Finding the Coordinates of a Point on a Curve with a Specific Gradient

In this article, we will explore how to find the coordinates of a point on the curve y frac{2x - 5}{sqrt{x_1}} where the gradient is frac{5}{4}. This process involves understanding how to calculate derivatives, solving equations, and applying the principles of calculus.

Understanding the Problem and Setup

Given the curve:

y frac{2x - 5}{sqrt{x_1}}

and the gradient (derivative) of the curve at a point being:

frac{5}{4}

Our task is to find the coordinates of the point on the curve where the gradient is frac{5}{4}.

Finding the Derivative of the Curve

Let's start by calculating the derivative of the curve using the quotient rule:

frac{dy}{dx} frac{d}{dx} left( frac{2x - 5}{sqrt{x_1}} right)

Using the quotient rule:

frac{dy}{dx} frac{(2)sqrt{x_1} - (2x - 5)frac{1}{2}(x_1)^{frac{-1}{2}}}{sqrt{x_1}}

Simplifying the expression:

frac{dy}{dx} frac{2sqrt{x_1} - (2x - 5)frac{1}{2sqrt{x_1}}}{sqrt{x_1}}

Further simplifying:

frac{dy}{dx} frac{2x_1 - (2x - 5)frac{1}{2sqrt{x_1}}}{x_1}

This can be rewritten as:

frac{dy}{dx} frac{4x_1 - (2x - 5)}{2sqrt{x_1}x_1} frac{4x_1 - 2x 5}{2x_1^{frac{3}{2}}}

Setting the derivative equal to the given gradient:

frac{4x_1 - 2x 5}{2x_1^{frac{3}{2}}} frac{5}{4}

Solving the Equation

Multiply both sides of the equation by 2x_1^{frac{3}{2}} to clear the denominator:

4x_1 - 2x 5 frac{5}{4}2x_1^{frac{3}{2}}

Multiplying both sides by 4 to clear the fraction:

16x_1 - 8x 20 5x_1^{frac{3}{2}}

Rearrange the equation to form a cubic equation:

5x_1^{frac{3}{2}} - 16x_1 8x - 20 0

Square both sides to eliminate the fractional exponent:

(5x_1^{frac{3}{2}} - 16x_1 8x - 20)^2 0

This results in a more complex polynomial equation. Using a computational tool like Wolfram Alpha, we can find the real roots of the equation. Through solving, we find that x 3.

Substitute x 3 back into the original equation to determine the y-coordinate:

y frac{2(3) - 5}{sqrt{3}} frac{6 - 5}{sqrt{3}} frac{1}{sqrt{3}}

Therefore, the coordinates of the point where the gradient is frac{5}{4} are:

(3, frac{1}{sqrt{3}})

Generalization and Application

The process demonstrated above can be generalized to other curves and gradients. By following these steps, one can accurately find the coordinates of any point on a curve where the gradient is specified.

This method is crucial in various fields such as engineering, physics, and economics where understanding the exact shape and behavior of curves is essential. Whether dealing with motion, optimization, or economic models, the ability to find specific points on functions is invaluable.

Conclusion

In conclusion, finding the coordinates of a point on a curve with a specific gradient involves a systematic approach that includes calculating derivatives, solving equations, and applying fundamental mathematical principles. By understanding and mastering these techniques, one can tackle more complex problems in various disciplines.

Whether you are a student, a professional in a technical field, or someone with a general interest in mathematics, the skills developed in this process are highly beneficial and widely applicable.