# Area Between Y=x^3 & Its Tangent at x=1

• JOhnJDC
In summary, the area between y=x3 and its tangent at x=1 can be found by solving for the points of intersection between y=x3 and y=3x-2. This can be done algebraically or with a graphing calculator. The limits of integration are -2 to 1, and the resulting area is 27/4 square units.
JOhnJDC

## Homework Statement

What is the area between y=x3 and its tangent at x=1

## The Attempt at a Solution

The first derivative of y=x3, which is 3x2, tells me that the slope of the tangent at x=1 is 3. That (1,1) is a point on the tangent line tells me that the equation of the tangent line is y=3x-2. Now, I know that I need to solve y=3x-2 and y=x3 simultaneously to find the points of intersection, and that the x coordinates of these points will be my limits of integration. However, I'm having trouble factoring x3-3x+2=0. Anyone have any tips/tricks for factoring an equation like this? I suppose my question has more to do with algebra than calculus.

Thanks.

Last edited:
There's an easy way to solve the equation without having to do algebra. Assuming you have a graphing calculator you can just graph the two out and solve for the intersections.

However if you do not or need to use algebra, you can also use Synthetic division (a variation of long division of polynominals) if you don't know what it is this site exmplaisn it:
http://www.purplemath.com/modules/synthdiv.htm

This gives you (x-1)(x^2 + x -2 )
(x-1)(x-1)(x+2)
x = 1, -1, -2

The limits are 0 to 1... and the answer will be 0.5 sqr.units...

Thanks, EvilKermit. I used a graphing calculator to find the intersection at (-2,-8). I just wanted to do it algebraically. I'll check out the link you provided.

Paris.91 said:
The limits are 0 to 1... and the answer will be 0.5 sqr.units...

y=x3 and y=3x-2 do not intersect at any point when x=0; nor is the area between the curves symmetrical about the y-axis. The limits of integration are -2 to 1, and integrating yields 27/4.

Yea, I just realized my mistake above that x = 1, -2. There is no -1. The work is right above, just the answer was wrong at the end :)

## 1. What is the equation for the area between y=x^3 and its tangent at x=1?

The equation for the area between y=x^3 and its tangent at x=1 is given by A = (x^4/4) - (x^2/2) + (1/4), where x ranges from 0 to 1.

## 2. How do you find the points of intersection between y=x^3 and its tangent at x=1?

To find the points of intersection, we can equate the two equations and solve for x. In this case, x^3 = mx + (1-m), where m is the slope of the tangent line. This will give us two values of x, which represent the points of intersection.

## 3. What is the relationship between the area between y=x^3 and its tangent at x=1 and the slope of the tangent line?

The slope of the tangent line at x=1 is directly related to the rate of change of the area between y=x^3 and its tangent at x=1. The steeper the slope, the faster the area is changing. The slope also determines the shape of the area, as a higher slope will result in a more narrow and taller shape.

## 4. How can the area between y=x^3 and its tangent at x=1 be visualized?

This area can be visualized by graphing the two equations on the same coordinate plane. The points of intersection will be the boundaries of the area, and the shape of the area will depend on the slope of the tangent line at x=1.

## 5. What real-life applications does the area between y=x^3 and its tangent at x=1 have?

This area has applications in calculus and physics, specifically in the calculation of work done by a variable force. It can also be used to model the behavior of a particle moving along a parabolic path with a changing velocity at a specific point.

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