Finding K For Tangency Line Y=2x+18 To The Function F(x)=k√x
In the realm of calculus, the concept of tangency holds significant importance. A tangent line to a curve at a particular point provides valuable information about the curve's behavior at that specific location. This article delves into the problem of finding the value of k such that a given line is tangent to the graph of a function involving a square root. Specifically, we will explore the function f(x) = k√x and the line y = 2x + 18, and determine the value(s) of k for which the line becomes a tangent to the curve. This problem combines algebraic manipulation with calculus concepts, offering a rich learning experience. Understanding this concept is crucial for various applications in physics, engineering, and economics, where rates of change and approximations are frequently used.
We are given the function f(x) = k√x and the line y = 2x + 18. Our goal is to find the value(s) of k such that the line is tangent to the graph of the function. This means that the line touches the curve at exactly one point, and at that point, the slope of the curve and the slope of the line are equal. The problem elegantly blends algebraic techniques with the principles of calculus, particularly derivatives, to arrive at a solution. The tangency condition is a critical aspect of calculus, with broad applications in optimization problems and curve analysis. Successfully solving this problem requires a solid understanding of both algebraic manipulations and the concept of derivatives.
Before diving into the solution, let's solidify our understanding of tangency. A line is tangent to a curve at a point if it touches the curve at that point and has the same slope as the curve at that point. This implies two key conditions:
- Intersection: The line and the curve must intersect at the point of tangency. This means that the y-values of the function and the line must be equal at the x-coordinate of the tangent point.
- Equal Slopes: The slope of the line must be equal to the derivative of the function at the point of tangency. The derivative gives the instantaneous rate of change of the function, which corresponds to the slope of the tangent line.
These two conditions form the foundation for solving problems involving tangency. We will use these conditions to set up equations and solve for the unknown variable k. The concept of tangency is fundamental in calculus, providing insights into the behavior of functions and their rates of change. Mastering this concept is essential for further studies in mathematics and its applications.
To solve the problem, we will follow these steps:
- Find the derivative of the function f(x) = k√x. The derivative, denoted as f'(x), will give us the slope of the tangent line to the curve at any point x. This step involves applying the power rule of differentiation to the square root function.
- Determine the slope of the line y = 2x + 18. The slope of a line in slope-intercept form (y = mx + b) is simply the coefficient of x, which is m. This step is straightforward and provides a constant value for the slope of the tangent line.
- Set the derivative equal to the slope of the line: f'(x) = 2. This equation represents the condition that the tangent line has the same slope as the given line. Solving this equation will give us the x-coordinate of the point of tangency.
- Substitute the x-coordinate into both the function and the line equations. This gives us two expressions for the y-coordinate of the point of tangency. Setting these expressions equal to each other represents the condition that the line and the curve intersect at the point of tangency.
- Solve the resulting equation for k. This will give us the value(s) of k for which the line is tangent to the curve.
This systematic approach ensures that we satisfy both conditions for tangency: intersection and equal slopes. Each step builds upon the previous one, leading us to the solution in a logical and clear manner. Careful attention to algebraic manipulation and calculus principles is crucial for success in this problem.
Let's now execute the solution step-by-step:
1. Find the derivative of f(x) = k√x
First, we rewrite the square root function using fractional exponents: f(x) = kx^(1/2). Now, we apply the power rule of differentiation, which states that if f(x) = ax^n, then f'(x) = nax^(n-1). Applying this rule, we get:
f'(x) = k * (1/2) * x^((1/2) - 1) = (k/2) * x^(-1/2) = k / (2√x)
Thus, the derivative of the function is f'(x) = k / (2√x). This derivative represents the slope of the tangent line to the curve at any point x in its domain. The ability to correctly differentiate functions is a fundamental skill in calculus, and this step is crucial for the rest of the solution.
2. Determine the slope of the line y = 2x + 18
The line is given in slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept. In this case, the slope of the line is the coefficient of x, which is 2. Therefore, the slope of the line is 2. This constant slope is a key piece of information for finding the point of tangency, as it allows us to equate the derivative of the function to a specific value.
3. Set the derivative equal to the slope of the line: f'(x) = 2
Now we equate the derivative we found in step 1 to the slope of the line:
k / (2√x) = 2
This equation represents the condition that the tangent line has the same slope as the given line. Solving for √x, we multiply both sides by 2√x and then divide by 2:
k = 4√x
√x = k/4
Squaring both sides to solve for x, we get:
x = (k/4)^2 = k^2 / 16
This value of x represents the x-coordinate of the point of tangency. The algebraic manipulation in this step is critical for expressing the x-coordinate in terms of k, which will be used in the subsequent steps.
4. Substitute the x-coordinate into both the function and the line equations
At the point of tangency, the y-values of the function and the line must be equal. So, we substitute x = k^2 / 16 into both equations:
Function: y = f(x) = k√x = k√(k^2 / 16) = k * (|k|/4)
Line: y = 2x + 18 = 2(k^2 / 16) + 18 = k^2 / 8 + 18
Equating the y-values, we get:
k * (|k|/4) = k^2 / 8 + 18
This equation represents the condition that the line and the curve intersect at the point of tangency. The substitution and equation setup are crucial for relating k to the tangency condition. The absolute value in the expression arises from the square root, and we must consider its implications in the next step.
5. Solve the resulting equation for k
We now have the equation k * (|k|/4) = k^2 / 8 + 18. To solve for k, we need to consider two cases: k ≥ 0 and k < 0.
Case 1: k ≥ 0
In this case, |k| = k, and the equation becomes:
k^2 / 4 = k^2 / 8 + 18
Multiplying both sides by 8 to eliminate fractions, we get:
2k^2 = k^2 + 144
k^2 = 144
k = ±12
Since we are considering k ≥ 0, we take the positive root, so k = 12.
Case 2: k < 0
In this case, |k| = -k, and the equation becomes:
k * (-k/4) = k^2 / 8 + 18
-k^2 / 4 = k^2 / 8 + 18
Multiplying both sides by 8, we get:
-2k^2 = k^2 + 144
-3k^2 = 144
k^2 = -48
Since k^2 cannot be negative, there are no solutions in this case.
Therefore, the only solution is k = 12. This value of k ensures that the line y = 2x + 18 is tangent to the curve f(x) = 12√x. The careful consideration of cases and algebraic manipulation in this step are crucial for arriving at the correct solution.
In conclusion, we have successfully found the value of k such that the line y = 2x + 18 is tangent to the graph of the function f(x) = k√x. By applying the principles of calculus, specifically differentiation, and algebraic manipulation, we determined that k = 12. This problem highlights the connection between algebra and calculus and demonstrates the importance of understanding tangency conditions. The ability to solve such problems is crucial for various applications in science and engineering. The step-by-step solution presented in this article provides a clear and logical approach to solving similar tangency problems, reinforcing the fundamental concepts of calculus and algebra. This comprehensive guide should serve as a valuable resource for students and anyone interested in mastering calculus problems involving tangency.
