Begin by finding the hypotenuse using the Pythagorean theorem:

$\displaystyle c^2=a^2+b^2$

$\displaystyle c^2=5^2in^2+12^2in^2$

$\displaystyle c^2=169in^2$

$\displaystyle c=13in$

Now, to relate the rates of growth, take the time derivative of the Pythagorean Theorem:

$\displaystyle 2c\frac{dc}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}$

$\displaystyle \frac{dc}{dt}=\frac{a\frac{da}{dt}+b\frac{db}{dt}}{c}$

$\displaystyle \frac{dc}{dt}=\frac{5in(0.6\frac{in}{s})+12in(0.6\frac{in}{s})}{13in}$

$\displaystyle \frac{dc}{dt}=0.79\frac{in}{s}$

The areas of the circle and square can be written in terms of the radius of the circle as follows:

$\displaystyle A_{circle}=\pi r^2 = 50\pi in^2$

$\displaystyle A_{square}=(\frac{2r}{\sqrt2})^2=2r^2=50in^2$

The area of the square follows from the fact that the diameter of the circle forms the diagonal of the square, and each side is equal to the diagonal divided by the square root of two. This can be verified via the Pythagorean Theorem $\displaystyle (s^2+s^2=d^2)$.

Now, the rate of change of areas can be found by taking the derivative of each function with respect to time:

$\displaystyle \frac{dA_{circle}}{dt}=2\pi r \frac{dr}{dt}$

$\displaystyle \frac{dA_{square}}{dt}=4r \frac{dr}{dt}$

Therefore, the rate of growth of the difference in the two areas can be found by finding the difference of these two quantities:

$\displaystyle 2r(\pi-2)\frac{dr}{dt}$

$\displaystyle 2(5\sqrt2in)(\pi-2)(0.7\frac{in}{s})$

$\displaystyle 7\sqrt2(\pi-2)\frac{in^2}{s}$

The area of a square is the square of its sides:

$\displaystyle A=s^2$

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

At time $\displaystyle t=2$

$\displaystyle A(2)=(5+4)^2$

$\displaystyle A=81$

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

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

Since $\displaystyle s=5+t^2$, we will use the power rule which states,

$\displaystyle x^n\ dx=nx^{n-1}$

therefore

$\displaystyle \frac{ds}{dt}=2t$, giving:

$\displaystyle \frac{dA(t)}{dt}=2(5+t^2)(2t)$

$\displaystyle \frac{dA(2)}{dt}=2(5+2^2)(2(2))$

$\displaystyle \frac{dA(2)}{dt}=72$

Note that the same result would be found if you took the derivative of the equation wherein the $\displaystyle 5+t^2$ was substituted in for $\displaystyle s$:

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

Use the power rule and chain rule to find the derivative of area with respect to time.

Chain Rule: $\displaystyle f(g(x)) dx=f'(g(x))\cdot g'(x)$

$\displaystyle \frac{dA(t)}{dt}=2(2t)(5+t^2)$

The area of the rectangle is given by the product of its sides:

$\displaystyle A=wl$

$\displaystyle A(t)=(2+t^3)(1+t^2)$

$\displaystyle A(t)=t^5+t^3+2t^2+2$

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

We will use the power rule which states,

$\displaystyle x^n\ dx=nx^{n-1}$

to find the following derivative.

$\displaystyle \frac{dA(t)}{dt}=5t^4+3t^2+4t$

$\displaystyle \frac{dA(3)}{dt}=5(81)+3(9)+4(3)$

$\displaystyle \frac{dA(3)}{dt}=405+27+12$

$\displaystyle \frac{dA(3)}{dt}=444$

The area of a circle is given by the equation:

$\displaystyle A=\pi r^2$

$\displaystyle A(t)=\pi(3+t^4)^2$

Find the time that satisfies the area given in the problem:

$\displaystyle 361\pi=\pi(3+t^4)^2$

$\displaystyle (3+t^4)^2=361$

$\displaystyle 3+t^4=19$

$\displaystyle t^4=16$

$\displaystyle t=2$

Now that the time is known, to find the rate of area growth, take the derivative of the area equation with respect to time:

We will use the power rule which states,

$\displaystyle x^n\ dx=nx^{n-1}$

and the chain rule,

$\displaystyle f(g(x))dx=f'(g(x))\cdot g'(x)$

to find the following derivative.

