The volume of a sphere is given by the equation

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

Since the radius is given as 

\(\displaystyle r(t)=6-4e^{-\frac{1}{2}t}\)

The volume can be expressed as:

\(\displaystyle V(t)=\frac{4}{3}\pi(6-4e^{-\frac{1}{2}t})^3\)

The rate of change of the volume can then, in turn, by taking the derivative of each side of the equation with respect to time.

Now, we could expand the function outwards, actually cubing it; however this is not necessary. We need only utilize the following derivative rules:

\(\displaystyle d(u^n)=nu^{n-1}du;(n=constant)\)

\(\displaystyle d(e^u)=e^udu\)

These rules hold even when u is a complex function; just don't assume du is 1!

Following these rules, we can find the derivative:

\(\displaystyle \frac{dV(t)}{dt}=\frac{4}{3}\pi(3)(2e^{-\frac{1}{2}t})(6-4e^{-\frac{1}{2}t})^2\)

\(\displaystyle \frac{dV(t)}{dt}=4\pi(2e^{-\frac{1}{2}t})(6-4e^{-\frac{1}{2}t})^2\)

At time \(\displaystyle t=\ln(9)\):

\(\displaystyle \frac{dV(\ln(9))}{dt}=4\pi(2e^{-\frac{1}{2}(\ln(9))})(6-4e^{-\frac{1}{2}(\ln(9))})^2\)

\(\displaystyle \frac{dV(ln(9))}{dt}=\frac{1568}{27}\pi\)

The hypotenuse of a triangle can be found in terms of the triangle's legs using the Pythagorean Theorem:

\(\displaystyle c^2=a^2+b^2\)

\(\displaystyle c=\sqrt{a^2+b^2}\)

This equation also enables us to find the rate of change of the hypotenuse; we need only take the derivative with respect to time.

The legs are given as \(\displaystyle a(t)=6-3e^{-2t}\) and \(\displaystyle b(t)=5-e^{-3t}\).

Plugging this into the equation for the hypotenuse, we find

\(\displaystyle c(t)=\sqrt{(6-3e^{-2t})^2+(5-e^{-3t})^2}\)

Now, to take the derivative of this equation, we'll utilize the following rules:

\(\displaystyle d(e^u)=du(e^u)\)

\(\displaystyle d(u^n)=n(u^{n-1})du; (n:\uptext{constant})\)

Note that u can represent a complicated function; the du term should not be dismissed!  Consideration of it is known as the chain rule.

With these rules in mind, let's consider the derivative. First, let's rewrite the hypotenuse equation in terms of exponents rather than the square root symbol:

\(\displaystyle c(t)=[(6-3e^{-2t})^2+(5-e^{-3t})^2]^{\frac{1}{2}}\)

Now let's take the derivative:

\(\displaystyle \frac{dc(t)}{dt}=\frac{1}{2}\frac{2(6e^{-2t})(6-3e^{-2t})+2(3e^{-3t})(5-e^{-3t})}{[(6-3e^{-2t})^2+(5-e^{-3t})^2]^{\frac{1}{2}}}\)

Notice how the chain rule was utilized for the functions inside of the square root as well, for example the first term: \(\displaystyle \frac{d}{dt}((6-3e^{-2t})^2)=2(6e^{-2t})(6-3e^{-2t})\)

It may seem complicated, but being mindful of order will serve well here.

Now, the rate of change of the hypotenuse at time \(\displaystyle t=\ln(2)\) is

\(\displaystyle \frac{dc(\ln(2)))}{dt}=\frac{1}{2}\frac{2(6e^{-2(\ln(2))})(6-3e^{-2(\ln(2))})+2(3e^{-3(\ln(2))})(5-e^{-3(\ln(2))})}{[(6-3e^{-2(\ln(2))})^2+(5-e^{-3(\ln(2))})^2]^{\frac{1}{2}}}\)

\(\displaystyle \frac{dc(\ln(2)))}{dt}=1.35\)

The area of a circle, in terms of its radius, is given by the function:

\(\displaystyle A=\pi r^2\)

Since the radius is given by the formula

\(\displaystyle r(t)=t^2+\cos(t)+4\)

The area in turn can be expressed as

\(\displaystyle A(t)=\pi(t^2+\cos(t)+4)^2\)

We're looking for the rate of change of this area now; this can be found by taking the derivative of each side of the equation with respect to time.

To do perform this derivative recall the following derivative properties:

\(\displaystyle d(u^n)=n(u^{n-1})du\)

\(\displaystyle d(\cos(u))=-du(\sin(u))\)

Where \(\displaystyle u\) can represent a complex function.

