Derivatives, Slopes, And Tangent Lines: Solved!
Hey guys! Let's dive into the fascinating world of calculus and tackle the problem of finding derivatives, slopes, and tangent line equations for some given functions. Don't worry, we'll break it down step by step so it's super easy to follow. We've got three functions to work with today: y = mx + b at point (a, b), y = -2x^3 at point (1, -2), and y = 10x^-3 at point (2, 1). So, buckle up and let's get started!
Understanding Derivatives, Slopes, and Tangent Lines
First, let's make sure we're all on the same page about what derivatives, slopes, and tangent lines actually mean.
- Derivatives: Think of the derivative as the instantaneous rate of change of a function. It tells us how much the function's output is changing with respect to its input at any given point. Mathematically, it's the limit of the difference quotient as the change in the input approaches zero. Whew, that's a mouthful! But basically, it's the slope of a line tangent to the curve at a specific point.
- Slopes: The slope is a measure of the steepness of a line. It's often described as "rise over run," which means the change in the vertical direction (y-axis) divided by the change in the horizontal direction (x-axis). A positive slope indicates an increasing line, a negative slope indicates a decreasing line, a zero slope indicates a horizontal line, and an undefined slope indicates a vertical line. Understanding slopes is crucial for visualizing the behavior of functions.
- Tangent Lines: A tangent line is a straight line that touches a curve at a single point and has the same slope as the curve at that point. Imagine zooming in on a curve at a specific point until it looks almost like a straight line – that line is the tangent line! The equation of a tangent line can be found using the point-slope form, which we'll use later in our calculations.
Understanding these concepts is fundamental to solving our problem. We'll use the power rule, point-slope form, and some basic algebra to find our answers. Remember, calculus is all about understanding how things change, and these tools help us quantify that change.
Function 1: y = mx + b at (a, b)
Our first function is the classic linear equation, y = mx + b, evaluated at the point (a, b). This is a great starting point because it's relatively straightforward and helps illustrate the basic principles. Let's break it down:
1. Finding the Derivative
The derivative of y = mx + b with respect to x is simply m. Why? Because 'm' represents the slope of the line, and for a linear function, the slope is constant everywhere. This means the rate of change is the same at every point on the line. To find this, we can use the power rule of differentiation. Remember, the power rule states that if y = ax^n, then dy/dx = nax^(n-1). In our case, 'mx' can be seen as mx^1, so applying the power rule gives us 1 * m * x^(1-1) = m * x^0 = m * 1 = m. The derivative of the constant 'b' is zero because constants don't change.
2. Determining the Slope
The slope of the line y = mx + b is, as we already know, 'm'. This is explicitly given in the equation itself. The beauty of a linear equation is that the slope is constant, meaning it's the same at any point along the line. So, at the point (a, b), the slope is still 'm'.
3. Calculating the Tangent Line Equation
To find the equation of the tangent line, we can use the point-slope form: y - y1 = m(x - x1), where (x1, y1) is the point of tangency and 'm' is the slope. In our case, (x1, y1) = (a, b) and the slope is 'm'. Plugging these values into the point-slope form, we get:
y - b = m(x - a)
This is the equation of the tangent line to y = mx + b at the point (a, b). Notice that this equation is actually the same as the original equation, just written in a slightly different form. This makes sense because the tangent line to a straight line is the line itself! This is a fundamental concept in calculus: the tangent line to a linear function at any point is the function itself. This is because the slope of the function is constant.
Function 2: y = -2x^3 at (1, -2)
Now, let's tackle a slightly more complex function: y = -2x^3, evaluated at the point (1, -2). This is a cubic function, so its graph is a curve, and the slope will change at different points. This means we'll need to use the derivative to find the slope at the specific point (1, -2).
1. Finding the Derivative
To find the derivative of y = -2x^3, we'll again use the power rule. Applying the power rule, we get:
dy/dx = 3 * (-2) * x^(3-1) = -6x^2
So, the derivative of y = -2x^3 is -6x^2. This derivative function gives us the slope of the tangent line at any point 'x' on the original curve. This is a powerful tool because it allows us to find the instantaneous rate of change.
2. Determining the Slope
To find the slope at the point (1, -2), we need to plug x = 1 into the derivative function:
Slope = -6(1)^2 = -6
Therefore, the slope of the tangent line at the point (1, -2) is -6. This tells us that the curve is decreasing at this point, and quite steeply!
3. Calculating the Tangent Line Equation
Now we can use the point-slope form to find the equation of the tangent line. We have the point (1, -2) and the slope -6. Plugging these values into y - y1 = m(x - x1), we get:
y - (-2) = -6(x - 1)
Simplifying, we get:
y + 2 = -6x + 6
y = -6x + 4
So, the equation of the tangent line to y = -2x^3 at the point (1, -2) is y = -6x + 4. This line touches the curve at the point (1, -2) and has a slope of -6, accurately representing the curve's direction at that specific location. Graphing the function and its tangent line can visually confirm this.
Function 3: y = 10x^-3 at (2, 1)
Our final function is y = 10x^-3, evaluated at the point (2, 1). This function involves a negative exponent, but don't worry, the power rule still applies! Let's see how it works.
1. Finding the Derivative
Again, we'll use the power rule to find the derivative. Applying the power rule to y = 10x^-3, we get:
dy/dx = -3 * 10 * x^(-3-1) = -30x^-4
So, the derivative of y = 10x^-3 is -30x^-4. Remember that x^-4 is the same as 1/x^4, so we could also write the derivative as -30/x^4. Negative exponents can sometimes look tricky, but they're just another way to express a reciprocal.
2. Determining the Slope
To find the slope at the point (2, 1), we need to plug x = 2 into the derivative function:
Slope = -30(2)^-4 = -30 / (2^4) = -30 / 16 = -15/8
Therefore, the slope of the tangent line at the point (2, 1) is -15/8. This is a negative slope, indicating that the curve is decreasing at this point, but not as steeply as in the previous example.
3. Calculating the Tangent Line Equation
Now, let's use the point-slope form to find the tangent line equation. We have the point (2, 1) and the slope -15/8. Plugging these values into y - y1 = m(x - x1), we get:
y - 1 = (-15/8)(x - 2)
To simplify, we can multiply both sides by 8 to get rid of the fraction:
8(y - 1) = -15(x - 2)
8y - 8 = -15x + 30
Now, let's rearrange the equation to get it in slope-intercept form (y = mx + b):
8y = -15x + 38
y = (-15/8)x + 38/8
y = (-15/8)x + 19/4
So, the equation of the tangent line to y = 10x^-3 at the point (2, 1) is y = (-15/8)x + 19/4. This line touches the curve at (2, 1) and has a slope of -15/8, accurately reflecting the curve's behavior at that location.
Conclusion
And there you have it, guys! We've successfully calculated the derivatives, slopes, and tangent line equations for three different functions. We started with a simple linear function, moved on to a cubic function, and finished with a function involving a negative exponent. Remember the key takeaways:
- The derivative tells us the instantaneous rate of change (the slope of the tangent line).
- The slope is a measure of the steepness of a line.
- The tangent line touches the curve at a single point and has the same slope as the curve at that point.
By understanding these concepts and applying the power rule and point-slope form, you can tackle a wide range of calculus problems. Keep practicing, and you'll become a derivative-finding, slope-calculating, tangent-line-equating machine in no time! If you have any questions, feel free to ask. Happy calculating!
