SOLUTION 9: Consider the given isosceles triangle of base $10$ and edges $x$, and assume that $x$ is a function of time $t$.

$\ \ \ \$ a.) The perimeter of the triangle is $$P= x+x+10 \ \ \ \ \longrightarrow$$ $$P= 2x+10$$ GIVEN: $\ \ \ \displaystyle{ dx \over dt } = 4 \ in/min.$

FIND: $\ \ \ \displaystyle{ dP \over dt }$ when $x=13 \ in.$

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

$$D \{ P \} = D \{ 2x+10 \} \ \ \ \longrightarrow$$ $$\displaystyle{ dP \over dt } = 2 \displaystyle{ dx \over dt } \ \ \ \longrightarrow$$

$\Big($ Now let $\displaystyle{ dx \over dt } = 4. \Big)$

$$\displaystyle{ dP \over dt } = 2 (4) \ \ \ \longrightarrow$$ $$\displaystyle{ dP \over dt } = 8 \ in^2/min.$$

$\ \ \ \$ b.) Consider the given isosceles triangle of base $10$, edges $x$, and height $h$, and assume that $x$ and $h$ are functions of time $t$. Divide the triangle into two right triangles of height $h$.

Using the Pythagorean Theorem we get the following height equation: $$5^2+h^2=x^2 \ \ \ \longrightarrow$$ $$h^2=x^2 - 25$$ GIVEN: $\ \ \ \displaystyle{ dx \over dt } = 4 \ in/min.$

FIND: $\ \ \ \displaystyle{ dh \over dt }$ when $x=13 \ in.$

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

$$D \{ h^2 \} = D \{ x^2 - 25 \} \ \ \ \ \longrightarrow$$ $$2h \displaystyle{ dh \over dt } = 2x \displaystyle{ dx \over dt } \ \ \ \ \longrightarrow$$

$\Big($ Multiply both sides of the equation by $1/2 . \Big)$

$$h \displaystyle{ dh \over dt } = x \displaystyle{ dx \over dt } \ \ \ \ \longrightarrow$$

$\Big($ Let $\ x=13 \$ in the height equation $\ 25+h^2=x^2 \ ,$ getting $\ 25 + h^2 = 13^2=169 \ \ \rightarrow \ \ \ h^2=144 \ \ \rightarrow \ h=12 .$ Now let $\displaystyle{ dx \over dt } = 4, x=13,$ and $h =12. \Big)$

$$(12) \displaystyle{ dh \over dt } = (13)(4) \ \ \ \longrightarrow$$ $$\displaystyle{ dh \over dt } = {52 \over 12 } \ \ \ \longrightarrow$$ $$\displaystyle{ dh \over dt } = { 13 \over 3 } \ in/min.$$

$\ \ \ \$ c.) Consider the given isosceles triangle of base $10$, edges $x$, and height $h$, and assume that $x$ and $h$ are functions of time $t$. The area of the isosceles triangle is $$A= (1/2)(base)(height) \ \ \ \ \longrightarrow$$ $$A= (1/2)(10)h \ \ \ \ \longrightarrow$$ $$A= 5h \ \ \ \ \longrightarrow$$

GIVEN: $\ \ \ \displaystyle{ dx \over dt } = 4 \ in/min.$

FIND: $\ \ \ \displaystyle{ dA \over dt }$ when $x=13 \ in.$

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

$$D \{ A \} = D \{ 5h \} \ \ \ \longrightarrow$$ $$\displaystyle{ dA \over dt } = 5 \displaystyle{ dh \over dt } \ \ \ \ \longrightarrow$$

$\Big($ See the solution to solution 9 part b. and let $\displaystyle{ dh \over dt } = { 13 \over 3 }. \Big)$

$$\displaystyle{ dA \over dt } = 5 \Big({ 13 \over 3 }\Big) \ \ \ \ \longrightarrow$$ $$\displaystyle{ dV \over dt } = { 65 \over 3 } \ in^2/min.$$