Following these rules, the rate of change of the area can be found to be

\(\displaystyle \frac{dA(t)}{dt}=2\pi (2t-\sin(t))(t^2+\cos(t)+4)\)

At time \(\displaystyle t=\frac{\pi}{2}\)

\(\displaystyle \frac{dA(\frac{\pi}{2})}{dt}=2\pi (2(\frac{\pi}{2})-\sin(\frac{\pi}{2}))((\frac{\pi}{2})^2+\cos(\frac{\pi}{2})+4)\)

\(\displaystyle \frac{dA(\frac{\pi}{2})}{dt}=87.03\)

The surface area of a cube is given by the formula

\(\displaystyle A=6s^2\)

For sides expressed by the formula

\(\displaystyle s(t)=5-2e^{-2t}\)

The area equation becomes

\(\displaystyle A(t)=6(5-2e^{-2t})^2\)

To  find the rate of change of the surface area, take the derivative of each side of equation with respect to time. We'll utilize the following rules:

\(\displaystyle d(u^n)=nu^{n-1}du; (n:constant)\)

\(\displaystyle d(e^u)=e^udu\)

Note that u can be a complex function.

Thus the derivative is 

\(\displaystyle \frac{dA(t)}{dt}=12(4e^{-2t})(5-2e^{-2t})\)

At time \(\displaystyle t=\ln(4)\)

\(\displaystyle \frac{dA(\ln(4))}{dt}=12(4e^{-2(\ln(4))})(5-2e^{-2(\ln(4))})\)

\(\displaystyle \frac{dA(\ln(4))}{dt}=\frac{117}{8}\)

The surface area of a sphere is given by the formula

\(\displaystyle A=4\pi r^2\)

We're told that \(\displaystyle r=3\) and \(\displaystyle \frac{dr}{dt}=0.9\)

Now, to find the rate of change of the surface area, take the derivative of the area equation with respect to time:

\(\displaystyle \frac{dA}{dt}=8\pi r \frac{dr}{dt}\)

\(\displaystyle \frac{dA}{dt}=8\pi (3)(0.9)\)

\(\displaystyle \frac{dA}{dt}=21.6\pi\)

We're given a rate of change for the circle's radius

\(\displaystyle \frac{dr(t)}{dt}=10e^{-\frac{1}{2}t}\)

The function for the radius can be found by integrating this function

\(\displaystyle \int e^{nt}dt=\frac{e^{nt}}{n}+C\)

\(\displaystyle r(t)=-20e^{-\frac{1}{2}t}+C\)

To find the constant of integration, use the initial condition

\(\displaystyle r(0)=-20e^{0}+C=6\)

\(\displaystyle -20+C=6\)

\(\displaystyle C=26\)

\(\displaystyle r(t)=-20e^{-\frac{1}{2}t}+26\)

Now, the formula for a circle's area is

\(\displaystyle A=\pi r^2\)

To find the rate of change, take the derivative with respect to time:

\(\displaystyle \frac{dA}{dt}=2\pi r \frac{dr}{dt}\)

\(\displaystyle \frac{dA(t)}{dt}=2\pi (26-20e^{-\frac{1}{2}t})(10e^{-\frac{1}{2}t})\)

\(\displaystyle \frac{dA(t)}{dt}=2\pi (26-20e^{-\frac{1}{2}(ln(4))})(10e^{-\frac{1}{2}(ln(4))})\)\(\displaystyle \frac{dA(t)}{dt}=160 \pi\)

We're told that the rate of change of the square's sides is \(\displaystyle 5-5e^{-t}\) which is to say

\(\displaystyle \frac{ds(t)}{dt}=5-5e^{-t}\)

The formula for the sides themselves can be found by integrating this equation

\(\displaystyle \int (a+e^{bt})dt=at+\frac{e^{bt}}{b}+C;(a,b,C:constants)\)

\(\displaystyle s(t)=5t+5e^{-t}+C\)

To find this constant of integration, we need to know the value of the equation at some point of time. We do not know the initial value of the sides, but we know the initial value of the area which enables us to calculate this:

\(\displaystyle A_0=s_0^2=36\)

\(\displaystyle s_0=6\)

\(\displaystyle s(0)=5(0)+5e^{0}+C=6\)

\(\displaystyle 5+C=6\)

\(\displaystyle C=1\)

\(\displaystyle s(t)=5t+5e^{-t}+1\)

Now let's return to the area equation

\(\displaystyle A=s^2\)

The rate of change of the area can be found by taking the derivative of this equation with respect to time:

\(\displaystyle \frac{dA}{dt}=2s\frac{ds}{dt}\)

\(\displaystyle \frac{dA(t)}{dt}=2(5t+5e^{-t}+1)(5-5e^{-t})\)

