Applications of Calculus Example

When a stone is dropped in a pond, the circular ripples that are formed spread out. The radius r grows at the rate dr/dt, and the disturbed area grows at the rate dA/dt. Let’s find the relationship between dA/dt and dr/dt.

1. Assuming dr/dt remains the same; if r is doubled dA/dt is ______ times greater

2. 2πrdr is the area of a rectangular ring of base 2πr and height dr. T/F 3. Why don’t the circular ripples continue to grow indefinitely?

Answer 1. Four 2. F, the opposite sides of a rectangle are equal 3. energy is dissipated Graph of C = 2𝜋r

C = 2πr

triangle dr

0 r radius

Answer 1. 2π, when r increases by 1, C increases by 2π 2. 2πrdr 3. 1/2(dr)2π(dr) = π(dr)² 4. 2𝜋r²/2 = πr² 5. 2πr Note area of the circle of radius = area of a triangle base 2𝜋r and height r

Remember If a sector has arc length s and radius r, its area A = (1/2)(s)t) = (1/2)sr.

Brain Twister

Since the slope of C = 2𝜋r is 2𝜋 when the radius increases by 1, the circumference C increases by 2𝜋 regardless of the radius!

Timeout

Junior got into real trouble with his mom and did not know what to do. He went to his room and tried to figure out a strategy. Finally, he came out and told his mom: “I will clean the kitchen”. No, thank you. He went back to his room and came out with a second strategy. ‘Mom I will clean the house.” No, thank you. Back to his room he went. “OK mom, I will clean the yard.” No, thank you. My strategy is not working! Was he getting the message? “Mom, I am sorry.” There was a big smile on mom’s face. “OK, now you can do all those chores.”

An expanding circle A = 𝜋r²is the area of a circle.

1. (A – A1)/(r – r1) = 𝜋(r² – r12)/(r – r1) measures______ .

2. Simplify (r² – r12)/(r – r1) = ______, and evaluate (r + r1) when r = r1

3. If r is very close to r1, (A – A1)/(r – r1) approaches the value ____ .

Answer 1. The average rate of growth of the area per unit increase in radius 2. r+ r1, 2r1 3. 2𝜋r1

Note If r is very close to r1, (A – A1)/(r – r1) = dA/dr.

Example 2

The position of a stone thrown up into the air is given by the equation s = – 5t² + 60t + 180, where every term is measured in meters.

O

In time dt the radius r increases by dr.

The increase in area ∆A = π[(r + dr)² – r²] = πr² + 2πrdr + π(dr)² – πr²

∆A = 2πrdr + (dr)² Note If dr is much smaller than 1, the term (dr)² is much smaller than 2𝜋rdr.

2πrdr approximates ∆A and is represented by dA.

The increase in area occurs in time dt

The rate at which the area increases dA/dt = (2πrdr)/dt = 2πrdr/dt

1. The slope of C = 2πr is ____

2. The area of the rectangle is _______

3. The area of the small triangle on top of the rectangle is ____

4. The area of the triangle with base r is ______ . 5. A = πr² and dA/dr = _____

The fence of the circle, 2πr, measures how rapidly the area, πr², grows with respect to the radius r

101 v, m/sec

g = – 10 m/sec² 0 t sec

Comments 1. The initial velocity v(0) = 60 m/sec. 2. v = 0 for all the points on the t axis. After 6 sec the stone stops rising.

3. The unit of the slope is (m/sec)/sec = m/sec² and measures the rate at which the velocity decreases. 4. The stone stops rising 6 sec after being released, that is when v = 0.

Evolution

Evolutionists always show pictures of the stages of the external evolution of a monkey, but never mention the internal evolution of the organs of the monkey. Do the vocal cords evolve enabling a monkey to talk and sing? How did a monkey learn to talk, sing, hate, love, etc. Why aren’t these topics discussed by evolutionists?

Draw a straight line through the points (0, 60) and (6, 0).

v, m/sec (0, 60)

vdt

(0, 0) (6, 0) (12, 0) t, sec

(0, – 60) •

Graph of s = – 5t² + 60t + 180

From s = – 5t² + 60t + 180, v = ds/dt = – 10t + 60, dv/dt = – 10 Finding a function when its slope is known

From dv/dt = – 10, v = ds/dt = – 10t + v0, s – ss = – 5t² + v0t

dv/dt = – 10 is the acceleration. The initial velocity v0 and initial position s0 are not known.

