/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Evaluate \(d z / d t,\) where \(... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate \(d z / d t,\) where \(z=x y^{2}, x=t^{2},\) and \(y=t,\) using Theorem \(15.7 .\) Check your work by writing \(z\) as a function of \(t\) and evaluating \(d z / d t\).

Short Answer

Expert verified
Question: Using the chain rule, find the derivative of z with respect to t for the given functions z = xy^2, x = t^2, and y = t. Answer: The derivative of z with respect to t is \(\frac{dz}{dt} = 4t^3\).

Step by step solution

01

Find the partial derivatives of z with respect to x and y

We are given z = xy^2. First, we need to find the partial derivatives of z with respect to x and y: \(\frac{\partial z}{\partial x} = y^2\) and \(\frac{\partial z}{\partial y} = 2xy\)
02

Express x and y as functions of t

We are given x = t^2 and y = t. So, x'(t) = 2t and y'(t) = 1.
03

Apply the chain rule

According to Theorem 15.7, the chain rule for partial derivatives states that: \(\frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt}\) Substitute the values from steps 1 and 2: \(\frac{dz}{dt} = (y^2) \cdot (2t) + (2xy) \cdot (1)\) As y = t and x = t^2, we can replace them in the equation: \(\frac{dz}{dt} = (t^2) \cdot (2t) + (2t^3) \cdot (1)\) Now simplify: \(\frac{dz}{dt} = 2t^3 + 2t^3\) \(\frac{dz}{dt} = 4t^3\)
04

Check the result

To confirm our result, we will write z as a function of t and evaluate its derivative with respect to t. z = x(t)y(t)^2 = t^2(t)^2 = t^4. Now we will find the derivative of z with respect to t: \(\frac{dz}{dt} = \frac{d(t^4)}{dt} = 4t^3\) As the result from Step 3 is the same as the result from this direct evaluation, our answer is correct. The derivative of z with respect to t is \(\frac{dz}{dt} = 4t^3\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
In multivariable calculus, partial derivatives represent the rate of change of a function with respect to one of its variables, while keeping the others constant. This is especially important in functions involving more than one variable. For instance, if you have a function \( z = f(x, y) \), the partial derivative with respect to \( x \), noted as \( \frac{\partial z}{\partial x} \), measures how \( z \) changes as \( x \) changes, with \( y \) held constant.
  • The partial derivative \( \frac{\partial z}{\partial x} \) in our exercise means deriving \( z = xy^2 \) while treating \( y \) as constant, resulting in \( y^2 \).
  • Similarly, \( \frac{\partial z}{\partial y} \) involves deriving with respect to \( y \), holding \( x \) constant, giving us \( 2xy \).
Understanding partial derivatives is crucial. They form the building blocks for applying the chain rule in multivariable calculus.
Multivariable Calculus
Multivariable calculus extends the principles of calculus to functions of several variables. Unlike single-variable calculus, where changes are often linear, in multivariable calculus changes can occur in multiple directions simultaneously. This allows us to explore more complex scenarios and applications.
  • In multivariable calculus, you interpret functions like \( z = f(x, y) \) as surfaces in three-dimensional space, which is excellent for modeling physical phenomena.
  • The calculation done in the exercise requires managing more than one variable since both \( x \) and \( y \) are expressed as functions of \( t \). This helps illustrate how input from two variables can affect a single output \( z \), demonstrating the interconnectedness in multivariable calculus.
  • Understanding this interconnectedness is vital for fields like physics, engineering, and economics, where systems with multiple influencing factors are common.
Calculus Theorems
Calculus theorems help simplify complex problems and facilitate finding solutions in calculus. In the context of our problem, we use Theorem 15.7, which is a specific instance of the chain rule. The chain rule is an essential tool that allows us to differentiate compositions of functions. By applying the chain rule for partial derivatives, we transfer changes from the inner functions \( x(t) \) and \( y(t) \) to the outer function \( z(x, y) \). This is executed by:
  • Multiplying the partial derivative \( \frac{\partial z}{\partial x} \) by the derivative \( \frac{dx}{dt} \).
  • Adding it to the product of \( \frac{\partial z}{\partial y} \) and \( \frac{dy}{dt} \). This process is summed up as \( \frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt} \)
This enables us to derive \( z \) in relation to \( t \), transforming our understanding from a complex multi-variable problem into a manageable single-variable form. Mastery of these theorems builds the foundational skills necessary for advanced calculus topics.

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Most popular questions from this chapter

Looking ahead- tangent planes Consider the following surfaces \(f(x, y, z)=0,\) which may be regarded as a level surface of the function \(w=f(x, y, z) .\) A point \(P(a, b, c)\) on the surface is also given. a. Find the (three-dimensional) gradient of \(f\) and evaluate it at \(P\). b. The set of all vectors orthogonal to the gradient with their tails at \(P\) form a plane. Find an equation of that plane (soon to be called the tangent plane). $$f(x, y, z)=e^{x+y-z}-1=0 ; P(1,1,2)$$

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