Finding Coefficients In A Cubic Equation Given A Tangent Line
Introduction
In the fascinating world of calculus, tangent lines play a crucial role in understanding the behavior of curves. A tangent line touches a curve at a single point, providing valuable information about the curve's slope and direction at that specific location. When we're given the equation of a tangent line to a curve, along with the curve's equation, we can often extract significant details about the curve itself. This article delves into a classic problem of this nature, where we're tasked with finding unknown coefficients in a polynomial equation given the tangent line at a specific point.
This problem perfectly demonstrates the interplay between differential calculus (finding derivatives and tangent lines) and algebra (solving equations). By combining these concepts, we can unravel the mysteries hidden within mathematical expressions. The problem we'll explore involves finding the values of 'a' and 'b' in the cubic equation y = x³ + ax² + bx + 4, given that the tangent to the curve at the point where x = 2 is y = x - 4. This involves a blend of finding the derivative, evaluating it at a specific point, and then setting up and solving equations to find the unknown values of 'a' and 'b'. Solving this problem requires a solid understanding of calculus, specifically the concept of derivatives and their relation to tangent lines. It also necessitates strong algebraic skills to manipulate and solve the resulting equations. This kind of problem is a staple in calculus courses and appears frequently in mathematics competitions, making it an important concept to master.
Setting up the problem
The heart of this problem lies in the relationship between the curve and its tangent. The curve is defined by the equation y = x³ + ax² + bx + 4, where 'a' and 'b' are constants we need to find. The tangent line, given by y = x - 4, touches the curve at the point where x = 2. This information provides us with two crucial pieces of the puzzle: the slope of the tangent and a point of intersection. The slope of the tangent line at any point on the curve is given by the derivative of the curve's equation. Therefore, our first step is to find the derivative of y = x³ + ax² + bx + 4 with respect to x. This will give us an expression for the slope of the tangent at any x value. Then, we'll evaluate this derivative at x = 2 to find the slope of the tangent at the specific point of interest. The derivative, denoted as dy/dx or y', is found using the power rule of differentiation, which states that the derivative of xⁿ is nxⁿ⁻¹. Applying this rule to each term in the equation, we get:
dy/dx = 3x² + 2ax + b
This expression tells us the slope of the curve at any point x. Since the line y = x - 4 is tangent to the curve at x = 2, the slope of the tangent line at this point must be equal to the derivative evaluated at x = 2. The slope of the line y = x - 4 is simply the coefficient of x, which is 1. Therefore, we have:
3(2)² + 2a(2) + b = 1
This simplifies to our first equation:
12 + 4a + b = 1
Or,
4a + b = -11
This is a linear equation in two unknowns, 'a' and 'b'. To solve for these unknowns, we need another independent equation. This second equation comes from the fact that the tangent line and the curve intersect at x = 2. Since the line is tangent to the curve at this point, it means that the y-values of the curve and the line must be equal when x = 2. In other words, the point (2, y) lies on both the curve and the tangent line. We can use this information to create our second equation. To find the y-coordinate of the point on the tangent line where x = 2, we substitute x = 2 into the equation of the tangent line y = x - 4. This gives us:
y = 2 - 4 = -2
So, the point of tangency is (2, -2). This point must also lie on the curve y = x³ + ax² + bx + 4. Substituting x = 2 and y = -2 into the curve's equation, we get:
-2 = (2)³ + a(2)² + b(2) + 4
This simplifies to:
-2 = 8 + 4a + 2b + 4
Further simplification yields our second equation:
4a + 2b = -14
Or, dividing by 2,
2a + b = -7
Now we have a system of two linear equations with two unknowns:
- 4a + b = -11
- 2a + b = -7
These equations represent two lines in the a-b plane. The solution to this system, which gives us the values of 'a' and 'b', is the point where these two lines intersect.
