The surface area of a cube is given in terms of its sides as

$\displaystyle A=6s^2$

This relation can be used to also relate rates of change. This can be done by taking the derivative of each side of the equation with respect to time:

$\displaystyle \frac{dA}{dt}=12s\frac{ds}{dt}$

We are given the rate of change of the sides $\displaystyle \frac{ds}{dt}=0.5$ and the lengths of the sides $\displaystyle s=4$. Thus the rate of growth of the surface area is

$\displaystyle \frac{dA}{dt}=12(4)(0.5)$

$\displaystyle \frac{dA}{dt}=24$

The area of a trapezoid is given by the formula

$\displaystyle A=\frac{1}{2}(a+b)h$

Let a represent the short side and b the long side.

For this problem, while not necessary, it may make things easier to expand this equation

$\displaystyle A=\frac{1}{2}ah+\frac{1}{2}bh$

Now, we can relate rates of change by taking the derivative of each side of the equation with respect to time:

$\displaystyle \frac{dA}{dt}=\frac{1}{2}a\frac{dh}{dt}+\frac{1}{2}h\frac{da}{dt}+\frac{1}{2}b\frac{dh}{dt}+\frac{1}{2}h\frac{db}{dt}$

Now recall that the height is constant. This means that $\displaystyle \frac{dh}{dt}=0$, which shortens the equation

$\displaystyle \frac{dA}{dt}=\frac{1}{2}h\frac{da}{dt}+\frac{1}{2}h\frac{db}{dt}$

$\displaystyle \frac{dA}{dt}=\frac{1}{2}h(\frac{da}{dt}+\frac{db}{dt})$

$\displaystyle \frac{dA}{dt}=\frac{1}{2}(4)(0.2+0.3)$

$\displaystyle \frac{dA}{dt}=1$

The surface area of a rectangular prism is given by the formula:

$\displaystyle A=2wl+2wh+2lh$

The rate of change of parameters can be found by taking the derivative of both sides of the equation with respect to time:

$\displaystyle \frac{dA}{dt}=2w\frac{dl}{dt}+2l\frac{dw}{dt}+2w\frac{dh}{dt}+2h\frac{dw}{dt}+2l\frac{dh}{dt}+2h\frac{dl}{dt}$

Now, we've been given some rates of change, namely $\displaystyle \frac{dw}{dt}=\frac{dl}{dt}=0.1$. Since the height is constant $\displaystyle \frac{dh}{dt}=0$, and the equation reduces to

$\displaystyle \frac{dA}{dt}=2w\frac{dl}{dt}+2l\frac{dw}{dt}+2h\frac{dw}{dt}+2h\frac{dl}{dt}$

Plugging in our other known values $\displaystyle w=2;l=3;h=8$, we can solve for the rate of change of the surface area

$\displaystyle \frac{dA}{dt}=2(2)(0.1)+2(3)(0.1)+2(8)(0.1)+2(8)(0.1)$

$\displaystyle \frac{dA}{dt}=0.4+0.6+1.6+1.6$

$\displaystyle \frac{dA}{dt}=4.2$

Although we are told that the sides of the cube have different rates of growth (meaning it won't stay a cube for long), at the time of this problem, since the shape is a cube it means all of the sides are equal:

$\displaystyle s_1=s_2=s_3=s$

The volume of a cube is given by the expression:

$\displaystyle V=s^3$

We'll be using this to find the current lengths of the sides

$\displaystyle s^3=64$

$\displaystyle s=4$

Now, rewrite the volume equation in terms of the three sides:

$\displaystyle V=s_1s_2s_3$

The rate of change of the volume can be found by taking the derivative of each side of the equation with respect to time. Remember to take the derivative of the right side with respect to each variable:

$\displaystyle \frac{dV}{dt}=s_2s_3\frac{ds_1}{dt}+s_1s_3\frac{ds_2}{dt}+s_1s_2\frac{ds_3}{dt}$

With the three rates of change $\displaystyle 0.3$, $\displaystyle 0.4$, and $\displaystyle 0.5$ and the known length of the sides, the rate of change is

$\displaystyle \frac{dV}{dt}=(4^2)(0.3)+(4^2)(0.4)+(4^2)(0.5)$

$\displaystyle \frac{dV}{dt}=19.2$

For this problem we're given the initial radius of a sphere and the rate of change of this radius. It will be useful to find an equation for the radius itself, however.

The rate, we are told, is $\displaystyle \frac{dr(t)}{dt}=5e^{-\frac{1}{2}t}$.

To find an expression for the radius, we need to integrate this function.

