2.6 Question 5

[Eigenvalue]

Precalculus
Michael Sullivan
Edition 4
Chapter 2, Section 2.6

A right triangle has one vertex on the graph of y=x³, x>0 at (x,y), another at the origin, and the third in the positive y-axis at (0,y). Express the area A of the triangle as a function of x .

See attachment.

[Eigenvalue]

(x,y) can be rewritten as (x, x³), as y=x³

The shorter leg of the right triangle, from (0,y) to (x,y) is x and the longer side of the triangle, from (0,0) to (0,y)=x³

Area=[tex]\frac{1}{2}[/tex]bh
=[tex]\frac{1}{2}[/tex]x*x³
=[tex]\frac{1}{2}[/tex]x⁴

[nycmath]

We want A(x).

Area of triangle = (1/2)(base)(heght)

Let x = base

Let y = height

Let y = [tex]x^{3 }[/tex]

A(x) = [tex]\frac{1}{2}[/tex]xy

A(x) = [tex]\frac{1}{2}[/tex]x•[tex]x^{3}[/tex]

A(x) = [tex]\frac{1}{2} x^{4 }[/tex]

[nycmath]

(x,y) can be rewritten as (x, x³), as y=x³

The shorter leg of the right triangle, from (0,y) to (x,y) is x and the longer side of the triangle, from (0,0) to (0,y)=x³

Area=[tex]\frac{1}{2}[/tex]bh
=[tex]\frac{1}{2}[/tex]x*x³
=[tex]\frac{1}{2}[/tex]x⁴

If we answer the same questions, section 2.6 will be a waste of time.