Notes Packet 3C - Kinematics, Optimization, Linear Approximation, Differentials, Newtons Method.pdf

Name: _______________________________________ Period: _________ CALCULUS NOTES

UNIT 3C – APPLICATIONS OF DIFFERENTIATION Kinematics/Rectilinear Motion

Ex. A particle moves along a horizontal line so that its position at any time t≥0is given by s t

( )

=2t3−7t2+ +4t 5, where s is measured in meters and t in seconds.

(a) Find the velocity at time t and at t = 1 second.

(b) When is the particle at rest? Moving left? Moving right? Justify your answers.

(c) Find the acceleration at time t and at t = 1 seconds.

(d) Find the displacement of the particle between t = 0 and t = 3 seconds. Explain the meaning of your answer.

(e) Find the distance traveled by the particle between t = 0 and t = 3 seconds.

(f) When is the particle speeding up? Slowing down? Justify your answer. Hint: Since speed is the absolute value of velocity, the particle is:

Ex. A particle moves along a horizontal line so that its position at any time t≥0 is given by s t

( )

= −t3 5t2+ −2t 3. Use the calculator.

(a) When is the particle moving right? Moving left? Justify your answer.

(b) Find the distance traveled by the particle from t = 0 to t = 5. Justify your answer.

(c) Find the intervals where the speed is increasing. Justify your answer.

Optimization

Ex. A particle moves along the x-axis so that its position is given by s t

( )

= −t3 5t2+12t for 0≤ ≤t 4. Find the minimum velocity of the particle and the time at which it occurs. Justify your answer.

First Method – Candidates Test

First find v t

( )

=3t2−10t+12 and v t′

( )

= −6t 10, and set v t′

( )

equal to zero to find the critical values.

( )

5

6 10 0 when 3

v t′ = −t = t= The minimum velocity may occur at 0, 5

3, or 4. t v(t)

0 12 5

3 3.667 4 20

The minimum velocity is 3.667, and it occurs when t = 5 3.

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Second Method – First Derivative Test

First find v t

( )

=3t2−10t+12 and v t′

( )

= −6t 10, and set v t′

( )

equal to zero to find the critical values.

( )

5

6 10 0 only when 3

v t′ = −t = t= .

( )

0 for 0 5 and

( )

0 for 5 4.

3 3

v t′ < ≤ <t v t′ > < ≤t

Therefore the minimum velocity is 3.667, and it occurs when t = 5 3.

***WARNING: The First Derivative Test only guarantees relative extrema. In order to conclude that a specific value is also an absolute extrema everything would have to fall perfectly into place with your sign charts. However, it’s rare when that happens.

EX sign charts:

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Third Method – Second Derivative Test

First find v t

( )

=3t2−10t+12 and v t′

( )

= −6t 10, and set v t′

( )

equal to zero to find the critical values.

( )

5

6 10 0 only when 3

v t′ = −t = t = . Now find v’’ (t) = 6 , so v’’ (5/3) = 6 > 0.

Therefore the minimum velocity is 3.667, and it occurs when t = 5 3.

***WARNING: Again, the Second Derivative Test will only guarantee a relative extrema. And it has the potential to ignore your endpoint values.

Ex. A particle moves along the x-axis so that its position is given by s t

( )

= −t3 5t2+12t for 0≤ ≤t 4. Find the maximum velocity of the particle and the time at which it occurs. Justify your answer.

Only Method – Candidates Test

First find v t

( )

=3t2−10t+12 and v t′

( )

= −6t 10, and set v t′

( )

equal to zero to find the critical values.

( )

5

6 10 0 when 3

v t′ = −t = t= The maximum velocity may occur at 0, 5

3, or 4. t v(t)

0 12 5

3 3.667 4 20

The maximum velocity is 20, and it occurs when t = 4.

***The other two methods will not work for this questions when trying to find the maximum. Why?

***So not only is the Candidates Test the best method, sometimes it is the ONLY method that will work! In summary, USE CANDIDATES TEST FOR ABSOLUTE EXTREMA!***

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Mr. Bell’s thinking process for optimization problems:

1 – What information does the problem give me? Variables? Units? 2 – Picture? – If not, draw one. Label everything possible.

3 – What quantity/value am I supposed to optimize? Do I maximize or minimize?

4 – What formulas (hopefully 2) do I need/can I use with the information I was given? How can I tie all of the variables together in the same equation(s)?

5 – Which variable can/should I make “go away”?

6 – Use algebra SKILLZ to make it go away. i.e. solve one of the equations (the extra one) for one of the variables, then plug in that expression in to the other equation (the one I want to optimize). 7 – Take the derivative of this “new and improved” equation.

8 – Set it equal to zero and/or undefined and solve for my critical values.

