Area of Rectangle Inscribed in Arc

[nycmathguy]

Section 5.3

How is this done?
Tomorrow morning we finish Section 5.3.

 

[MathLover1]

the area of rectangle inscribed in one arc of the graph A=2x*cos(x), 0<x<pi/2

let P be a point on the curve where rectangle touches curve

coordinates are P(x,2x*cos(x))

sides of the rectangle are
2x-> the length
y=2x*cos(x)->the height

to find max area, you need take derivative of A=2x*cos(x)

(d/dx)(2x*cos(x))=2 (cos(x) - x sin(x))

equal to zero
2 (cos(x) - x sin(x))=0
cos(x) - x* sin(x)=0....use calculator
x=0.86
then area is
A=2*0.86*cos(0.86)=1.1221924452421692
A=1.122

b.

find x for what A>=1
2x*cos(x)>=1/(2x), 0<x<pi/2
cos(x)>=1/(2x)...use calculator

0.610031<= x <=1.09801

Interval notation
[0.610031, 1.09801]

 

[nycmathguy]

the area of rectangle inscribed in one arc of the graph A=2x*cos(x), 0<x<pi/2

let P be a point on the curve where rectangle touches curve

coordinates are P(x,2x*cos(x))

sides of the rectangle are
2x-> the length
y=2x*cos(x)->the height

to find max area, you need take derivative of A=2x*cos(x)

(d/dx)(2x*cos(x))=2 (cos(x) - x sin(x))

equal to zero
2 (cos(x) - x sin(x))=0
cos(x) - x* sin(x)=0....use calculator
x=0.86
then area is
A=2*0.86*cos(0.86)=1.1221924452421692
A=1.122

b.

find x for what A>=1
2x*cos(x)>=1/(2x), 0<x<pi/2
cos(x)>=1/(2x)...use calculator

0.610031<= x <=1.09801

Interval notation
[0.610031, 1.09801]

Impressive as always. My question remains:

How did you know what to do?

 

[MathLover1]

To find the value of x that gives an area A maximum, we need to find the first derivative dA/dx (A is a function of x). If A has a maximum value, it happens at x such that dA/dx = 0

 

[nycmathguy]

To find the value of x that gives an area A maximum, we need to find the first derivative dA/dx (A is a function of x). If A has a maximum value, it happens at x such that dA/dx = 0

The derivative? This problem is not from a calculus textbook. Perhaps there's a precalculus approach to find the answer.

 

[MathLover1]

you can approximate using a graph
as you can see, x is half way between 0 and pi/2
so x=pi/4
the length of rectangle is 2x=2(pi/4)=pi/2, the height is x=pi/4

A=(pi/2)(pi/4)=1.2337005501

 

[MathLover1]

also, you have derivative in pre calculus

take a look
https://philschatz.com/precalculus-book/contents/m49455.html

 

[nycmathguy]

also, you have derivative in pre calculus

take a look
https://philschatz.com/precalculus-book/contents/m49455.html

Ok. I will take a look. Thanks.

 

[nycmathguy]

you can approximate using a graph
as you can see, x is half way between 0 and pi/2
so x=pi/4
the length of rectangle is 2x=2(pi/4)=pi/2, the height is x=pi/4

A=(pi/2)(pi/4)=1.2337005501

We can only approximate the area via graphing. I will keep this information for my calculus study time next year, hopefully, by April.