First, we want the slope, so we have to take a derivative of $\displaystyle f(t)$. We will need to use the power rule which is, 

$\displaystyle y=x^n \rightarrow \frac{d}{dx}=nx^{n-1}$ on the first term. We need to also recall that the derivative of $\displaystyle e^u$ is$\displaystyle e^u\cdot \frac{du}{dx}$.

Applying these rules we get the following derivative.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}(t^2-e^{2t})=2t-2e^{2t}$ 

We're looking for the time the slope is $\displaystyle -10$, so we have to set the derivative (which gives you slope) equal to $\displaystyle -10$. 

$\displaystyle 2t-2e^{2t}=-10 \therefore t-e^{2t}=-5$.

At this point you can use a graphing calculator to graph the function $\displaystyle t-e^{2t}$, and trace the graph to find the x value that results in a y value of $\displaystyle -5$. The positive solution rounded to the nearest hundredth is $\displaystyle 0.89$.

In this problem we are given the length and width of a rectangle as well as the rate at which the width is increasing. We are asked to find the rate of change of the area of a rectangle. The equation for finding the area of a rectangle is given as

$\displaystyle A=l\cdot w$.  

By taking the derivative of this equation with respect to time, we can find how the area changes with respect to time.  To take the derivative of an equation with two variables, we must use the product rule,

$\displaystyle (uv)^{'}=u^{'}v+uv^{'}$.

Applying the product rule to the equation we obtain 

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

Because the width of the rectangle is increases at a rate of $\displaystyle 2\frac{ft}{s}$, $\displaystyle \frac{dw}{dt}=2$. 

Since the length of the rectangle does not change with respect to time, $\displaystyle \frac{dl}{dt}=0$.  

$\displaystyle l$ and $\displaystyle w$ are given to us as 4 feet and 6 feet respectively $\displaystyle \frac{dA}{dt}=8$.  

Therefore the area of this rectangle changes at a rate of $\displaystyle 8 ft^2/s$ when the width of the rectangle is increasing by $\displaystyle 2\frac{ft}{s}$.

We can solve by utilizing the formula for the average rate of change: $\displaystyle \frac{ f(x_2)-f(x_1)}{x_2-x_1}$.  Solving for $\displaystyle f(x)$ at our given points:

$\displaystyle f(x_1)=f(2)=7(2)-4=14-4=10$

 $\displaystyle f(x_2)=f(3)=7(3)-4=21-4=17$

Plugging our values into the average rate of change formula, we get:

$\displaystyle \frac{(17-10)}{(3-2)}=\frac{7}{1}=7$.

We can solve by utilizing the formula for the average rate of change: $\displaystyle \frac{ f(x_2)-f(x_1)}{x_2-x_1}$.  Solving for $\displaystyle f(x)$ at our given points:

$\displaystyle f(x_1)=f(4)=9(4)+12=36+12=48$

 $\displaystyle f(x_2)=f(6)=9(6)+12=54+12=66$

Plugging our values into the average rate of change formula, we get:

$\displaystyle \frac{(66-48)}{(6-4)}=\frac{18}{2}=9$.

We can solve by utilizing the formula for the average rate of change: $\displaystyle \frac{ f(x_2)-f(x_1)}{x_2-x_1}$.  Solving for $\displaystyle f(x)$ at our given points:

$\displaystyle f(x_1)=f(3)=2(3)+15=6+15=21$

 $\displaystyle f(x_2)=f(5)=2(5)+15=10+15=25$

Plugging our values into the average rate of change formula, we get:

$\displaystyle \frac{(25-21)}{(5-3)}=\frac{4}{2}=2$.

Using right triangles we know that  

$\displaystyle \tan(\theta(t))= \frac{h(t)}{300} \right$.

Solving for $\displaystyle \theta(t)$ we get 

$\displaystyle \theta(t)=\arctan\left ( \frac{h(t)}{300} \right )$.

