Question

If \(z=x^{2} y, x=2 t+s\), and \(y=1-s t^{2}\), find $$ \left.\frac{\partial z}{\partial t}\right|_{s=1, t=-2} $$

Short answer

The partial derivative is 72.

Step-by-step solution

Understanding the Problem

We are given the function \( z = x^2 y \), where \( x = 2t + s \) and \( y = 1 - st^2 \), and we need to find the partial derivative \( \left.\frac{\partial z}{\partial t}\right|_{s=1, t=-2} \). This means we need to differentiate \( z \) with respect to \( t \) while treating \( s \) as a constant, and evaluate the derivative at \( s = 1 \) and \( t = -2 \).

Expressing z in terms of t and s

Substitute \( x = 2t + s \) and \( y = 1 - st^2 \) into \( z = x^2 y \). This gives us:\[z = (2t + s)^2 (1 - st^2)\]Now, expand the expression to simplify it.

Expanding the expression

First, expand \((2t + s)^2\):\[(2t + s)^2 = 4t^2 + 4ts + s^2\]Then, substitute back into the expression for \( z \):\[z = (4t^2 + 4ts + s^2)(1 - st^2)\]Multiply to expand \( z \):\[z = (4t^2 + 4ts + s^2) - st^2(4t^2 + 4ts + s^2)\]Which leads to:\[z = 4t^2 + 4ts + s^2 - 4st^4 - 4st^3s - st^2s^2\]Simplify this further.

Simplifying the expression

From the previous step, we have:\[z = 4t^2 + 4ts + s^2 - 4st^4 - 4s^2t^3 - s^2t^2\]This simplifies the expression completely in preparation for taking the derivative.

Taking the partial derivative

Differentiate \( z \) with respect to \( t \) (treat \( s \) as a constant):\[\frac{\partial z}{\partial t} = \frac{\partial}{\partial t}(4t^2 + 4ts + s^2 - 4st^4 - 4s^2t^3 - s^2t^2)\]This gives us:\[\frac{\partial z}{\partial t} = 8t + 4s - 16st^3 - 12s^2t^2 - 2s^2t\]

Evaluating the derivative at specified values

Substitute \( s = 1 \) and \( t = -2 \) into the derivative:\[\frac{\partial z}{\partial t} = 8(-2) + 4(1) - 16(1)(-2)^3 - 12(1)^2(-2)^2 - 2(1)^2(-2)\]Calculate the values step-by-step:- First term: \( 8(-2) = -16 \)- Second term: \( 4(1) = 4 \)- Third term: \( -16(-8) = 128 \) (since \((-2)^3 = -8\))- Fourth term: \( -12(4) = -48 \) (since \((-2)^2 = 4\))- Fifth term: \( -2(-2) = 4 \)Combine all:\[-16 + 4 + 128 - 48 + 4 = 72\]

Conclusion

The value of \( \left.\frac{\partial z}{\partial t}\right|_{s=1, t=-2} \) is \( 72 \).

Key concepts

calculus problem solving

To tackle calculus problems effectively, we must follow a structured approach. Start by fully understanding the problem's requirements. Identify what function you are working with and the variables involved, especially in multivariable settings. In our exercise, the task is to find the partial derivative of a composite function with specific conditions for the variables involved. Understanding the setup is crucial for accurate problem-solving.

Next, translate the problem into mathematical expressions that make it easier to manipulate. Substitute given expressions, such as variables in terms of other variables, into the main function. This step simplifies the problem and prepares it for differentiation or integration. Do not rush through this step; an error here can complicate the whole solving process.

Once simplified, perform the required calculus operations like differentiation or integration. Double-check each derivation step to prevent simple errors. In our case, finding the partial derivative involves careful differentiation, considering each term's impact. Then finally, substitute any specific values given in the problem for evaluation. Always verify your final value to ensure accuracy.

multivariable calculus

Multivariable calculus deals with functions that depend on two or more variables. This branch extends the methods of single-variable calculus to higher dimensions. A key concept here is understanding how variables interact within a function. For example, in our exercise, the function \(z = x^2 y\) involves both \(x\) and \(y\), each in turn a function of \(t\) and \(s\).

In multivariable calculus, you often deal with partial derivatives which measure how a function changes as one variable changes, holding the other constant. This is distinct from total derivatives, which consider changes in all variables at once. Practicing partial derivatives helps you predict and understand changes in multivariable systems, a reason it's crucial in fields like physics and engineering.

Another dimension of multivariable calculus is the notion of gradients and tangent planes at points of interest on surfaces defined by such functions. Each partial derivative contributes to the gradient vector, offering insights into the function's behavior in different directions.

chain rule in calculus

The chain rule is a pivotal calculus tool, especially in problems involving composite functions like ours. In essence, the chain rule helps us differentiate functions nested inside each other. When a function is the composition of two or more functions, the chain rule systematically breaks down the differentiation task.

For our exercise, we have nested functions where \(z = x^2 y, x = 2t + s\), and \(y = 1 - st^2\). The chain rule allows us to differentiate \(z\) efficiently with respect to \(t\) by considering how changes in \(t\) affect \(z\) through \(x\) and \(y\). This involves calculating derivatives of \(x\) and \(y\) concerning \(t\) and then assessing their contribution to the change in \(z\).

The chain rule can be visualized as a flow of changes in variables. If you change one variable slightly, how does it impact the overall function? By dissecting each layer of a compound function, the chain rule provides a clear and organized layout for finding derivatives, making it essential for solving complex calculus problems.