$\displaystyle \frac{dA(t)}{dt}=2\pi(4t^3)(3+t^4)$

$\displaystyle \frac{dA(2)}{dt}=2\pi(4(2)^3)(3+(2)^4)$

$\displaystyle \frac{dA(2)}{dt}=2\pi(32)(19)$

$\displaystyle \frac{dA(2)}{dt}=1216\pi$

The length of the hypotenuse of a right triangle is given by the Pythagorean Theorem:

$\displaystyle c^2=a^2+b^2$

$\displaystyle c^2(t)=(2+t^2)^2+(t^3)^2$

$\displaystyle c(t)=\sqrt{t^6+t^4+4t^2+4}$

Therefore the rate of change of the hypotenuse can be found by taking the derivative of the equation:

We will use the power rule which states,

$\displaystyle x^n\ dx=nx^{n-1}$ and the chain rule,

$\displaystyle f(g(x))dx=f'(g(x))\cdot g'(x)$

to find the following derivative.

$\displaystyle \frac{dc(t)}{dt}=\frac{6t^5+4t^3+8t}{2\sqrt{t^6+t^4+4t^2+4}}$

$\displaystyle \frac{dc(2)}{dt}=\frac{192+32+16}{2\sqrt{64+16+16+4}}$

$\displaystyle \frac{dc(2)}{dt}=\frac{240}{2\sqrt{100}}$

$\displaystyle \frac{dc(2)}{dt}=12$

The volume of a sphere is given by the expression:

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

$\displaystyle V(t)=\frac{4}{3}\pi (1+2t^2+t^3)^3$

The rate of change of the volume can be find by taking the derivative of each side of the equation with respect to time:

We will use the chain rule,

$\displaystyle f(g(x))dx=f'(g(x))\cdot g'(x)$

to find the following derivative.

$\displaystyle \frac{dV(t)}{dt}=4\pi (4t+3t^2)(1+2t^2+t^3)^2$

$\displaystyle \frac{dV(2)}{dt}=4\pi (20)(17)^2$

$\displaystyle \frac{dV(2)}{dt}=23120\pi$

We are given the rate of change of the cube's volume:

$\displaystyle \frac{dV(t)}{dt}=3+t^2$

The volume equation can be found by taking the integral of this function with respect to time:

We will use the following rule which states,

$\displaystyle \int x^n\ dx=\frac{x^{n+1}}{n+1}$ to find the following integral. 

$\displaystyle V(t)=3t+\frac{1}{3}t^3+C$

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

$\displaystyle V(0)=3(0)+\frac{1}{3}(0^3)+C=9$

$\displaystyle C=9$

Giving the volume as a funciton of time:

$\displaystyle V(t)=3t+\frac{1}{3}t^3+9$

Now, find the volume at time $\displaystyle t=3$:

$\displaystyle V(3)=3(3)+\frac{1}{3}3^3+9$

$\displaystyle V(3)=9+9+9$

$\displaystyle V(3)=27$

From this we can find the length of a side at this time:

$\displaystyle V(t)=s(t)^3$

$\displaystyle s(3)^3=V(3)=27$

$\displaystyle s(3)=3$

Now, let's return the volume equation $\displaystyle V(t)=s(t)^3$; we can relate rates of change by taking the derivative of each side with respect to time:

$\displaystyle \frac{dV(t)}{dt}=3s(t)^2\frac{ds(t)}{dt}$

At time $\displaystyle t=3$

$\displaystyle \frac{dV(3)}{dt}=3s(3)^2\frac{ds(3)}{dt}$

$\displaystyle \frac{ds(3)}{dt}=\frac{3+t^2}{3s(3)^2}$

$\displaystyle \frac{ds(3)}{dt}=\frac{12}{27}$ (Recall that $\displaystyle s(3)=3$)

$\displaystyle \frac{ds(3)}{dt}=\frac{4}{9}$

To find the rate of change of a line connected by two points.  We will use the following equation for slope.

$\displaystyle m=\frac{\Delta y}{\Delta x}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\displaystyle \frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{28-7}{4-1}=\frac{21}{3}=7$

The rate of change of the line is $\displaystyle 7$

The rate at which $\displaystyle z$ changes is $\displaystyle \frac{dz}{dt}$.  The find $\displaystyle \frac{dz}{dt}$, we will find the first derivative of the equation

$\displaystyle z=x^{3}-2y^{2}$  

$\displaystyle \frac{d}{dt}(z)=\frac{d}{dt}\left [x^{3}-2y^{2} \right ]=3x^{2}\frac{dx}{dt}-4y\frac{dy}{dt}$

We are given the following parameters

$\displaystyle x=3$, $\displaystyle y=4$, $\displaystyle \frac{dx}{dt}=5 units/s$, and $\displaystyle \frac{dy}{dt}=7 units/s$.

Subsituting these values into our derivative equation

$\displaystyle \frac{dz}{dt}=3x^{2}\frac{dx}{dt}-4y\frac{dy}{dt}$

$\displaystyle \frac{dz}{dt}=3(3)^{2}(5)-4(4)(7)=135-112=23$