Curve Gradient: Find The Point With Gradient 3

When we talk about curves in mathematics, one of the most fascinating aspects is their gradient. The gradient, essentially, tells us how steep a curve is at any given point. It's the instantaneous rate of change of the function, and understanding it unlocks a world of possibilities in calculus. Today, we're going to dive deep into a specific problem: finding the coordinates of a point on the curve defined by the equation $y = x^2 + 5x - 4$ where the gradient is exactly 3. This isn't just about crunching numbers; it's about understanding the relationship between a function, its derivative, and the geometric interpretation of that derivative. We'll break down the process step-by-step, making sure you grasp each concept, from differentiation to solving for the unknown coordinates. So, grab your favorite thinking cap, and let's get started on this mathematical adventure!

Understanding the Gradient

The gradient of a curve at a particular point is a fundamental concept in calculus. It represents the slope of the tangent line to the curve at that point. Imagine a roller coaster; the gradient at any point tells you how steep the track is right there – whether you're going uphill, downhill, or on a flat section. Mathematically, the gradient of a function $y = f(x)$ is given by its derivative, denoted as $dy/dx$ or $f'(x)$. The derivative is found by applying the rules of differentiation to the function. For our specific curve, $y = x^2 + 5x - 4$, we need to find the derivative to express the gradient as a function of $x$. The power rule is our best friend here: for any term of the form $ax^n$, its derivative is $anx^{n-1}$. Applying this to our equation:

• The derivative of $x^2$ is $2x^{2-1} = 2x$.
• The derivative of $5x$ (which is $5x^1$) is $5 imes 1x^{1-1} = 5x^0 = 5 imes 1 = 5$.
• The derivative of a constant, $-4$, is 0.

So, the derivative of $y = x^2 + 5x - 4$ is $dy/dx = 2x + 5$. This expression, $2x + 5$, represents the gradient of the curve at any point $x$. It's a powerful tool because it allows us to calculate the steepness of the curve for any value of $x$ we choose. For instance, if we wanted to know the gradient at $x=1$, we'd simply substitute 1 into $2x+5$, giving us $2(1)+5 = 7$. This means at the point where $x=1$ on the curve, the tangent line has a slope of 7. Now, our task is to find the specific point where this gradient is not just any value, but a predetermined value: 3.

Setting Up the Equation

We've established that the gradient of the curve $y = x^2 + 5x - 4$ at any point $x$ is given by the derivative $dy/dx = 2x + 5$. The problem asks us to find the point where this gradient is equal to 3. This is where we bridge the gap between the general expression for the gradient and the specific condition given in the problem. We simply set our gradient expression equal to the desired gradient value:

$2x + 5 = 3$

This is a straightforward linear equation in $x$. Our goal is to solve this equation for $x$. Once we find the value of $x$ that satisfies this condition, we will have found the x-coordinate of the point we are looking for. This equation represents the core of the problem – it's the mathematical statement that translates the problem's requirement into something we can solve. It signifies that we are looking for the specific location on the curve where its instantaneous steepness matches the value of 3. This is a common strategy in calculus problems: use the derivative to define the gradient and then set that derivative equal to a given value to pinpoint specific points of interest on the curve. The simplicity of this resulting equation, $2x + 5 = 3$, belies the underlying calculus that got us here. It's a beautiful example of how abstract mathematical concepts like derivatives can be applied to solve concrete problems about the shape and behavior of functions. The next step, of course, is to solve this equation to find that crucial $x$-value.

Solving for the x-coordinate

Now that we have our equation, $2x + 5 = 3$, we can proceed to solve for $x$. This is a basic algebraic manipulation. Our aim is to isolate $x$ on one side of the equation. First, we'll subtract 5 from both sides of the equation to get the term with $x$ by itself:

$2x + 5 - 5 = 3 - 5$

This simplifies to:

$2x = -2$

Next, to find the value of $x$, we need to divide both sides of the equation by 2:

$(2x) / 2 = -2 / 2$

Which gives us:

$x = -1$

So, we have found the x-coordinate of the point where the gradient of the curve $y = x^2 + 5x - 4$ is 3. It is $x = -1$. This means that at the point on the curve where the horizontal position is -1, the curve is sloping upwards with a gradient of 3. It's important to note that this is only half of the information we need. The problem asks for the coordinates of the point, which means we need both the $x$ and the $y$ values. Finding the $x$-coordinate is often the first step, and it's driven by the derivative. The next logical step is to use this $x$-coordinate to find the corresponding $y$-coordinate on the original curve.

Finding the y-coordinate

We have successfully found the x-coordinate of the point where the gradient is 3, and it is $x = -1$. To find the complete coordinates of this point, we need to determine its corresponding y-coordinate. We do this by substituting our found $x$-value back into the original equation of the curve, which is $y = x^2 + 5x - 4$. It's crucial to use the original equation, not the derivative, because the original equation defines the actual position of the points on the curve, whereas the derivative defines the gradient at those points.

Substituting $x = -1$ into the equation:

$y = (-1)^2 + 5(-1) - 4$

Now, let's calculate the value of $y$ step-by-step:

• $(-1)^2 = 1$ (a negative number squared is positive).
• $5(-1) = -5$.

So, the equation becomes:

$y = 1 - 5 - 4$

Performing the subtraction:

$y = -4 - 4$

$y = -8$

Therefore, the y-coordinate of the point where the gradient is 3 is -8. We now have both the x-coordinate and the y-coordinate. The point on the curve $y = x^2 + 5x - 4$ where the gradient is 3 is $(-1, -8)$. This is the final answer to our problem. It's a specific location on the parabola defined by the equation, and at this precise spot, the slope of the tangent line is exactly 3. This process illustrates the power of calculus in linking the rate of change (gradient) to the actual points on a curve. It's a fundamental technique used in many areas of mathematics, physics, and engineering, from optimization problems to analyzing motion.

Conclusion

We've successfully navigated the process of finding a specific point on a curve based on its gradient. By understanding that the derivative of a function gives us its gradient at any point, we were able to set up and solve an equation to find the x-coordinate. Subsequently, substituting this x-coordinate back into the original curve equation allowed us to determine the corresponding y-coordinate. The point on the curve $y = x^2 + 5x - 4$ where the gradient is 3 is indeed (-1, -8). This method is a cornerstone of differential calculus and has wide-ranging applications. Whether you're studying physics, engineering, economics, or pure mathematics, the ability to relate a curve's shape to its rate of change is invaluable.

For further exploration into the fascinating world of calculus and its applications, I highly recommend visiting websites like Khan Academy which offers comprehensive resources on calculus topics, or the MathWorld website by Wolfram Research, for in-depth mathematical explanations and formulas.