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

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

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

.

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

substitute:

Simplify:

By trigonometric identities we get:

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

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

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 , where  is the slope of the line and .  

To find the slope, we need to evaluate the derivative at :

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

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 , where m is the slope and  is the y adjustment. To get the slope, find the derivative of  and plug in the desired point  for , giving us an answer of  for the slope.

To find the y adjustment pick a point 0 in the original function. For simplicity, let's plug in , which gives us a y of 1, so an easy point is . Next plug in those values into the equation of a line, . The new equation with all parameters plugged in is

Now you simply solve for , which is .

Final equation of the line tangent to  at  is 

To get the slope, find the derivative of  and plug in the desired point  for , giving us an answer of  for the slope.

Remember that the derivative of .

To find the  adjustment pick a point  (for example) in the original  function. For simplicity, let's plug in , which gives us a  of , so an easy point is . Next plug in those values into the equation of a line, . The new equation with all parameters plugged in is

The coefficient in front of the  is the slope.

Now you simply solve for , which is .

Final equation of the line tangent to  at is .

The equation of the tangent line is  To find , the slope, calculate the derivative and plug in the desired point.

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

Let's choose .

The  value at this point is .

Plugging in those values we can solve for .

Solving for  we get =.

First we need to find the slope of  at . To do this we need the derivative of . To take the derivative we need to use the power rule for the first term and recognize that the derivative of sine is cosine.

At,

Now we need to know

.

Now we have a slope,  and a point 

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

Plugging in and rearranging we find

.

First, evaluate  when .

Thus, we need a line that contains the point 

Next, find the derivative of  and evaluate it at .

To find the derivative we will use the power rule,

.

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

In point-slope form, , a line with the point  and a slope of 8 will be:

The tangent line to  at  must have the same slope as .

Applying the chain rule we get

.

Therefore the slope of the line is, 

.

In addition, the tangent line touches the graph of  at . Since , the point  lies on the line.

Plugging in the slope and point we get .