Taking the derivative, we need to remember to apply the chain rule to $\displaystyle h(t)$ since the height depends on time,

$\displaystyle \frac{d\theta}{dt}=\frac{1}{1+\left(\frac{h}{300} \right )^2}\frac{1}{300}\frac{dh}{dt}$.

We are asked to find $\displaystyle \frac{d\theta}{dt}|_{t=5}$. We are given $\displaystyle \frac{dh}{dt}=50$and since $\displaystyle \frac{dh}{dt}$ is constant, we know that the height of the balloon is given by $\displaystyle h(t)=\frac{dh}{dt}t$.

Therefore, at $\displaystyle t=5$ we know that the height of the balloon is  $\displaystyle h(5)=5\cdot 50=250$.

Plugging these numbers into $\displaystyle \frac{d\theta}{dt}$ we find

$\displaystyle \frac{d\theta}{dt}|_{t=5} =\frac{1}{1+\left(\frac{250}{300} \right )^2}\frac{1}{300}50=0.0098$ radians.

This scenario describes a right triangle where the hypotenuse is the distance between the two boats. Let $\displaystyle x$ denote the distance boat $\displaystyle A$ is from the port, $\displaystyle y$ denote the distance boat $\displaystyle B$ is from the port, $\displaystyle D$ denote the distance between the two boats, and $\displaystyle t$ denote the time since they left the port. Applying the Pythagorean Theorem we have,

$\displaystyle D(t)^2=x(t)^2+y(t)^2$.

Implicitly differentiating this equation we get

$\displaystyle 2D(t)\frac{dD}{dt}=2x(t)\frac{dx}{dt}+2y(t)\frac{dy}{dt}$.

We need to find $\displaystyle \frac{dD}{dt}$ when $\displaystyle t=3$.

We are given 

$\displaystyle \frac{dx}{dt}=30, \frac{dy}{dt}=20$ which tells us 

$\displaystyle x(3)=3\cdot 30=90, y(3)=3\cdot 20=60,$

$\displaystyle D(3)=\sqrt{90^2+60^2}=30\sqrt{13}$. 

Plugging this in we have

$\displaystyle 2\cdot 30\sqrt{13}\frac{dD}{dt}|_{t=3}=2\cdot90\cdot 30+2\cdot60\cdot20$.  

Solving we get 

$\displaystyle \frac{dD}{dt}|_{t=3}=10\sqrt{13}$.

The volume function, in terms of a radius $\displaystyle r$, is given as

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

The change in volume over the change in time, or

$\displaystyle \frac{dV}{dt}$ is given as

$\displaystyle \frac{dV}{dt} = \frac{d}{dt}[V(r)]$

and by implicit differentiation, the chain rule, and the power rule, 

$\displaystyle \frac{d}{dt}[V(r)]=\frac{d}{dt}[\frac{4}{3}\pi r^3]=4\pi r^2 \frac{dr}{dt}$.

Setting $\displaystyle \frac{dr}{dt}=2$ we get

$\displaystyle \frac{d}{dt}[V(r)]= 4\pi r^2(2)=8\pi r^2$.

As such, 

$\displaystyle \frac{dV}{dt} =8\pi r^2$.

We can solve by utilizing the formula for the average rate of change: $\displaystyle \frac{f(x_2)-f(x_1)}{x_2-x_1}$.

Solving for f(x) at our given points:

$\displaystyle f(x_1)=f(2)=5(2)+11=10+11=21$

$\displaystyle f(x_2)=f(3)=5(3)+11=15+11=26$

Plugging our values into the average rate of change formula, we get:

$\displaystyle \frac{(26-21)}{(3-2)}=\frac{5}{1}=5$.

We can solve by utilizing the formula for the average rate of change: $\displaystyle \frac{f(x_2)-f(x_1)}{x_2-x_1}$.

Solving for f(x) at our given points:

$\displaystyle f(x_1)=f(5)=7(5)-8=35-8=27$

$\displaystyle f(x_2)=f(6)=7(6)-8=42-8=34$

Plugging our values into the average rate of change formula, we get:

$\displaystyle \frac{(34-27)}{(6-5)}=\frac{7}{1}=7$.