Volumes Question1

Solution!

A.I.) Exploration of Functions

4x2 = 2x3 < find points of intersection

*To find the points of intersection, make each function equal to each other to find points where each function coexist


4x2 = 2x3 < find points of intersection0 = 2x3 4x2 < solve for the roots0 = 2x2 (x 2) < factor out 2x2x = 0, 2 < points of intersection

*Through simple algebra, we find two points of intersection. These points are the boundaries of which the area coincides.


4x2 = 2x3 < find points of intersection0 = 2x3 4x2 < solve for the roots0 = 2x2 (x 2) < factor out 2x2x = 0, 2 < points of intersection

0 2

4(1)2 = 4 < input value within (0, 2) interval2(1)3 = 2

4x2 > 2x3

*By inputting some value between the points of intersection, we can find the positions of the functions: more specifically, which function is "on top" of the other.

Number Line

*This diagram depicts the points of intersection shared by the two functions 4x2 and 2x3. To find the volume of the solid, we find the area between each function, then rotate it around a vertical axis x = 2.


x

y

x = 2

dx2 x

[f(x)g(x)]

*When the area between the functions are rotated around a vertical axis, the result is a washerlike solid shown above. Now if we make a vertical slice down the center, we could unravel the solid to form a rectangular prism, which volume is found by the equation: V=lwh

A.II.) Volume of Revolution

x

y

x = 2

dx2 x

[f(x)g(x)]


V = lwhl = 2 x < circumference of a circlew = dx < infinitely small piecesh = [4x2 2x3] < distance between two functions


V = lwhl = 2 x < circumference of a circlew = dx < infinitely small piecesh = [4x2 2x3] < distance between two functions

V = 2 [(x + 2)(4x2 2x3)] dx < integral expression of volume0

2

*In this integral expression, the constant 2 is brought outside of the integral. Note1: "x", representing the radius of the circumference expression is replaced by (x+2) because the vertical line we are revolving about is x = 2, 2 units away from the yaxis, thus increasing radius by 2.Note2: the integral of (4x2 2x3) represents the area between the functions



2

V = 2 [4x3 2x4 + 8x2 4x3] dx0

2

< expand



2

V = 2 [4x3 2x4 + 8x2 4x3] dx0

2

V = 2 [2x4 + 8x2] dx0

2

< simplify

< expand



2

V = 2 [4x3 2x4 + 8x2 4x3] dx0

2

V = 2 [2x4 + 8x2] dx0

2

< simplify

< expand

V = 2 [ + ]2x55

8x33

2

0< antidifferentiate with power rule



2

V = 2 [4x3 2x4 + 8x2 4x3] dx0

2

V = 2 [2x4 + 8x2] dx0

2

< simplify

< expand

V = 2 [ + ]2x55

8x33

2

0< antidifferentiate with power rule

V = u325615 < exact volume solution

B.I.) Interpretation

V = u325615

t = 25615

23

t = 35.7443 minutes

< multiply by the reciprocal of (3/2)

< time remaining of battery life

*Note that in the context of the question, the numerical value of the volume generated was (3/2) greater than that of the time remaining in minutes of the battery life

B.II.) Differentiation

s(t) = esin(3x) < distance function of RC car

*Note that this function is composed of 3 functions: ex, sin(x), and 3x. Therefore, we will use a double chain rule to differentiate this compound function.

Chain Rule: F'(x) = f'(g(x))g'(x)Double Chain Rule: F'(x) = f'(g(h(x)))g'(h(x))h'(x)

*Note that the second application of the chain rule comes into play when g'(x) is required.



f(x) = ex g(x) = sin(x) h(x) = 3xf'(x) = exg'(x) = cos(x)h'(x) = 3

Assigned Inner and Outer Functions

Differentiate each Inner and Outer Function







v(t) = f'(g(h(x)))g'(h(x))h'(x)


*Since we already differentiated each sub function and possess the double chain rule template, the work is much simplified!






v(t) = f'(g(h(x)))g'(h(x))h'(x)v(t) = esin(3x)(cos(3x)(3))


< input functions





v(t) = 3cos(3x)esin(3x)< input functions

*Simplify the expression to make life easier! This function represents the velocity function of the RC car.






v(35.7443) = 3cos(3(35.7443))esin(3(35.7443))

*From section B.I. of the solution, we take the discovered, remaining battery life of the RC car in minutes and input it into the velocity function.






v(35.7443) = 3cos(3(35.7443))esin(3(35.7443))

v(35.7443) = 4.1159 ft/min

At the end of the RC car's battery life, the instantaneous velocity was 4.1159 ft/min.

< don't forget units!

Finish!

Technology

Volumes Question1