In order to solve this problem, we must first know that the area for a rectangle is \(\displaystyle A=l*w\).  In order to find the rate at which the area of the rectangle changes with respect to time, we simply take the derivative of the formula for the area of a rectangle with respect to time.  In order to do so, we must apply the product rule, 

\(\displaystyle \frac{d}{dt}u*v=u'v+uv'\).  

Applying the product rule, the derivative becomes 

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

Because we know that  

\(\displaystyle l=50,w=40,\frac{dw}{dt}=-0.5,\frac{dl}{dt}=2\), we simply plug those into the equation to solve for \(\displaystyle \frac{dA}{dt}\).  

\(\displaystyle \frac{dA}{dt}=-0.5*50+2*40=55\frac{ft}{sec}\)

In order to solve this problem, we must know that the circumfrence of a circle is equivalent to \(\displaystyle C=2\pi r\).  

Therefore by taking the derivative with respect to time, we obtain 

\(\displaystyle \frac{dC}{dt}=2\pi\frac{dr}{dt}\).  

Given that we know that 

\(\displaystyle \frac{dC}{dt}=0.6\), we can solve for \(\displaystyle \frac{dr}{dt}\).

\(\displaystyle 0.6=2\pi\frac{dr}{dt}\)

\(\displaystyle \frac{dr}{dt}=\frac{3}{10\pi}\)

In order to solve for how fast the area of the circle changes, we must know that the area of a circle is defined as \(\displaystyle A=\pi r^2\).  

Taking the derivative with respect to time, we obtain 

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

We know that \(\displaystyle r=40,\frac{dr}{dt}=\frac{3}{10\pi}\), so by plugging in all the variables, we can solve for \(\displaystyle \frac{dA}{dt}\).

\(\displaystyle \frac{dA}{dt}=2\pi(40)\left(\frac{3}{10\pi}\right)\)

\(\displaystyle \frac{dA}{dt}=24 \frac{ft}{sec}\).

The rate of change of a function is its derivative.

Recall the following rules of differentiation to help solve this problem.

Power Rule: 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}(at^{n}) = nat^{n-1}\) 

 Differentiation rule for tangent: 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}(atan(bt)) = absec^2(bt)\)

Therefore, the rate of change of \(\displaystyle f(t)\), by the power rule and the differentiation rule for tangent, is 

\(\displaystyle f'(t) = sec^2(t) + 8t\).

The area of a square is given by the formula:

\(\displaystyle A=l^2\)

The rate of growth of the area can be related to the rate of growth of sides by differentiating each side with respect to time:

\(\displaystyle \frac{dA}{dt}=\frac{d}{dt}l^2=2l\frac{dl}{dt}\)

Therefore, the rate of growth of the square is:

\(\displaystyle \frac{dA}{dt}=2(10in)(-0.01\frac{in}{s})=-0.2\frac{in^2}{s}\)

The area of circle in terms of its radius is given as:

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

This can be rewritten to find the radius:

\(\displaystyle r=\sqrt{\frac{A}{\pi}}\)

In this problem, the radius can be found to be:

\(\displaystyle r=\sqrt{\frac{49\pi}{\pi}}=7\)

Now, relate rates of change by deriving each side of the area equation with respect to time:

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

Solve then for \(\displaystyle \frac{dr}{dt}\):

\(\displaystyle \frac{dr}{dt}=\frac{1}{2\pi r}\frac{dA}{dt}\)

\(\displaystyle \frac{dr}{dt}=\frac{1}{14\pi in}(0.14\pi\frac{in^2}{s})\)

\(\displaystyle \frac{dr}{dt}=0.01\frac{in}{s}\)

The area of a triangle in terms of its base and height is given by the formula:

\(\displaystyle A=\frac{1}{2}bh\)

To find how the rates of change of each term relate, derive each side of the equation with respect to time:

\(\displaystyle \frac{dA}{dt}=\frac{d}{dt}(\frac{1}{2}bh)\)

\(\displaystyle \frac{dA}{dt}=\frac{1}{2}b\frac{dh}{dt}+\frac{1}{2}h\frac{db}{dt}\)

The base isn't widening, so \(\displaystyle \frac{db}{dt}=0\)

Therefore:

\(\displaystyle \frac{dA}{dt}=\frac{1}{2}(6m)(1\frac{m}{s})+\frac{1}{2}(1m)(0)\)

\(\displaystyle \frac{dA}{dt}=3\frac{m^2}{s}\)

The area of a rectangle is given in terms of its length and width as:

\(\displaystyle A=lw\).

The rates of change of each parameter with respect to time can be found by deriving each side of the equation with respect to time:

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

Therefore, with our known values, it's possible to find the inexplicable rate of change of the rectangle. Remember that the width is shrinking, so its rate should be treated as negative:

\(\displaystyle \frac{dA}{dt}=20in(-2\frac{in}{s})+15in(3\frac{in}{s})\)

\(\displaystyle \frac{dA}{dt}=5\frac{in^2}{s}\)

Note for this problem, we're looking for a change in a rate, so this is dealing with second time derivatives:

Relating the area of a square to length:

\(\displaystyle A=l^2\)

Leads to

\(\displaystyle \frac{d^2A}{dt^2}=\frac{d^2}{dt^2}l^2\)

\(\displaystyle \frac{d^2A}{dt^2}=\frac{d}{dt}(2l\frac{dl}{dt})\)

Now, it's important to note that the \(\displaystyle \frac{dl}{dt}\) term cannot be ignored in this second derivation; treat it like another variable!

\(\displaystyle \frac{d^2A}{dt^2}=2l\frac{d^2l}{dt^2}+2\frac{dl}{dt}\frac{dl}{dt}\)

\(\displaystyle \frac{d^2A}{dt^2}=2l\frac{d^2l}{dt^2}+2(\frac{dl}{dt})^2\)

Plugging in the given values, this gives:

\(\displaystyle \frac{d^2A}{dt^2}=2(4in)(0.2\frac{in}{s^2})+2(0.01\frac{in}{s})^2\)

\(\displaystyle \frac{d^2A}{dt^2}=1.6\frac{in^2}{s^2}+0.02\frac{in^2}{s^2}\)

\(\displaystyle \frac{d^2A}{dt^2}=1.62\frac{in^2}{s^2}\)

Begin this problem by solving for the radius. The volume of a sphere is given as

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

Solving for the radius:

\(\displaystyle r=\sqrt[3]{\frac{3}{4\pi}\frac{32}{3}\pi cm^3}=2cm\)

Now time rate of change between quantities can be found by deriving each side with respect to time:

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

Solving for the radius rate of time then gives:

\(\displaystyle \frac{dr}{dt}=\frac{1}{4\pi r^2}\frac{dV}{dt}\)

\(\displaystyle \frac{dr}{dt}=\frac{1}{16\pi cm^2}(\pi \frac{cm^3}{s})\)

\(\displaystyle \frac{dr}{dt}=\frac {1}{16}\frac{cm}{s}\)

The area of a triangle is given by the function:

\(\displaystyle A=\frac{1}{2}bh\)

The relationship between the rate of change of each parameter can be found by deriving each side of the equation with respect to time:

\(\displaystyle \frac{dA}{dt}=\frac{1}{2}b\frac{dh}{dt}+\frac{1}{2}h\frac{db}{dt}\)

\(\displaystyle \frac{dA}{dt}=\frac{1}{2}\left(6m(2\frac{m}{s})+20m(-1\frac{m}{s})\right)\)

\(\displaystyle \frac{dA}{dt}=-4\frac{m^2}{s}\)