$\displaystyle \int e^{at}dt=\frac{e^{at}}{a}+C; [a,C:constants]$

$\displaystyle r(t)=-10e^{-\frac{1}{2}t}+C$

To find C, the constant of integration, utilize the initial condition:

$\displaystyle r(0)=-10e^{-\frac{1}{2}(0)}+C=4$

$\displaystyle C-10=4$

$\displaystyle C=14$

So our radius expression is

$\displaystyle r(t)=14-10e^{-\frac{1}{2}t}$

Now to find the maximum radius, take the limit as time goes towards infinity:

$\displaystyle lim_{t\rightarrow \infty }14-10e^{-\frac{1}{2}t}=14-0=14$

The volume of a sphere is given by the equation:

$\displaystyle V=\frac{4}{3}\pi r^3$

Therefore, the maximum volume is

$\displaystyle V_{max}=\frac{4}{3}\pi 14^3$

$\displaystyle V_{max}=\frac{10976}{3}\pi$

The area of a rectangle in relation to its sides is given by the formula

$\displaystyle A=wl$

This rule holds even when the the sides are expressed in more complicated terms such as those of this problem.

Since $\displaystyle w(t)=2e^{2t}$ and $\displaystyle l(t)=5+e^t$

$\displaystyle A(t)=2e^{2t}(5+e^t)$

$\displaystyle A(t)=10e^{2t}+2e^{3t}$

The rate of change of the area can be found by taking the derivative of both sides of the equation with respect to time.

Recall that the rule of derivation for an exponential:

$\displaystyle d(e^{u})=du(e^u)$, where $\displaystyle u$ is any function in the exponent.

For this problem, the derivative is then

$\displaystyle \frac{dA(t)}{dt}=20e^{2t}+6e^{3t}$

This is the formula for the rate of growth of the rectangle's area.

The area of a square is given in terms of its sides as

$\displaystyle A=s^2$

For an area of $\displaystyle 9$

$\displaystyle s^2=9$

$\displaystyle s=3$

The area equation can also be used to find rates of change; take the derivative of each side of the equation with respect to time:

$\displaystyle A=s^2$

$\displaystyle \frac{dA}{dt}=2s\frac{ds}{dt}$

We're told that $\displaystyle \frac{ds}{dt}=2.5$

Therefore

$\displaystyle \frac{dA}{dt}=2(3)(2.5)$

$\displaystyle \frac{dA}{dt}=15$

Since it is known that $\displaystyle g(x)h(y)=f(y)$, differentiate implicitly on both sides w.r.t. $\displaystyle x$ (it is also helpful to notice that, since cosine is an even function, $\displaystyle \cos(-y)=\cos (y)$:

$\displaystyle -3\pi (-2e^{-2x}y+e^{-2x}\frac{dy}{dx})=-\sin y\frac{dy}{dx}$    USE PRODUCT RULE

$\displaystyle 6\pi e^{-2x}y-3\pi e^{-2x}\frac{dy}{dx}=-\sin y\frac{dy}{dx}$       FACTOR IN THE CONSTANT

$\displaystyle 6\pi e^{-2x}y = (3\pi e^{-2x}-\sin y)\frac{dy}{dx}$          COMBINE  $\displaystyle \frac{dy}{dx}$ TERMS

$\displaystyle \frac{dy}{dx}= \frac{6\pi e^{-2x}y}{3\pi e^{-2x}-\sin y}$                           DIVIDE

The volume of sphere, in terms of its radius, is defined as

$\displaystyle V=\frac{4}{3}\pi r^3$

However, in the case of the problem, we're given the lengths of the sides of a cube in which the sphere fits. Since the outside of the sphere is touching the sides walls of the cube, the length of the cube is the diameter of the sphere:

$\displaystyle s=d=2r$

$\displaystyle r=\frac{s}{2}$

$\displaystyle r=\frac{6}{2}=3$

Furthermore, the rate of growth of a sphere's radius will be half the rate of growth of it's diameter

$\displaystyle \frac{dr}{dt}=\frac{1}{2}\frac{ds}{dt}$

$\displaystyle \frac{dr}{dt}=\frac{1}{2}(0.4)=0.2$

Now returning to the volume equation

$\displaystyle V=\frac{4}{3}\pi r^3$

The rate of growth can be found by taking the derivative with respect to time:

$\displaystyle \frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

$\displaystyle \frac{dV}{dt}=4\pi (6^2)(0.2)$

$\displaystyle \frac{dV}{dt}=28.8\pi$

The area of an equilateral triangle, in terms of its side, is

$\displaystyle A=\frac{\sqrt{3}}{4}s^2$

We're told that the length of a side is $\displaystyle 2\sqrt{3}$ and the rate of change is $\displaystyle 0.5$. Now, to utilize the fact, let's find an equation for the rate of change of the area. Find this by taking the derivative of each side of the equation with respect to time:

$\displaystyle \frac{dA}{dt}=\frac{\sqrt{3}}{2}s\frac{ds}{dt}$

$\displaystyle \frac{dA}{dt}=\frac{\sqrt{3}}{2}(2\sqrt{3})(0.5)$

$\displaystyle \frac{dA}{dt}=1.5$