Brain Twister

An object is thrown upwards. If v is positive upwards, why is s positive upwards and g negative?

1. Find the intercept of v = – 10t + 60 with the v axis.

2. Find the intercept of v = – 10t + 60 with the t axis.

3. The slope or rate at which v decreases is _____

4. The stone comes to a stop when t = ____

Answer 1. let t = 0, v = 60 m/sec 2. let v = 0, t = 6 3. – 10 4. t = 6

1. What is the unit of the area of the graph?

2. Find the area of the triangle whose vertices are 0, 0), (6, 0), (0, 60).

3. Find the area of the triangle whose vertices are (6, 0), (12, 0), (0, – 60).

4. Why is the area between v = – 10t + 60 and v = 0, from t = 0 to t = 12, cero?

5. v = – 10t + 60, is a. s = – 5t² + 60t b. s – s0 = – 5t² + 60t

Answer 1. m 2. 180 m 3. – 180 m 4. after t = 12 sec the object returns to its starting position 5. b.

1. What is the initial position?

2. If s = – 5t² + 60t + 180, find ds/dt.

3. The slope of the graph at t = 0 is ______

4. v = ds/dt = – 10t + 60. When is the slope cero?

5. If v = – 10t + 60t, dv/dt = ____ .

6. The slope of the graph is always decreasing. T/F 7. Gravity is pulling the object down. T/F

Answer 1. s = 180 2. – 10t + 60 3. 60, the initial velocity 4. t = 6 5. – 10 6. T 7. T

Maybe

People who want everything free must like slavery. They don’t want to work.

The slope of the slope – Graph of y = x³ – x

Graphs of y = x³ – x, its slope is 6x² – 1, and 6x is the slope of the slope

When the value of y depends on the value of x, we say that y is a function of x, y = f(x). The function consists of the set of points (x, f(x)).

Rate of growth of A = xy

1. Show that (x + dx)(y + dy) – xy = ydx + xdy + dxdy Note, three rectangles

2. If the growth of x and y depend on t, where x = 3t and y = 2t, dAX,Y = y__ + x__ + 6dt² 3. dAX,Y/dt = ___________

Answer 1. (x + dx)(y + dy) = xy + xdy + ydx + dxdy – xy 2. y(3dt) + x(2dt) 3. 3y + 2x + 6dt Note if dt is very small, dAX,Y / dt = 3y + 2x; measures the rate of growth of x plus rate of growth of y.

Since the area of the rectangle depends on the values of the variables x and y, A(x, y) = xy

If only x increases, the increase in area dAX = ydx If only y increases, the increase in area dAY = xdy If x increases first and then y dAX,Y = ydx + xdy

If both x and y are increased simultaneously

∆A = ydx + xdy + dxdy y = x³ – x = x(x² – 1) = x(x – 1)(x + 1) 1. For which values of x is y = 0?

2. The slope dy/dx = 3x² – 1 is cero when x = _______

3. If x < – (1/3)^1/2, the function is increasing T/F 4. For x > (1/3)^1/2 the slope is ______

For x < 0 the curve concave down and concave up for x > 0.

5. When is the slope of the slope cero?

Answer 1. – 1, 0, 1 2. ∓ (1/3)^1/2 3. T 4. positive 5. 6x = 0, x = 0

If y = f(x), dy/dx is called the first derivative of f(x), and

𝑑 𝑑𝑥(^𝑑𝑦

𝑑𝑥)= d²y/dx² the second derivative

The slope of x³ – x is 3x² – 1.

The slope of 3x² – 1 is 6x.

6x is the slope of the slope of x³ – x

6x is negative for x less than 0 and positive for x greater than 0.

At x = 0 the curve changes from concave down to concave up.

(0, 0) is a point of inflection; from left to right the curve changes concavity.

X³ – x 3x² – 1

6x –(1/3)^1/2

(1/3)^1/2

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