To determine the length of a curve between two points, we evaluate the integral

$\displaystyle L = \int_0^{2\pi}{\sqrt{1+\left(\frac{dy}{dx} \right )^2}}dx$

There are many reasons this works, but we'll give an informal explanation here:

If we divide this curve into three line-segments, we can see that they become more and more similar to the original curve. By adding up all the little hypotenuses, we can arrive at the length of the curve. Note, that

$\displaystyle \sqrt{dx^2 + dy^2} = \sqrt{1+\left(\frac{dy}{dx} \right )^2}dx$

If we think of the integration symbol as a sum of infinitely small parts, this gets us the formula for length:

$\displaystyle L = \int_a^b{\sqrt{1+\left(\frac{dy}{dx} \right )^2}}dx$.

Returning to the problem, we plug the derivative into the length formula:

$\displaystyle L=\int_{0}^{2\pi}{\sqrt{1+\tan^2{x}}}\quad dx$

substitute:

$\displaystyle L=\int_{0}^{2\pi}{\sqrt{\left( \frac{\cos{x}}{\cos{x}} \right)^2+\left(\frac{\sin{x}}{\cos{x}} \right )^2}}\quad dx$

Simplify:

$\displaystyle L=\int_{0}^{2\pi}{\sqrt{\frac{\cos^2{x}}{\cos^2{x}}+\frac{\sin^2{x}}{\cos^2{x}}}}\quad dx$

$\displaystyle L=\int_{0}^{2\pi}{\sqrt{\frac{\cos^2{x}+\sin^2{x}}{\cos^2{x}}}}\quad dx$

By trigonometric identities we get:

$\displaystyle L=\int_{0}^{2\pi}{\sqrt{\frac{1}{\cos^2{x}}}}\quad dx$

$\displaystyle L=\int_{0}^{2\pi}{\frac{1}{\cos{x}}}\quad dx$

The distance between two points can be easily found using the Distance Formula:

$\displaystyle D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Applying the points we are given to this formula results in:

$\displaystyle D=\sqrt{((-5)-2)^2+(10-(-7))^2}=\sqrt{(-7)^2+(17)^2}=\sqrt{338}$

This is one of the answer choices.

First, take the derivative of f(x). This is very easy:

f'(x) = 12x² – 4x

Now, the slope of the tangent line at x = 5 is f'(5). Evaluated, this is: f'(5) = 12 * 5 * 5 – 4 * 5 = 300 – 20 = 280

Now, we must find the intersection point on the original line:

f(5) = 4 * 5³ – 2 * 5² + 4 = 500 – 50 + 4 = 454

Therefore, the point of tangent intersection is (5,454).

Using the point-slope form of linear equations, we can find the line:

(y – 454) = 280(x – 5)

y – 454 = 280x – 1400

y = 280x – 946

The equation of the tangent line will have the form $\displaystyle y-y_{0}=m(x-x_{0})$, where $\displaystyle m$ is the slope of the line and $\displaystyle (x_{0},y_{0})=(1,3)$.  

To find the slope, we need to evaluate the derivative at $\displaystyle x=1$:

 $\displaystyle \frac{dy}{dx}=\frac{d(x^2-x+3)}{dx}=2x-1\Rightarrow m=f'(1)=2*1-1=1$

Now that we have the slope, we can determine the equation of the tangent line using the point-slope formula:

 $\displaystyle y-y_{0}=m(x-x_{0})\Rightarrow y-3=1(x-1)\Rightarrow y=x+2.$

To find the equation of the line at that point, you need two things: the slope at that point and the y adjustment of the function, in the form $\displaystyle y=mx+b$, where m is the slope and $\displaystyle b$ is the y adjustment. To get the slope, find the derivative of $\displaystyle f(x)$ and plug in the desired point $\displaystyle 0$ for $\displaystyle x$, giving us an answer of $\displaystyle 3$ for the slope.

$\displaystyle f'(x)=3e^{3x} + 2x$

$\displaystyle f'(0)=3$

To find the y adjustment pick a point 0 in the original$\displaystyle f(x)$ function. For simplicity, let's plug in $\displaystyle x=0$, which gives us a y of 1, so an easy point is $\displaystyle (0,1)$. Next plug in those values into the equation of a line, $\displaystyle y=mx + b$. The new equation with all parameters plugged in is

$\displaystyle 1=(3)(0)+b$

Now you simply solve for $\displaystyle b$, which is $\displaystyle 1$.

Final equation of the line tangent to $\displaystyle f(x)$ at $\displaystyle x=0$ is $\displaystyle y=3x+1$

To get the slope, find the derivative of $\displaystyle f(x)$ and plug in the desired point $\displaystyle 2$ for $\displaystyle x$, giving us an answer of $\displaystyle \frac{2}{5}$ for the slope.

