If $u = X^5 Y + Y^2 Z^3$, Where $x = R \sin T$, \$y = R^2 E^{-t}$[/tex\], And $z = R^2 \sin T$, Then Find The Value Of $\frac{\partial U}{\partial T}$ When:$\begin{array}{l} r = 2 \\ s = 1

**If $u = x^5 y + y^2 z^3$, where $x = r \sin t$, $$y = r^2 e^{-t}$[/tex], and $z = r^2 \sin t$, then find the value of $\frac{\partial u}{\partial t}$ when:$\begin{array}{l} r = 2 \ s = 1

Introduction

In this article, we will explore the concept of partial derivatives and how to find the value of $\frac{\partial u}{\partial t}$ when given a function $u = x^5 y + y^2 z^3$ and its variables $x = r \sin t$, $y = r^2 e^{-t}$, and $z = r^2 \sin t$. We will use the chain rule to find the partial derivative of $u$ with respect to $t$.

The Chain Rule

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 $u = x^5 y + y^2 z^3$ that depends on three variables $x$, $y$, and $z$, which in turn depend on two variables $r$ and $t$. To find the partial derivative of $u$ with respect to $t$, we need to use the chain rule.

Finding the Partial Derivative of $u$ with Respect to $t$

To find the partial derivative of $u$ with respect to $t$, we need to find the partial derivatives of $x$, $y$, and $z$ with respect to $t$ and then substitute them into the expression for $u$.

Finding the Partial Derivative of $x$ with Respect to $t$

We are given that $x = r \sin t$. To find the partial derivative of $x$ with respect to $t$, we can use the product rule:

$$\frac{\partial x}{\partial t} = r \cos t$$

Finding the Partial Derivative of $y$ with Respect to $t$

We are given that $y = r^2 e^{-t}$. To find the partial derivative of $y$ with respect to $t$, we can use the chain rule:

$$\frac{\partial y}{\partial t} = -r^2 e^{-t}$$

Finding the Partial Derivative of $z$ with Respect to $t$

We are given that $z = r^2 \sin t$. To find the partial derivative of $z$ with respect to $t$, we can use the product rule:

$$\frac{\partial z}{\partial t} = r^2 \cos t$$

Substituting the Partial Derivatives into the Expression for $u$

Now that we have found the partial derivatives of $x$, $y$, and $z$ with respect to $t$, we can substitute them into the expression for $u$:

$$\frac{\partial u}{\partial t} = \frac{\partial}{\partial t} (x^5 y + y^2 z^3)$$

$$= 5x^4 \frac{\partial x}{\partial t} y + x^5 \frac{\partial y}{\partial t} + 2y \frac{\partial y}{\partial t} z^3 + y^2 3z^2 \frac{\partial z}{\partial t}$$

$$= 5x^4 r \cos t y + x^5 (-r^2 e^{-t}) + 2y (-r^2 e^{-t}) z^3 + y^2 3(r^2 \sin t)^2 r^2 \cos t$$

Simplifying the Expression for $\frac{\partial u}{\partial t}$

Now that we have substituted the partial derivatives into the expression for $u$, we can simplify the expression:

$$\frac{\partial u}{\partial t} = 5x^4 r \cos t y - x^5 r^2 e^{-t} - 2r^4 e^{-2t} z^3 + 3r^6 \sin^2 t \cos t$$

Evaluating the Expression for $\frac{\partial u}{\partial t}$ when $r = 2$ and $s = 1$

Now that we have simplified the expression for $\frac{\partial u}{\partial t}$, we can evaluate it when $r = 2$ and $s = 1$.

Finding the Values of $x$, $y$, and $z$ when $r = 2$ and $t = 1$

We are given that $x = r \sin t$, $y = r^2 e^{-t}$, and $z = r^2 \sin t$. To find the values of $x$, $y$, and $z$ when $r = 2$ and $t = 1$, we can substitute these values into the expressions for $x$, $y$, and $z$:

$$x = 2 \sin 1$$

$$y = 4 e^{-1}$$

$$z = 4 \sin 1$$

Evaluating the Expression for $\frac{\partial u}{\partial t}$ when $r = 2$ and $t = 1$

Now that we have found the values of $x$, $y$, and $z$ when $r = 2$ and $t = 1$, we can evaluate the expression for $\frac{\partial u}{\partial t}$:

$$\frac{\partial u}{\partial t} = 5(2 \sin 1)^4 2 \cos 1 4 e^{-1} - (2 \sin 1)^5 2^2 e^{-1} - 2(4 e^{-1})^2 (4 \sin 1)^3 + 3(4 \sin 1)^2 (2 \cos 1)$$

$$= 5(16 \sin^4 1) 2 \cos 1 4 e^{-1} - (32 \sin^5 1) 2^2 e^{-1} - 2(16 e^{-2}) (64 \sin^3 1) + 3(16 \sin^2 1) (2 \cos 1)$$

$$= 320 \sin^4 1 \cos 1 e^{-1} - 256 \sin^5 1 e^{-1} - 2048 e^{-2} \sin^3 1 + 96 \sin^2 1 \cos 1$$

