Proof of the Fundamental Theorem of Calculus

Oct 13, 2025 by thesweetdesigns.com 130 views

If a Particle Moves Along a Curve: Finding Derivatives

In this article, we will explore the concept of finding derivatives of a function that represents the position of a particle moving along a curve. Specifically, we will consider the curve $$y=x^{\frac{2}{3}}$$ and find the derivative of $$y$$ with respect to time $$t$$ when $$\frac{d x}{d t}=3$$ for all $$x$$. This problem requires the application of the chain rule and the fundamental theorem of calculus.

The chain rule is a fundamental concept in calculus that allows us to find the derivative of a composite function. In this case, we have a function $$y=x^{\frac{2}{3}}$$, which is a composite function of $$x$$ and $$t$$. To find the derivative of $$y$$ with respect to $$t$$, we need to use the chain rule.

The chain rule states that if we have a composite function $$y=f(g(x))$$, then the derivative of $$y$$ with respect to $$x$$ is given by:

\frac{d y}{d x}=\frac{d f}{d g}\cdot\frac{d g}{d x}

In our case, we have $$y=x^{\frac{2}{3}}$$, which can be written as $$y=(x^{{\frac{1}{3}})}2$$. Therefore, we can apply the chain rule as follows:

\frac{d y}{d x}=\frac{d}{d x}((x^{\frac{1}{3}})^2)=2x^{\frac{1}{3}}\cdot\frac{d}{d x}(x^{\frac{1}{3}})=2x^{\frac{1}{3}}\cdot\frac{1}{3}x^{-\frac{2}{3}}=\frac{2}{3}x^{-\frac{1}{3}}

Now that we have found the derivative of $$y$$ with respect to $$x$$, we can use the chain rule to find the derivative of $$y$$ with respect to $$t$$. We are given that $$\frac{d x}{d t}=3$$ for all $$x$$. Therefore, we can substitute this expression into the derivative of $$y$$ with respect to $$x$$ to get:

\frac{d y}{d t}=\frac{d y}{d x}\cdot\frac{d x}{d t}=\frac{2}{3}x^{-\frac{1}{3}}\cdot3=\frac{2}{x^{\frac{1}{3}}}

Now that we have found the derivative of $$y$$ with respect to $$t$$, we can use this expression to find the value of $$\frac{d y}{d t}$$ when $$x=-1$$. Substituting $$x=-1$$ into the expression for $$\frac{d y}{d t}$$, we get:

\frac{d y}{d t}=\frac{2}{(-1)^{\frac{1}{3}}}

Since $$(-1)^{\frac{1}{3}}=-1$$, we have:

\frac{d y}{d t}=\frac{2}{-1}=-2

Therefore, the value of $$\frac{d y}{d t}$$ when $$x=-1$$ is $$-2$$.

Similarly, we can use the expression for $$\frac{d y}{d t}$$ to find the value of $$\frac{d y}{d t}$$ when $$x=1$$. Substituting $$x=1$$ into the expression for $$\frac{d y}{d t}$$, we get:

\frac{d y}{d t}=\frac{2}{(1)^{\frac{1}{3}}}

Since $$(1)^{\frac{1}{3}}=1$$, we have:

\frac{d y}{d t}=\frac{2}{1}=2

Therefore, the value of $$\frac{d y}{d t}$$ when $$x=1$$ is $$2$$.

In this article, we have used the chain rule to find the derivative of a function that represents the position of a particle moving along a curve. We have found the derivative of $$y$$ with respect to $$t$$ and used this expression to find the value of $$\frac{d y}{d t}$$ when $$x=-1$$ and $$x=1$$. This problem requires the application of the chain rule and the fundamental theorem of calculus.

[1] Calculus, 3rd edition, Michael Spivak
[2] Calculus, 2nd edition, James Stewart

Proof of the Chain Rule: The chain rule can be proved using the definition of a derivative. Let $$y=f(g(x))$$ be a composite function. Then, we can write:

\frac{d y}{d x}=\lim_{h\to0}\frac{f(g(x+h))-f(g(x))}{h}

Using the definition of a derivative, we can rewrite this expression as:

\frac{d y}{d x}=\lim_{h\to0}\frac{f(g(x+h))-f(g(x))}{h}=\lim_{h\to0}\frac{f(g(x+h))-f(g(x))}{g(x+h)-g(x)}\cdot\frac{g(x+h)-g(x)}{h}

Since $$\lim_{h\to0}\frac{g(x+h)-g(x)}{h}=\frac{d g}{d x}$$, we have:

\frac{d y}{d x}=\frac{d f}{d g}\cdot\frac{d g}{d x}

This proves the chain rule.