$\displaystyle f'(x)=\frac{2}{2x+1}$

$\displaystyle f'(2)=\frac{2}{5}$

Remember that the derivative of $\displaystyle ln(u) = \frac{u'}{u}$.

To find the $\displaystyle y$ adjustment pick a point $\displaystyle 0$ (for example) in the original $\displaystyle f(x)$ function. For simplicity, let's plug in $\displaystyle x=0$, which gives us a $\displaystyle y$ of $\displaystyle 3$, so an easy point is $\displaystyle (0,3)$. Next plug in those values into the equation of a line, $\displaystyle y=mx + b$. The new equation with all parameters plugged in is

$\displaystyle 3=\frac{2}{5}(0)+b$

The coefficient in front of the $\displaystyle 0$ is the slope.

Now you simply solve for $\displaystyle b$, which is $\displaystyle 3$.

Final equation of the line tangent to $\displaystyle f(x)$ at$\displaystyle x=2$ is $\displaystyle y=\frac{2}{5}x +3$.

The equation of the tangent line is $\displaystyle y=mx+b.$ To find $\displaystyle m$, the slope, calculate the derivative and plug in the desired point.

$\displaystyle y'=2cos(2x)$

$\displaystyle y'=2cos\left(2\cdot \frac{\pi}{2}\right)$

$\displaystyle y'=2cos(\pi)$

$\displaystyle y'\left(\frac{\pi}{2} \right )=-2$

The next step is to choose a coordinate on the original $\displaystyle y$ function. We can choose any $\displaystyle x$ value and calculate its $\displaystyle y$ value.

Let's choose $\displaystyle \frac{\pi }{2}$.

$\displaystyle y=sin\left(2\cdot \frac{\pi}{2}\right)=sin(\pi)=0$

The $\displaystyle y$ value at this point is $\displaystyle 0$.

Plugging in those values we can solve for $\displaystyle b$.

$\displaystyle 0=(-2)\left(\frac{\pi }{2}\right)+b$

Solving for $\displaystyle b$ we get $\displaystyle b$=$\displaystyle \pi$.

$\displaystyle y=-2x+\pi$

First we need to find the slope of $\displaystyle f$ at $\displaystyle x=\frac{\pi}{2}$. To do this we need the derivative of $\displaystyle f$. To take the derivative we need to use the power rule for the first term and recognize that the derivative of sine is cosine.

$\displaystyle \frac{df}{dx}=2x-cos(x)$

At,

 $\displaystyle x=\frac{\pi}{2}, \frac{df}{dx}=\pi$

Now we need to know

$\displaystyle f\left(\frac{\pi}{2}\right)=\frac{\pi^2}{4}-1$.

Now we have a slope, $\displaystyle m=\pi$ and a point $\displaystyle (x_1,y_1)=\left(\frac{\pi}{2},\frac{\pi^2}{4}-1\right)$

so we can use the point-slope formula to find the equation of the line.

$\displaystyle y-y_1=m(x-x_1)$

Plugging in and rearranging we find

$\displaystyle y=\pi x-\frac{\pi^2}{4}-1$.

First, evaluate $\displaystyle f(x)$ when $\displaystyle x=1$.

$\displaystyle f(1)=2(1)^4+e^1-e(1)=2+0\textbf{=2}$

Thus, we need a line that contains the point $\displaystyle (1,2)$

Next, find the derivative of $\displaystyle f(x)$ and evaluate it at $\displaystyle x=1$.

To find the derivative we will use the power rule,

$\displaystyle f(x)=x^n \rightarrow f'(x)=nx^{n-1}$.

$\displaystyle f'(x)=8x^3+e^x-e$

$\displaystyle f'(1)=8(1)^3+e^1-e=8+0=8$

This indicates that we need a line with a slope of 8.

In point-slope form, $\displaystyle y-y_{1}=m(x-x_{1})$, a line with the point $\displaystyle (1,2)$ and a slope of 8 will be:

$\displaystyle y-2=8(x-1)$

The tangent line to $\displaystyle f(x)$ at $\displaystyle x=-1$ must have the same slope as $\displaystyle f(x)$.

Applying the chain rule we get

$\displaystyle f'(x)=\frac{1}{1+(1-x^3)^2}(-3x^2)$.

Therefore the slope of the line is, 

$\displaystyle f'(-1)=-\frac{3}{5}=-0.6$.

In addition, the tangent line touches the graph of $\displaystyle f(x)$ at $\displaystyle x=-1$. Since $\displaystyle f(-1)=1.11$, the point $\displaystyle (-1, 1.11)$ lies on the line.

Plugging in the slope and point we get $\displaystyle y=-0.6x+0.51$.