Problem 1
Question
In Problems 1-6, find \(d w / d t\) by using the Chain Rule. Express your final answer in terms of \(t\). $$ w=x^{2} y^{3}, x=t^{3}, y=t^{2} $$
Step-by-Step Solution
Verified Answer
\( \frac{dw}{dt} = 12t^{11} \).
1Step 1: Identify Variables
First, we identify the functions involved. We have the function \( w = x^2 y^3 \), with \( x = t^3 \) and \( y = t^2 \).
2Step 2: Determine Derivatives of x and y
Calculate the derivatives of \( x \) and \( y \) with respect to \( t \). We have \( x = t^3 \) so \( \frac{dx}{dt} = 3t^2 \), and \( y = t^2 \) so \( \frac{dy}{dt} = 2t \).
3Step 3: Apply the Chain Rule to dw/dt
The Chain Rule for \( w = x^2 y^3 \) is \( \frac{dw}{dt} = \frac{\partial w}{\partial x} \frac{dx}{dt} + \frac{\partial w}{\partial y} \frac{dy}{dt} \).
4Step 4: Partial Derivative with Respect to x
Find \( \frac{\partial w}{\partial x} \) by treating \( y \) as a constant: \( \frac{\partial}{\partial x} (x^2 y^3) = 2x y^3 \).
5Step 5: Partial Derivative with Respect to y
Find \( \frac{\partial w}{\partial y} \) by treating \( x \) as a constant: \( \frac{\partial}{\partial y} (x^2 y^3) = 3y^2 x^2 \).
6Step 6: Substitute and Simplify
Substitute \( x = t^3 \), \( y = t^2 \), \( \frac{dx}{dt} = 3t^2 \), and \( \frac{dy}{dt} = 2t \) into the expression for \( \frac{dw}{dt} \):\( \frac{dw}{dt} = (2x y^3)(3t^2) + (3y^2 x^2)(2t) \).
7Step 7: Final Expression in Terms of t
Substitute back \( x = t^3 \) and \( y = t^2 \), and simplify:\( \frac{dw}{dt} = (2(t^3)(t^2)^3)(3t^2) + (3(t^2)^2(t^3)^2)(2t) \).Expand and simplify to obtain the final form in terms of \( t \).
8Step 8: Simplify Further
Simplifying further, we have:\( \frac{dw}{dt} = 6t^{11} + 6t^{11} = 12t^{11} \).
Key Concepts
Partial DerivativesParametric DifferentiationCalculus Problem Solving
Partial Derivatives
Partial derivatives are a fundamental concept in calculus. They involve finding the derivative of a function with respect to one variable while keeping other variables constant.
In the given exercise, we have a function of two variables, namely, \( w = x^2y^3 \). To differentiate \( w \) partially with respect to \( x \), we assume \( y \) is constant. Thus, the partial derivative is computed as \( \frac{\partial w}{\partial x} = 2x y^3 \).
Similarly, when differentiating \( w \) with respect to \( y \), \( x \) is treated as constant. The result is \( \frac{\partial w}{\partial y} = 3y^2 x^2 \).
In the given exercise, we have a function of two variables, namely, \( w = x^2y^3 \). To differentiate \( w \) partially with respect to \( x \), we assume \( y \) is constant. Thus, the partial derivative is computed as \( \frac{\partial w}{\partial x} = 2x y^3 \).
Similarly, when differentiating \( w \) with respect to \( y \), \( x \) is treated as constant. The result is \( \frac{\partial w}{\partial y} = 3y^2 x^2 \).
- Partial derivatives help understand how a function changes as one variable varies while others stay fixed.
- They are essential for applying the Chain Rule in multivariable calculus.
Parametric Differentiation
Parametric differentiation involves expressing variables in terms of another parameter. Here, both \( x \) and \( y \) are functions of \( t \). This way, we can understand the behavior of \( w \) as \( t \) changes.
In our exercise, the parameter is \( t \), with \( x = t^3 \) and \( y = t^2 \). Hence, differentiating these with respect to \( t \) gives us \( \frac{dx}{dt} = 3t^2 \) and \( \frac{dy}{dt} = 2t \).
This approach provides a way to analyze dynamic systems, showing how a change in one independent variable affects dependent variables connected through parametric relations.
In our exercise, the parameter is \( t \), with \( x = t^3 \) and \( y = t^2 \). Hence, differentiating these with respect to \( t \) gives us \( \frac{dx}{dt} = 3t^2 \) and \( \frac{dy}{dt} = 2t \).
This approach provides a way to analyze dynamic systems, showing how a change in one independent variable affects dependent variables connected through parametric relations.
- Parametric differentiation translates variables in terms of a common parameter.
- It's particularly useful when variables are interlinked with respect to another changing quantity.
Calculus Problem Solving
Calculus problem solving often requires a combination of methods like applying the Chain Rule and using partial derivatives. In our example, the Chain Rule is used to compute the total derivative \( \frac{dw}{dt} \) based on known connections due to the parameter \( t \).
According to the Chain Rule, \( \frac{dw}{dt} = \frac{\partial w}{\partial x} \frac{dx}{dt} + \frac{\partial w}{\partial y} \frac{dy}{dt} \). By plugging in the derivatives we computed earlier, we find \( \frac{dw}{dt} = 6t^{11} + 6t^{11} \).
After simplification, we get \( \frac{dw}{dt} = 12t^{11} \). This showcases the power of calculus in breaking down complex relationships into manageable steps and providing a clear solution.
According to the Chain Rule, \( \frac{dw}{dt} = \frac{\partial w}{\partial x} \frac{dx}{dt} + \frac{\partial w}{\partial y} \frac{dy}{dt} \). By plugging in the derivatives we computed earlier, we find \( \frac{dw}{dt} = 6t^{11} + 6t^{11} \).
After simplification, we get \( \frac{dw}{dt} = 12t^{11} \). This showcases the power of calculus in breaking down complex relationships into manageable steps and providing a clear solution.
- Combining different calculus techniques aids in solving intricate problems efficiently.
- Understanding the interdependencies between variables allows for constructing a comprehensive solution.
Other exercises in this chapter
Problem 1
Find the minimum of \(f(x, y)=x^{2}+y^{2}\) subject to the constraint \(g(x, y)=x y-3=0\)
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Find all critical points. Indicate whether each such point gives a local maximum or a local minimum, or whether it is a saddle point. Hint: Use Theorem \(\mathr
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Find the equation of the tangent plane to the given surface at the indicated point. \(x^{2}+y^{2}+z^{2}=16 ;(2,3, \sqrt{3})\)
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Find the gradient \(\nabla f\). $$ f(x, y)=x^{2} y+3 x y $$
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