Finding Dx/dt Given Y = 4x^2 + 3 And Dy/dt = 1 At X = -5

Introduction

In the realm of calculus, understanding the relationships between rates of change is paramount. This article delves into a specific problem involving related rates, a fundamental concept in differential calculus. We are tasked with finding the rate of change of $x$ with respect to $t$, denoted as $\frac{dx}{dt}$, at a specific point $x = -5$, given the relationship between $y$ and $x$ as $y = 4x^2 + 3$ and the rate of change of $y$ with respect to $t$, $\frac{dy}{dt} = 1$. This problem exemplifies how the chain rule and implicit differentiation are applied to connect seemingly disparate rates of change. By carefully differentiating the given equation and substituting the provided values, we can unravel the intricate dance between these variables and their rates of change. Let's embark on this mathematical journey to unveil the solution, step by meticulous step.

Problem Statement

We are given the equation $y = 4x^2 + 3$, which establishes a direct relationship between the variables $y$ and $x$. Additionally, we know that the rate of change of $y$ with respect to $t$ is $\frac{dy}{dt} = 1$. Our objective is to determine the rate of change of $x$ with respect to $t$, $\frac{dx}{dt}$, specifically at the point where $x = -5$. This problem is a classic example of related rates, where we use calculus to find the relationship between the rates of change of different variables that are related to each other. To solve this, we will employ implicit differentiation, a powerful technique that allows us to differentiate equations where one variable is not explicitly defined as a function of the other. By applying the chain rule, we can connect the derivatives of $y$ and $x$ with respect to $t$, and ultimately solve for the desired rate of change, $\frac{dx}{dt}$.

Differentiating the Equation

To find the relationship between $\frac{dx}{dt}$ and $\frac{dy}{dt}$, we need to differentiate the given equation $y = 4x^2 + 3$ with respect to $t$. This requires the application of implicit differentiation, a technique that allows us to differentiate equations where one variable is not explicitly defined as a function of the other. Differentiating both sides of the equation with respect to $t$, we get:

$$\frac{d}{dt}(y) = \frac{d}{dt}(4x^2 + 3)$$

On the left-hand side, we have $\frac{dy}{dt}$. On the right-hand side, we apply the chain rule to differentiate $4x^2$ with respect to $t$. The chain rule states that if $y$ is a function of $x$, and $x$ is a function of $t$, then $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$. Applying this to our equation, we get:

$$\frac{dy}{dt} = 8x \frac{dx}{dt} + 0$$

The derivative of the constant 3 with respect to $t$ is 0. Thus, our differentiated equation simplifies to:

$$\frac{dy}{dt} = 8x \frac{dx}{dt}$$

This equation now provides a direct link between the rates of change of $y$ and $x$ with respect to $t$. We can now use this equation along with the given information to solve for $\frac{dx}{dt}$. The next step involves substituting the given values into this equation and isolating the desired rate of change.

Substituting Given Values

We are given that $\frac{dy}{dt} = 1$ and we want to find $\frac{dx}{dt}$ when $x = -5$. Substituting these values into the equation we derived in the previous section, $\frac{dy}{dt} = 8x \frac{dx}{dt}$, we get:

$$1 = 8(-5) \frac{dx}{dt}$$

This equation now contains only one unknown, $\frac{dx}{dt}$. We can solve for this unknown by isolating it on one side of the equation. This involves a simple algebraic manipulation, dividing both sides of the equation by $8(-5)$. This will give us the value of $\frac{dx}{dt}$ at the specified point $x = -5$. The subsequent step involves performing this calculation and stating the final answer, which represents the rate of change of $x$ with respect to $t$ at the given condition.

Solving for dx/dt

To isolate $\frac{dx}{dt}$, we divide both sides of the equation $1 = 8(-5) \frac{dx}{dt}$ by $8(-5)$, which is $-40$:

$$\frac{1}{-40} = \frac{dx}{dt}$$

Therefore, we have:

$$\frac{dx}{dt} = -\frac{1}{40}$$

This result tells us that at the point where $x = -5$, the rate of change of $x$ with respect to $t$ is $-\frac{1}{40}$. This means that as $t$ increases, $x$ is decreasing at a rate of $\frac{1}{40}$ units per unit of $t$. The negative sign indicates that $x$ is decreasing, which is consistent with the relationship between $y$ and $x$ and the fact that $y$ is increasing at a rate of 1. This concludes the solution to the problem, providing a clear and concise answer to the question posed.

Conclusion

In conclusion, by applying the principles of implicit differentiation and the chain rule, we have successfully determined that $\frac{dx}{dt} = -\frac{1}{40}$ when $x = -5$, given $y = 4x^2 + 3$ and $\frac{dy}{dt} = 1$. This problem highlights the power of calculus in analyzing related rates and understanding how changes in one variable affect the rate of change of another. The systematic approach of differentiating the equation, substituting known values, and solving for the unknown rate is a common strategy in related rates problems. This exercise not only reinforces the application of fundamental calculus concepts but also provides a deeper appreciation for the interconnectedness of variables and their rates of change in dynamic systems. The negative value of $\frac{dx}{dt}$ indicates an inverse relationship in this specific scenario, further enriching our understanding of the problem's context and implications. This journey through related rates serves as a valuable stepping stone for tackling more complex problems in calculus and its applications.

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• Related Rates
• Implicit Differentiation
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• dy/dt
• Calculus
• Rate of Change
• Derivatives
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• Differential Calculus