Problem 1
Question
Let \(f(x, y)=x^{2}+y^{2}\) with \(x(t)=3 t\) and \(y(t)=e^{t} .\) Find the derivative of \(w=f(x, y)\) with respect to \(t\) when \(t=\ln 2\).
Step-by-Step Solution
Verified Answer
The derivative is \(18(\ln 2) + 8\) when evaluated at \(t = \ln 2\).
1Step 1: Define the Composite Function
The function \(w\) is defined as a composite function: \(w(t) = f(x(t), y(t))\), where \(f(x, y)=x^2 + y^2\), \(x(t)=3t\), and \(y(t)=e^t\). Substitute the expressions for \(x(t)\) and \(y(t)\) into \(f(x, y)\) to get \(w(t) = (3t)^2 + (e^t)^2\). Simplify to find \(w(t) = 9t^2 + e^{2t}\).
2Step 2: Differentiate with Respect to t
Take the derivative of \(w(t) = 9t^2 + e^{2t}\) with respect to \(t\). The derivative \(\frac{dw}{dt} = \frac{d}{dt}(9t^2) + \frac{d}{dt}(e^{2t})\). Use the power rule for \(9t^2\) to get \(18t\). For \(e^{2t}\), use the chain rule: the derivative is \(2e^{2t}\). Combining these, \(\frac{dw}{dt} = 18t + 2e^{2t}\).
3Step 3: Evaluate the Derivative at t=ln(2)
Substitute \(t = \ln 2\) into the derivative equation: \(\frac{dw}{dt} = 18(\ln 2) + 2e^{2(\ln 2)}\). Simplify \(2e^{2(\ln 2)}\) using the property \(e^{2(\ln 2)} = (e^{\ln 2})^2 = 2^2 = 4\). Therefore, \(\frac{dw}{dt} = 18(\ln 2) + 8\).
4Step 4: Simplify the Expression
The expression to evaluate is \(18(\ln 2) + 8\). Since \(\ln 2\) is a constant, this is the final form of the derivative. Calculate numerically if needed, recognizing that \(\ln 2\) is approximately 0.693.
Key Concepts
Composite FunctionChain RuleDifferentiation
Composite Function
In the realm of multivariable calculus, a composite function is essentially a function within a function. Imagine you're baking a complex cake. Each layer is made from specific ingredients, but they all come together to form the final delicious dessert. Similarly, here is the concept of a composite function.
- A composite function involves an outer function and an inner function, such that the output of the inner function becomes the input to the outer function.
- In the problem given, the composite function is defined as \( w(t) = f(x(t), y(t)) \).
- This means that the variables \( x(t) \) and \( y(t) \) are plugged into another function \( f(x, y) \) to generate a new function depending only on \( t \).
Chain Rule
The chain rule is a pivotal technique in calculus used when differentiating composite functions. It is like a magician's secret trick for unveiling the derivative of nestled functions.
- The chain rule helps to compute the derivative of a composition by multiplying the derivative of the outer function by the derivative of the inner function.
- In our exercise, when dealing with \( e^{2t} \), the chain rule comes into play because the exponent itself is a function of \( t \).
- This means the derivative is calculated by taking the derivative of \( e^{u} \) with respect to \( u \) (which is \( e^{u} \)), and then multiplying by the derivative of \( u \) with respect to \( t \), where \( u = 2t \).
- As a result, the derivative becomes \( 2e^{2t} \).
Differentiation
Differentiation is a fundamental concept in calculus that is used to compute the derivative of a function. This is akin to measuring how a shadow lengthens and shortens as the sun moves across the sky.
- The derivative represents the rate at which a variable quantity changes with respect to another variable.
- In our example, the function \( w(t) = 9t^2 + e^{2t} \) is differentiated with respect to \( t \). The goal is to find how \( w \) changes as \( t \) changes.
- Different rules apply during differentiation, such as the power rule for \( 9t^2 \), where the exponent is brought in front and the power itself is reduced by one, resulting in \( 18t \).
- Similarly, exponential functions like \( e^{2t} \) require careful application of the chain rule as previously described.
Other exercises in this chapter
Problem 1
The tangent plane at the indicated point \(\left(x_{0}, y_{0}, z_{0}\right)\) exists. Find its equation. $$ f(x, y)=2 x^{3}+y^{2} ;(1,2,6) $$
View solution Problem 1
Cardiac output (CO) is a physiological quantity that is calculated as the product of heart rate (HR) and stroke volume (SV). Write cardiac output as a function
View solution Problem 1
In Problems 1-16, find \(\partial f / \partial x\) and \(\partial f / \partial y\) for the given functions. $$ f(x, y)=x^{2} y+x y^{2} $$
View solution