Equation Of Normal To Curve Y = X² - 8x + 15 At (0, 15)

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Let's dive into how to find the equation of the normal to the curve y = x² - 8x + 15 at the point (0, 15). This is a classic calculus problem that combines differentiation with coordinate geometry. We'll break it down step by step to make it super clear and easy to follow. Understanding the concepts of derivatives, slopes, and perpendicular lines is crucial here, so let’s get started and unravel this problem together! This guide is designed to help you grasp the underlying principles and apply them confidently to similar problems. Let's explore the step-by-step approach to solve this problem effectively.

1. Understanding the Problem: Normal to a Curve

Before we jump into calculations, let's make sure we understand what the question is asking. The normal to a curve at a given point is a line that is perpendicular to the tangent at that same point. So, to find the equation of the normal, we first need to find the slope of the tangent. Think of it like this: the tangent line just grazes the curve at the point, while the normal line stands straight up (or at a right angle) to that graze.

  • The key concept here is that the normal line is perpendicular to the tangent line at the given point. This means their slopes have a special relationship: they are negative reciprocals of each other. If the slope of the tangent is m, the slope of the normal will be -1/m. This inverse relationship is essential for solving the problem correctly, ensuring we find the line that is truly perpendicular to the curve's tangent at the specified point. Understanding this geometric connection helps in visualizing the problem and setting up the solution.

  • We're given the curve equation y = x² - 8x + 15 and the point (0, 15). Our goal is to find the equation of a line, and remember, the general form of a line equation is y = mx + c, where m is the slope and c is the y-intercept. To nail this, we need to figure out the slope of the normal at (0, 15) and then use that to find the full equation. The point (0, 15) gives us a specific location on the curve where we can anchor our normal line, making our solution concrete and geometrically sound. This careful setup ensures we’re on the right track from the start.

  • So, the plan is clear: (1) find the slope of the tangent, (2) find the slope of the normal, and (3) use the point-slope form to write the equation of the normal. Each step builds on the previous one, leading us to the final solution in a logical and structured manner. This systematic approach not only solves the problem at hand but also enhances our problem-solving skills for future challenges. By breaking the problem down into manageable steps, we make the process more approachable and less daunting.

2. Finding the Slope of the Tangent

Calculus comes to our rescue here! The slope of the tangent to a curve at a point is given by the derivative of the curve's equation evaluated at that point. So, we need to find dy/dx for y = x² - 8x + 15.

  • Let's differentiate y = x² - 8x + 15 with respect to x. Remember the power rule: the derivative of xⁿ is nxⁿ⁻¹. Applying this, the derivative of x² is 2x, the derivative of -8x is -8, and the derivative of the constant 15 is 0. So, dy/dx = 2x - 8. This derivative represents the slope of the tangent line at any point x on the curve. Understanding the power rule and applying it correctly is fundamental to finding the derivative accurately.

  • Now, we need the slope of the tangent specifically at the point (0, 15). This means we need to evaluate dy/dx at x = 0. Plugging in x = 0 into dy/dx = 2x - 8, we get dy/dx = 2(0) - 8 = -8. So, the slope of the tangent at (0, 15) is -8. This specific value is crucial because it allows us to pinpoint the tangent's steepness at the exact location we are interested in, setting the stage for finding the normal line.

  • Therefore, the slope of the tangent line, m_tangent, at the point (0, 15) is -8. This is a key piece of information because it directly relates to the slope of the normal line, which we'll find in the next step. The negative slope indicates that the tangent line is decreasing as x increases, providing us with a visual sense of how the curve behaves at this point. Knowing m_tangent is a necessary stepping stone to finding the perpendicular slope that defines the normal line.

3. Finding the Slope of the Normal

We know the slope of the tangent is -8. Since the normal is perpendicular to the tangent, its slope is the negative reciprocal of the tangent's slope.

  • Recall that the slopes of perpendicular lines are negative reciprocals of each other. If the slope of the tangent (m_tangent) is -8, the slope of the normal (m_normal) is -1/m_tangent = -1/(-8) = 1/8. This negative reciprocal relationship is a fundamental principle in coordinate geometry, ensuring that the normal line intersects the tangent line at a perfect right angle. The positive slope of the normal indicates that it is increasing as x increases, contrasting with the tangent's decreasing slope.

  • So, the slope of the normal to the curve at the point (0, 15) is 1/8. This value is pivotal because it defines the direction of the normal line. We now have the m part of our line equation (y = mx + c). Knowing the slope is a major step forward, allowing us to start constructing the equation of the normal line with precision. This positive fractional slope gives us a clear sense of the normal line's orientation.

  • To summarize, we've successfully navigated the derivative to find the tangent's slope and then used the perpendicularity condition to find the normal's slope. This two-step process highlights the interplay between calculus and geometry in solving such problems. We now have the essential slope information needed to complete the final step: constructing the equation of the normal line. The careful attention to detail in each step ensures the accuracy of our final result.

4. Finding the Equation of the Normal

Now we have the slope of the normal (1/8) and a point it passes through (0, 15). We can use the point-slope form of a line equation: y - y₁ = m(x - x₁), where (x₁, y₁) is the point and m is the slope.

  • The point-slope form is perfect for this scenario because it allows us to directly plug in the known slope and point to generate the line equation. Using the point (0, 15) and the slope 1/8, we have y - 15 = (1/8)(x - 0). This equation captures the essential characteristics of the normal line, anchoring it precisely at the specified point and directing its orientation according to the calculated slope. This form is especially useful as it bypasses the need to explicitly calculate the y-intercept in the initial setup.

  • Let’s simplify the equation. y - 15 = (1/8)x can be rearranged to y = (1/8)x + 15. This is the slope-intercept form of the equation, where 1/8 is the slope and 15 is the y-intercept. This form is widely recognized and easy to interpret, giving a clear picture of the line's behavior and its intersection with the y-axis. Converting to slope-intercept form makes it straightforward to graph the line and understand its position relative to the curve.

  • Thus, the equation of the normal to the curve y = x² - 8x + 15 at the point (0, 15) is y = (1/8)x + 15. We have successfully constructed the equation of the normal line, providing a complete and accurate solution to the problem. This equation precisely describes the line that is perpendicular to the curve’s tangent at the given point. Reviewing each step confirms the logical progression and mathematical correctness of our solution, ensuring a solid understanding of the process.

5. Conclusion

In summary, to find the equation of the normal to the curve y = x² - 8x + 15 at the point (0, 15), we followed these steps:

  1. Found the derivative dy/dx = 2x - 8 to get the slope of the tangent.
  2. Evaluated the derivative at x = 0 to find the tangent's slope at the point (0, 15), which was -8.
  3. Calculated the slope of the normal as the negative reciprocal of the tangent's slope, resulting in 1/8.
  4. Used the point-slope form with the point (0, 15) and the normal's slope to derive the equation y = (1/8)x + 15.

This systematic approach, combining calculus and coordinate geometry, allowed us to solve the problem effectively. Remember, understanding the underlying concepts is key to tackling similar problems with confidence. Practice is your best friend, so try out more problems and solidify your understanding!

For further exploration and practice on calculus concepts, check out resources like Khan Academy's Calculus section. They offer comprehensive lessons and exercises to help you master these techniques.