Conclusion

In this article, we have used the chain rule to find the partial derivative of $u$ with respect to $t$ when given a function $u = x^5 y + y^2 z^3$ and its variables $x = r \sin t$, $y = r^2 e^{-t}$, and $z = r^2 \sin t$. We have then evaluated the expression for $\frac{\partial u}{\partial t}$ when $r = 2$ and $t = 1$. The final answer is $\boxed{320 \sin^4 1 \cos 1 e^{-1} - 256 \sin^5 1 e^{-1} - 2048 e^{-2} \sin^3 1 + 96 \sin^2 1 \cos 1}$.
**Q&A: If $u = x^5 y + y^2 z^3$, where $x = r \sin t$, $$y = r^2 e^{-t}$[/tex], and $z = r^2 \sin t$, then find the value of $\frac{\partial u}{\partial t}$ when:$\begin{array}{l} r = 2 \ s = 1

Q&A

Q: What is the chain rule and how is it used in this problem?

A: The chain rule is a fundamental concept in calculus that allows us to find the derivative of a composite function. In this problem, we have a function $u = x^5 y + y^2 z^3$ that depends on three variables $x$, $y$, and $z$, which in turn depend on two variables $r$ and $t$. To find the partial derivative of $u$ with respect to $t$, we need to use the chain rule.

Q: How do we find the partial derivative of $x$ with respect to $t$?

A: We are given that $x = r \sin t$. To find the partial derivative of $x$ with respect to $t$, we can use the product rule:

$$\frac{\partial x}{\partial t} = r \cos t$$

Q: How do we find the partial derivative of $y$ with respect to $t$?

A: We are given that $y = r^2 e^{-t}$. To find the partial derivative of $y$ with respect to $t$, we can use the chain rule:

$$\frac{\partial y}{\partial t} = -r^2 e^{-t}$$

Q: How do we find the partial derivative of $z$ with respect to $t$?

A: We are given that $z = r^2 \sin t$. To find the partial derivative of $z$ with respect to $t$, we can use the product rule:

$$\frac{\partial z}{\partial t} = r^2 \cos t$$

Q: How do we substitute the partial derivatives into the expression for $u$?

A: Now that we have found the partial derivatives of $x$, $y$, and $z$ with respect to $t$, we can substitute them into the expression for $u$:

$$\frac{\partial u}{\partial t} = \frac{\partial}{\partial t} (x^5 y + y^2 z^3)$$

$$= 5x^4 \frac{\partial x}{\partial t} y + x^5 \frac{\partial y}{\partial t} + 2y \frac{\partial y}{\partial t} z^3 + y^2 3z^2 \frac{\partial z}{\partial t}$$

Q: How do we simplify the expression for $\frac{\partial u}{\partial t}$?

A: Now that we have substituted the partial derivatives into the expression for $u$, we can simplify the expression:

$$\frac{\partial u}{\partial t} = 5x^4 r \cos t y - x^5 r^2 e^{-t} - 2r^4 e^{-2t} z^3 + 3r^6 \sin^2 t \cos t$$

Q: How do we evaluate the expression for $\frac{\partial u}{\partial t}$ when $r = 2$ and $t = 1$?

A: Now that we have simplified the expression for $\frac{\partial u}{\partial t}$, we can evaluate it when $r = 2$ and $t = 1$.

Q: What is the final answer for $\frac{\partial u}{\partial t}$ when $r = 2$ and $t = 1$?

A: The final answer is $\boxed{320 \sin^4 1 \cos 1 e^{-1} - 256 \sin^5 1 e^{-1} - 2048 e^{-2} \sin^3 1 + 96 \sin^2 1 \cos 1}$.

Additional Resources

• For more information on the chain rule, see this article.
• For more information on partial derivatives, see this article.
• For more information on the product rule, see this article.

Conclusion

In this article, we have used the chain rule to find the partial derivative of $u$ with respect to $t$ when given a function $u = x^5 y + y^2 z^3$ and its variables $x = r \sin t$, $y = r^2 e^{-t}$, and $z = r^2 \sin t$. We have then evaluated the expression for $\frac{\partial u}{\partial t}$ when $r = 2$ and $t = 1$. The final answer is $\boxed{320 \sin^4 1 \cos 1 e^{-1} - 256 \sin^5 1 e^{-1} - 2048 e^{-2} \sin^3 1 + 96 \sin^2 1 \cos 1}$.