Solving for a and b
With our two equations in hand, we can now solve for the unknowns 'a' and 'b'. We have a system of linear equations that can be solved using various methods, such as substitution, elimination, or matrix methods. The elimination method is particularly convenient in this case because the coefficient of 'b' is the same in both equations. To use the elimination method, we simply subtract the second equation from the first equation. This will eliminate 'b', leaving us with an equation solely in terms of 'a'. Subtracting the equation 2a + b = -7 from the equation 4a + b = -11, we get:
(4a + b) - (2a + b) = -11 - (-7)
This simplifies to:
2a = -4
Dividing both sides by 2, we find:
a = -2
Now that we have the value of 'a', we can substitute it back into either of the original equations to solve for 'b'. Let's use the second equation, 2a + b = -7. Substituting a = -2, we get:
2(-2) + b = -7
This simplifies to:
-4 + b = -7
Adding 4 to both sides, we find:
b = -3
Therefore, we have found the values of 'a' and 'b':
a = -2 and b = -3
These values are the solution to our problem. They represent the coefficients in the original cubic equation that satisfy the given conditions: the tangent to the curve at x = 2 is the line y = x - 4. To verify our solution, we can substitute these values back into the original equation and check that the tangent condition is indeed satisfied. Substituting a = -2 and b = -3 into the equation y = x³ + ax² + bx + 4, we get the specific cubic equation:
y = x³ - 2x² - 3x + 4
Now, let's find the derivative of this equation:
dy/dx = 3x² - 4x - 3
Evaluating the derivative at x = 2, we get:
dy/dx |_(x=2) = 3(2)² - 4(2) - 3 = 12 - 8 - 3 = 1
This confirms that the slope of the tangent at x = 2 is indeed 1, which matches the slope of the given tangent line y = x - 4. Next, we need to check that the point (2, -2) lies on both the curve and the tangent line. We already know that it lies on the tangent line. Substituting x = 2 into the equation y = x³ - 2x² - 3x + 4, we get:
y = (2)³ - 2(2)² - 3(2) + 4 = 8 - 8 - 6 + 4 = -2
This confirms that the point (2, -2) also lies on the curve. Therefore, our solution a = -2 and b = -3 is correct. This verification step is crucial in any mathematical problem-solving process. It ensures that our calculated values not only satisfy the equations we set up but also make sense in the context of the original problem. In this case, verifying that the slope of the tangent matches and that the point of tangency lies on both the curve and the line gives us high confidence in the correctness of our solution.
Conclusion
In conclusion, by skillfully applying the principles of differential calculus and algebraic manipulation, we successfully determined the values of 'a' and 'b' in the given cubic equation. The key to solving this problem was understanding the relationship between a curve, its tangent line, and the derivative. By finding the derivative, evaluating it at the point of tangency, and setting it equal to the slope of the tangent line, we established our first equation. The second equation was derived from the fact that the point of tangency lies on both the curve and the tangent line. Solving the resulting system of linear equations gave us the desired values of 'a' and 'b'.
This problem serves as an excellent illustration of how calculus and algebra work together to solve complex mathematical problems. The ability to find derivatives, understand their geometric interpretation as slopes of tangent lines, and manipulate algebraic equations is crucial for success in calculus and related fields. Problems like this often appear in calculus courses, examinations, and mathematics competitions, highlighting the importance of mastering these concepts. Moreover, the techniques used to solve this problem have broad applications in various areas of science and engineering. For example, finding tangent lines is essential in optimization problems, where we seek to maximize or minimize a certain quantity. It also plays a vital role in curve sketching, where understanding the slope of the tangent helps us visualize the shape of the curve. The concepts explored in this problem are fundamental building blocks for more advanced topics in calculus and beyond. By tackling such problems, students not only enhance their mathematical skills but also develop a deeper appreciation for the power and beauty of mathematics.
In summary, solving this problem required a multi-faceted approach, involving calculus concepts such as derivatives and their geometric interpretation, as well as algebraic techniques for solving systems of equations. The problem underscores the importance of a strong foundation in both calculus and algebra for tackling mathematical challenges. By successfully navigating this problem, we've reinforced our understanding of these fundamental concepts and honed our problem-solving skills.
