Geometry

[nycmathguy]

Section 1.8
Question 61

See attachment.

Let me see.

For part (a), I know that the area of a square is side squared or A = s^2, where s = x for this problem. I think the answer is r(x) = x^2.

For part (b), the area of the circular base is pi•r^2.
I think the answer is A(r) = pi•r^2.

For part (c), I must find (A ° r)(x).

I know that (A ° r)(x) = A(r(x)).

I know that r(x) = x^2 from part (a).

I conclude that A(r(x)) = A(x^2).

If I am correct, how do I interpret A(x^2)?

 

[MathLover1]

r(x)=x/2

A=r^2*pi

(A \circ r)(x)=A(r(x))
(A \circ r)(x)=A(x/2)
(A \circ r)(x)=(x/2)^2*pi
(A \circ r)(x)=(x^2/4)*pi
(A \circ r)(x)=(1/4)x^2*pi

 

[nycmathguy]

r(x)=x/2

A=r^2*pi

(A \circ r)(x)=A(r(x))
(A \circ r)(x)=A(x/2)
(A \circ r)(x)=(x/2)^2*pi
(A \circ r)(x)=(x^2/4)*pi
(A \circ r)(x)=(1/4)x^2*pi

What did I do wrong?

(A\circ r)(x) = (A ° r)(x). Yes?

 

[MathLover1]

r(x) = x^2 => that was wrong
should be
r(x)=x/2

 

[nycmathguy]

r(x) = x^2 => that was wrong
should be
r(x)=x/2

Why is r(x) = x/2, which means (1/2) times x?

 

[MathLover1]

take a look at picture, diameter of the circle is equal to the side x
=> radius is half of it

 

[nycmathguy]

take a look at picture, diameter of the circle is equal to the side x
=> radius is half of it

Yes, I see it now.