\(\displaystyle \frac{dA(ln(5))}{dt}=2(5ln(5)+5e^{-ln(5)}+1)(5-5e^{-ln(5)})\)

\(\displaystyle \frac{dA(ln(5))}{dt}=16+40ln(5)\)

The rates of change of the width and length of a rectangle are \(\displaystyle 3e^{-t}\) and \(\displaystyle 6e^{-t}\), which is to say

\(\displaystyle \frac{dw(t)}{dt}=3e^{-t}\)

\(\displaystyle \frac{dl(t)}{dt}=6e^{-t}\)

These equations can be used to find formulae for the width and length by taking the integral of each:

\(\displaystyle \int e^{nt}dt=\frac{e^{nt}}{n}+C\)

\(\displaystyle w(t)=-3e^{-t}+C_w\)

\(\displaystyle l(t)=-6e^{-t}+C_l\)

These constants of integration can be found by using the initial conditions:

\(\displaystyle w(0)=-3e^{0}+C_w=4\)

\(\displaystyle -3+C_w=4\)

\(\displaystyle C_w=7\)

\(\displaystyle l(0)=-6e^{0}+C_l=8\)

\(\displaystyle -6+C_l=8\)

\(\displaystyle C_l=14\)

\(\displaystyle w(t)=-3e^{-t}+7\)

\(\displaystyle l(t)=-6e^{-t}+14\)

Now the area of a rectangle is given by the formula

\(\displaystyle A=wl\)

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

\(\displaystyle \frac{dA}{dt}=w\frac{dl}{dt}+l\frac{dw}{dt}\)

\(\displaystyle \frac{dA(t)}{dt}=(7-3e^{-t})(6e^{-t})+(14-6e^{-t})(3e^{-t})\)

\(\displaystyle \frac{dA(ln(3))}{dt}=(7-3e^{-ln(3)})(6e^{-ln(3)})+(14-6e^{-ln(3)})(3e^{-ln(3)})\)

\(\displaystyle \frac{dA(ln(3))}{dt}=24\)

The rate of change of the sides of a square is \(\displaystyle 5-3sin(t)\), which is to say

\(\displaystyle \frac{ds(t)}{dt}=5-3sin(t)\).

A formula for the length of the sides themselves can be found by integrating this equation with respect to time:

\(\displaystyle \int a +bsin(ct)=at-\frac{bcos(ct)}{c}+C;(a,b,c,C:constants)\)

\(\displaystyle s(t)=5t+3cos(t)+C\)

To find this constant of integration, use the initial condition given

\(\displaystyle s(0)=5(0)+3cos(0)+C=8\)

\(\displaystyle 3+C=8\)

\(\displaystyle C=5\)

\(\displaystyle s(t)=5t+3cos(t)+5\)

Now, the area of a square is given by the equation

\(\displaystyle A=s^2\)

The rate of change of the area can be found by taking the derivative of this equation with respect to time:

\(\displaystyle \frac{dA}{dt}=2s\frac{ds}{dt}\)

\(\displaystyle \frac{dA(t)}{dt}=2(5t+3cos(t)+5)(5-3sin(t))\)

\(\displaystyle \frac{dA(\pi)}{dt}=2(5\pi+3cos(\pi)+5)(5-3sin(\pi))\)

\(\displaystyle \frac{dA(\pi)}{dt}=20+50\pi\)

The rate of change of the radius of a circle is given by the equation \(\displaystyle 2+cos(t)\) which is to say

\(\displaystyle \frac{dr(t)}{dt}=2+cos(t)\)

The equation for the length of the radius can be found by taking the integral with respect to time:

\(\displaystyle \int (a+bcos(ct))dt=at+\frac{bsin(ct)}{c}+C;(a,b,c,C:constants)\)

\(\displaystyle r(t)=2t+sin(t)+C\)

This constant of integration can be found by using the initial condition:

\(\displaystyle r(0)=2(0)+sin(0)+C=3\)

\(\displaystyle C=3\)

\(\displaystyle r(t)=2t+sin(t)+3\)

The area of a circle is given by the equation

\(\displaystyle A=\pi r^2\)

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

\(\displaystyle \frac{dA}{dt}=2\pi r \frac{dr}{dt}\)

\(\displaystyle \frac{dA(t)}{dt}=2\pi (2t+sin(t)+3)(2+cos(t))\)

\(\displaystyle \frac{dA(2\pi)}{dt}=2\pi (2(2\pi)+sin(2\pi)+3)(2+cos(2\pi))\)

\(\displaystyle \frac{dA(2\pi)}{dt}=18\pi+24\pi^2\)