9 – Take these values for my first variable and plug them back into the expression for the other(s). 10 – Test the critical value points AND THE ENDPOINTS to see which one/set produces the desired optimization result.

Ex. A swimmer is 2 miles in the ocean and wishes to get to 3 mi. Town a town 3 miles down the coast which is very rocky. The

swimmer needs to swim to the shore and then walk along 2 the shore. He can swim at 2 mph and walk at 4 mph. mi. To what point should he swim along the shoreline so that

the time it takes to get to town is a minimum?

______________________________________________________________________________ Ex. Find the dimensions of a 12-oz. can that can be constructed with the least amount of metal (12 oz. ≈ 355 mL = 355cm ). Justify your answer. 3

Ex. The sum of the perimeter of a square and the circumference of a circle is 20.

(a) Find the dimensions of the square and the circle that produce a minimum area. Justify. (b) Find the dimensions of the square and the circle that produce a maximum area. Justify.

Using Tangent Lines: Linear Approximation (Linearization) and Differentials

The tangent line to a curve gives us a useful way to approximate a value on the curve. We usually write our tangent line equations in point-slope form: y−y₁=m x

(

−x₁

)

or for a particular value of x1 named a, we could

write y− f a

( )

= f′

( )(

a x−a

)

.

When we are using a tangent line to find an approximation of a value on the curve, we might want to rearrange the terms of the tangent line equation to write: y= f a

( )

+ f′

( )(

a x−a

)

.

Note: y represents the y-coordinate of a on the tangent line. This gives us an approximation of f a

( )

. The y-coordinate of a on the function f x

( )

is represented by f a

( )

. f a

( )

is the actual value.

Ex. Given f x

( )

=x2.

(a) Write the equation of the tangent line to f at x = 1

(b) Use your answer to (a) to find an approximation for f

(

1.01 .

)

(c) Find the actual value of f

(

1.01 .

)

(d) Find the amount of error of your approximation.

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Leibniz used the notation dy

dx to represent the derivative of y with respect to x. The notation looks like a

quotient of real numbers, but it is really a limit of quotients in which both numerator and denominator go to zero (without actually equaling zero). Leibniz did most of his calculus using dy and dx as separate entities, but he never quite settled the issue of what they were. To him, they were “infinitesimals” ---nonzero numbers but infinitesimally small.

It is tricky to define dy and dx as separate entities, but since we really only need to define dy and dx as formal variables, textbooks define them in terms of each other so that their quotient must be the derivative.

Definition: Let y= f x

( )

be a differentiable function. The differential dx is an independent variable. The differential dy is dy= f′

( )

x dx. The variable dyis always a dependent variable; it depends on both x and

dx.

Ex. Find the differential dy, given y=5 .x3

__________________________________________________________________________________________ Ex. The radius r of a circle increases from 10 m to 10.1 m.

(a) Use the differential dA to estimate the increase in the circle’s area A.

(b) Compare the estimate you found in (a) with the true change ∆A, and find the error in your approximation.

_________________________________________________________________________________________ Ex. Suppose the earth were a perfect sphere and we determined its radius to be 3959±0.1 miles. What

Newton’s Method of Finding Zeros

Suppose f is a differentiable function on (a, b). If f a

( )

( )

and f b differ in sign, then f has at least one zero in (a, b). Newton made a guess that this zero occurs at some x₁ in (a, b), and he discovered that the zero of the tangent line was close to the zero of the function.

Write the equation of the tangent line to f at

( )

(

x1, f x1

)

, and solve for x, the x-intercept of the tangent line.

Each successive approximation is called an iteration.

Formula for Newton’s Method:

( )

( )

1

n n

n n

f x

x x

f x

Calculator: Let y1 function and 2= y =derivative

On the home screen, type the x₁ you were given and press Enter.

Then type:

(

)

(

)

1 Answer

Answer Enter Enter Enter ...

2 Answer

y

y

−

The result after the first “Enter” is x2. The result after the second “Enter” is x3. The result after the third “Enter” is x₄.

Keep pressing Enter until your iterations give you as accurate an answer as the directions ask you for.

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Ex. Use Newton’s Method to approximate the zero of f x

( )

=2x3+x2− +x 1. Continue the iterations until two successive approximations differ by less than 0.001. Let x₁= −1.2.

( )

( )

3 2 3 2

2 1 so let 1 2 1

so let 2

f x x x x y x x x

f x y

= + − + = + − +

′ = =

Then go to the home screen and type – 1.2 Enter and then

(

)

(

)

1 Answer

Answer Enter Enter Enter ...

2 Answer y y − 2 3 4 x x x = = =

By Newton’s Method, the zero is approximately _____________.

__________________________________________________________________________________ Ex. Use Newton’s Method to estimate the x-coordinate of the positive point of intersection