### All Calculus 1 Resources

## Example Questions

### Example Question #1 : How To Find Length Of Line By Graphing Functions

Consider a function with first-derivative:

.

Which integral can calculate the length along this curve between and ?

**Possible Answers:**

This is not possible with the information given.

**Correct answer:**

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:

### Example Question #2 : Length Of Line

### Find the length of the line segment between points A and B:

**Possible Answers:**

**Correct answer:**

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.

### Example Question #1 : Lines

What is the equation of the line tangent to f(x) = 4x^{3} – 2x^{2} + 4 at x = 5?

**Possible Answers:**

y = 220x – 550

y = 85x + 24

y = 280x – 946

None of the other answers

y = 44x + 245

**Correct answer:**

y = 280x – 946

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

f'(x) = 12x^{2} – 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 * 5^{3} – 2 * 5^{2} + 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

### Example Question #221 : Graphing Functions

Find the equation of the line tangent to at the point .

**Possible Answers:**

**Correct answer:**

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:

### Example Question #3 : How To Find Equation Of Line By Graphing Functions

Find the equation of the line tangent to at .

**Possible Answers:**

**Correct answer:**

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

### Example Question #1 : How To Find Equation Of Line By Graphing Functions

Find the equation of the line tangent to at .

**Possible Answers:**

**Correct answer:**

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 .

### Example Question #5 : How To Find Equation Of Line By Graphing Functions

Find the equation of the line tangent to at .

**Possible Answers:**

**Correct answer:**

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 =.

### Example Question #6 : How To Find Equation Of Line By Graphing Functions

A function, , is given by

.

Find the line tangent to at .

**Possible Answers:**

**Correct answer:**

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

.

### Example Question #7 : How To Find Equation Of Line By Graphing Functions

Let .

Find the equation for a line tangent to when .

**Possible Answers:**

**Correct answer:**

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:

### Example Question #1 : How To Find Equation Of Line By Graphing Functions

What is the equation of the line tangent to at ? Round to the nearest hundreth.

**Possible Answers:**

**Correct answer:**

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 .

Certified Tutor

Certified Tutor