function

*s*of a point**P**on a coordinate line is:*s*(*t*)*= t*³*–*9*t*+ 1 ,*t*in*seconds*,*s*(*t*)*cm.*a) Find the

**velocity**and**acceleration**at time*t*.b) Describe the motion of

**P**during the time interval [–3, 3].c) Draw a diagram of the motion of

**P**during the time interval [–3, 3]..

5) Show that the rate of change of the volume of a sphere with respect to its radius is numerically equal to the surface area of the sphere.

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Derivative = slope velocity and acceleration Commerce applications practice solutions .

1) Our company estimates that the cost in dollars of producing

*x*boblets per day is:*C(x) = 2600 + 2x + 0.001x*^{ 2}a) Find the

**cost**,**average cost**and**marginal cost**of producing**1000**boblets, and**2000**boblets.*c(x) = C(x)/x = 2600 / x + 2 + 0.001x*(average cost function)^{ }*C '(x) = 2 + 0.002x*(marginal cost function)*x**C(x)**c(x)**C '(x)*1,000 5,600,000 5.60 4.00 2,000 10,600,000 5.30 6.00 .

b) What production level (

*x*) will**minimize the average cost**?Set marginal cost

*C '(x)*= average cost*c(x)*to minimize average cost.*2 + 0.002x*=*2600 / x + 2 + 0.001x*Producing 1612 boblets per day will minimize the average cost.

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c) What is the

**minimum average cost**per boblet?*c(1612) = $5.22 per boblet.*.

2) The

**average cost**of producing*x*units of a commodity is:*c(x) = 21.4 - 0.002x*Find the

**marginal cost function**and evaluate it at a production level of 1000 units.Since

*c(x) = C(x)/ x*, we will multiply*c(x)*by*x*to get*C(x)*.*C(x) = 21.4x - 0.002x*^{ 2}Now, marginal cost function =

*C '(x) = 21.4 - 0.004x**C '(1000) = 21.4 - 0.004(1000) = 17.40*The marginal cost of 1000 units is $17.40.

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3) The cost in dollars of producing

*x*meters of fine silk fabric is:*C(x) = 1200 + 12x - 0.1x*^{ 2}+ 0.0005x^{ 3}The demand function

*p*-- (the selling price in dollars per meter) -- is:*p(x) = 29 - 0.00021x*What production level will

**maximize the profits**?The Profit function

*P(x)*is*R(x) - C(x)*, so we must find*R(x)*.*R(x) = px = 29x - 0.00021x*^{ 2}So,

*P(x) = (29x - 0.00021x*^{ 2}) -*(1200 + 12x - 0.1x*^{ 2}+ 0.0005x^{ 3})*P(x) = - 0.0005x*.^{ 3}+ 0.09979x^{ 2}+ 17x - 1200To maximize this function, we set

*P'(x) = 0**P'(x) = - 0.0015x*^{ 2}+ 0.19958x + 17 = 0The zeros are at - 59 and 192 so the answer is

**192 meters**.

4) The position function

*s*of a point**P**on a coordinate line is:*s(t) = t*,^{ 3}- 9t + 1*t*in*seconds*,*s(t)*in*cm.*a) Find the

**velocity**and**acceleration**at time*t*.*v(t) = s'(t) = 3t*^{ 2}- 9*a(t) = s''(t) = 6t*b) Describe the motion of

**P**during the time interval [–3, 3].*v(t)= 3t*=^{ 2}– 9 = 3(t^{ 2}– 3)*v(t) = 0 at**v(t) < 0*so**P**moves left on*v(t) > 0*so**P**moves right on*a*(*t*) = 0 when*t*= 0*a*(*t*) < 0 when*t*< 0, and*a*(*t*) > 0 when*t*> 0*t =*–3 0 3 *s(t) =*1 11.39 1 –9.39 1 *s'(t) = v(t) =*18 0 –9 0 18 *s''(t) = a(t) =*–18 0 18 c) Draw a diagram of the motion of

**P**during the time interval [ – 3, 3]..

5) Show that the rate of change of the volume of a sphere with respect to its radius is numerically equal to the surface area of the sphere.

*4**o**r*is the surface area of the sphere.^{ 2}.

Derivative = slope velocity and acceleration Commerce applications practice solutions .

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