Problem 217 - Calculus 3

Question

For the following exercises, use the information provided to solve the problem. If \(w=5 x^{2}+2 y^{2}, x=-3 s+t, \quad\) and \(y=s-4 t\) find \(\frac{\partial w}{\partial s}\) and \(\frac{\partial w}{\partial t}\)

Step-by-Step Solution

Answer

\(\frac{\partial w}{\partial s} = 94s - 46t\) and \(\frac{\partial w}{\partial t} = -46s + 74t\).

Step 1: Substitute expressions for x and y into w

First, we need to express \(w\) in terms of \(s\) and \(t\) by substituting \(x = -3s + t\) and \(y = s - 4t\) into the equation for \(w\). This gives us: \[ w = 5(-3s + t)^2 + 2(s - 4t)^2 \].

Step 2: Expand the expressions

Next, expand the expressions \((-3s + t)^2\) and \((s - 4t)^2\). For \((-3s + t)^2\): \((-3s + t)^2 = 9s^2 - 6st + t^2\) For \((s - 4t)^2\): \((s - 4t)^2 = s^2 - 8st + 16t^2\).

Step 3: Simplify w in terms of s and t

Substitute the expanded forms back into \(w\): \[ w = 5(9s^2 - 6st + t^2) + 2(s^2 - 8st + 16t^2) \] Next, distribute the coefficients: \[ w = 45s^2 - 30st + 5t^2 + 2s^2 - 16st + 32t^2 \] Combine like terms: \[ w = 47s^2 - 46st + 37t^2 \].

Step 4: Find the partial derivative with respect to s

Differentiate \(w = 47s^2 - 46st + 37t^2\) with respect to \(s\) while treating \(t\) as a constant: \[ \frac{\partial w}{\partial s} = \frac{d}{ds}(47s^2) - \frac{d}{ds}(46st) + \frac{d}{ds}(37t^2) \] Calculating each term gives: \[ \frac{\partial w}{\partial s} = 94s - 46t \].

Step 5: Find the partial derivative with respect to t

Differentiate \(w = 47s^2 - 46st + 37t^2\) with respect to \(t\) while treating \(s\) as a constant: \[ \frac{\partial w}{\partial t} = \frac{d}{dt}(47s^2) - \frac{d}{dt}(46st) + \frac{d}{dt}(37t^2) \] Calculating each term gives: \[ \frac{\partial w}{\partial t} = -46s + 74t \].

Key Concepts

Chain Rule

In calculus, the Chain Rule is a fundamental method used when dealing with composite functions. A composite function involves one function nested inside another, and the Chain Rule helps us differentiate such a function. Picture it as peeling layers of an onion to get to the inside.

When you differentiate a composite function, you work from the outside in. Suppose you have a function \(f(g(x))\). Using the Chain Rule, the derivative is found by first taking the derivative of the outer function \(f\) with respect to the inside \(g\), and then multiplying it by the derivative of the inner function \(g\). Mathematically, it is expressed as\[\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)\]This principle becomes valuable in multivariable calculus when functions depend on more than one variable, like in our exercise with \(w\). Here, we use the Chain Rule when substituting \(x = -3s + t\) and \(y = s - 4t\) into \(w\) to re-express it in terms of \(s\) and \(t\).

• It ensures you account for all variable changes.
• Don't forget about the linkage between variables.
• Helps in keeping your calculations on track.

Multivariable Calculus

Multivariable Calculus extends the principles of single-variable calculus to functions of multiple variables. This branch of calculus allows us to explore how functions behave when they depend on two or more variables, such as \(x\), \(y\), \(s\), and \(t\) in our problem.

In multivariable calculus, one crucial aspect is analyzing how changing one variable affects the entire function while holding others constant. This is where the concept of partial derivatives comes in. When you see \(\frac{\partial w}{\partial s}\) or \(\frac{\partial w}{\partial t}\), it means calculating how \(w\) changes with respect to \(s\) or \(t\), treating the other as a constant.

• It allows us to explore functions in multiple dimensions.
• Provides tools to examine real-world scenarios where many variables play a role.
• Use partial derivatives to pinpoint effects isolated to one variable at a time.

Practicing problems in this field helps develop a deeper understanding of how variables interrelate and influence outcomes.

Differentiation

Differentiation is the calculus technique used to compute derivatives. Derivatives represent rates of change; they are like how fast something is happening. For single-variable functions, differentiation finds how the function changes as its input changes. But with multivariable functions, it gets a bit more involved.

When dealing with functions that depend on more than one variable, we use partial derivatives. Each variable's partial derivative shows how that variable specifically influences the function. To differentiate with respect to a particular variable, you treat all other variables as constants. This was demonstrated in our problem by finding \( \frac{\partial w}{\partial s}\) and \(\frac{\partial w}{\partial t}\).

• Differentiation allows for precise examination of changes in function values.
• It is crucial for understanding dynamic processes across various scientific fields.
• Enables optimization and finding local extrema.

Mastering differentiation and partial derivatives empowers you to tackle complex problems involving change across multiple dimensions.