Proof of the Fundamental Theorem of Calculus: The fundamental theorem of calculus states that if $$F(x)=\int_{a}^{x}f(t)dt$$, then $$F'(x)=f(x)$$. To prove this, we can use the definition of a derivative. Let $$F(x)=\int_{a}^{x}f(t)dt$$. Then, we can write:

F'(x)=\lim_{h\to0}\frac{F(x+h)-F(x)}{h}

Using the definition of an integral, we can rewrite this expression as:

F'(x)=\lim_{h\to0}\frac{\int_{a}^{x+h}f(t)dt-\int_{a}^{x}f(t)dt}{h}

Since $$\int_{a}^{{x+h}f(t)dt=\int_{a}}{x}f(t)dt+\int_{x}^{x+h}f(t)dt$$, we have:

F'(x)=\lim_{h\to0}\frac{\int_{x}^{x+h}f(t)dt}{h}

Using the definition of an integral, we can rewrite this expression as:

F'(x)=\lim_{h\to0}\frac{1}{h}\int_{x}^{x+h}f(t)dt

Since $$\lim_{h\to0}\frac{1}{h}\int_{x}^{x+h}f(t)dt=f(x)$$, we have:

F'(x)=f(x)

This proves the fundamental theorem of calculus.
Q&A: If a Particle Moves Along a Curve

A: The curve that the particle is moving along is given by the equation $$y=x^{\frac{2}{3}}$$.

A: The derivative of $$y$$ with respect to $$x$$ is given by:

\frac{d y}{d x}=\frac{2}{3}x^{-\frac{1}{3}}

A: The derivative of $$y$$ with respect to $$t$$ is given by:

\frac{d y}{d t}=\frac{2}{x^{\frac{1}{3}}}

A: The value of $$\frac{d y}{d t}$$ when $$x=-1$$ is $$-2$$.

A: The value of $$\frac{d y}{d t}$$ when $$x=1$$ is $$2$$.

A: The chain rule is a fundamental concept in calculus that allows us to find the derivative of a composite function. It states that if we have a composite function $$y=f(g(x))$$, then the derivative of $$y$$ with respect to $$x$$ is given by:

\frac{d y}{d x}=\frac{d f}{d g}\cdot\frac{d g}{d x}

A: The fundamental theorem of calculus states that if $$F(x)=\int_{a}^{x}f(t)dt$$, then $$F'(x)=f(x)$$. This theorem allows us to find the derivative of an integral.

A: To apply the chain rule, you need to identify the outer and inner functions of the composite function. Then, you need to find the derivatives of the outer and inner functions with respect to the inner variable. Finally, you need to multiply the derivatives of the outer and inner functions to get the derivative of the composite function.

A: To apply the fundamental theorem of calculus, you need to identify the integral that you want to find the derivative of. Then, you need to use the definition of an integral to rewrite the integral as a function of the upper limit of integration. Finally, you need to use the chain rule to find the derivative of the function with respect to the upper limit of integration.

A: The chain rule and the fundamental theorem of calculus have many applications in physics, engineering, and economics. Some common applications include:

Finding the velocity and acceleration of an object moving along a curve
Finding the maximum and minimum values of a function
Finding the area and volume of a region
Finding the work done by a force on an object
Finding the energy of a system

A: To practice using the chain rule and the fundamental theorem of calculus, you can try the following exercises:

Find the derivative of a composite function using the chain rule
Find the derivative of an integral using the fundamental theorem of calculus
Use the chain rule and the fundamental theorem of calculus to solve problems in physics, engineering, and economics
Practice finding the maximum and minimum values of a function using the chain rule and the fundamental theorem of calculus

By practicing using the chain rule and the fundamental theorem of calculus, you can develop your skills and become more confident in your ability to solve problems in calculus.