SOLUTION 1: Draw a square with edges labeled $x$, and assume each edge is a function of time $t$.

$\ \ \ \$ a.) The perimeter of the square is $$P=x+x+x+x \ \ \ \ \longrightarrow$$ $$P=4x$$ GIVEN: $\ \ \ \displaystyle{ dx \over dt } = 3 \ cm/sec.$

FIND: $\ \ \ \displaystyle{ dP \over dt }$ when $x=10 \ cm.$

Now differentiate the perimeter equation with repect to time $t$, getting

$$D \{ P \} = D \{4x\} \ \ \ \longrightarrow$$ $$\displaystyle{ dP \over dt } = 4 \displaystyle{ dx \over dt } \ \ \ \longrightarrow$$

$\Big($ Now let $\displaystyle{ dx \over dt } = 3. \Big)$

$$\displaystyle{ dP \over dt } = 4 (3) = 12 \ cm/sec.$$

$\ \ \ \$ b.) The area of the square is $$A=(base)(height)= x \cdot x \ \ \ \ \longrightarrow$$ $$A=x^2$$ GIVEN: $\ \ \ \displaystyle{ dx \over dt } = 3 \ cm/sec.$

FIND: $\ \ \ \displaystyle{ dA \over dt }$ when $x=10 \ cm.$

Now differentiate the area equation with respect to time $t$, getting

$$D \{ A \} = D \{x^2 \} \ \ \ \longrightarrow$$ $$\displaystyle{ dA \over dt } = 2x \displaystyle{ dx \over dt } \ \ \ \longrightarrow$$

$\Big($ Now let $\displaystyle{ dx \over dt } = 3$ and $x=10. \Big)$

$$\displaystyle{ dA \over dt } = 2 (10) (3) = 60 \ cm^2/sec.$$