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Chapter 4: Problem 228

Let \(z=x^{2} y,\) where \(x=t^{2}\) and \(y=t^{3} .\) Find \(\frac{d z}{d t}\).

Short Answer

Expert verified
The derivative \( \frac{dz}{dt} = 7t^6 \).

Step by step solution

01

Substitute Expressions for x and y

Given that \(x = t^2\) and \(y = t^3\), we can substitute these into the expression for \(z\), which is \(z = x^2 y\). This gives us \(z = (t^2)^2 (t^3)\).
02

Simplify the Expression for z

Simplify \(z = (t^2)^2 (t^3)\) to get \(z = t^4 \, t^3 = t^{7}\).
03

Differentiate z with Respect to t

Now we differentiate \(z = t^7\) with respect to \(t\). Use the power rule which states that the derivative of \(t^n\) is \(n t^{n-1}\). So, \(\frac{dz}{dt} = 7 t^6\).

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Key Concepts

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

Chain Rule
In multivariable calculus, the chain rule is a fundamental tool that allows us to differentiate composite functions. It's especially helpful when dealing with functions that are expressed in terms of different variables. The chain rule connects the rates of change of the variables.
  • If we have a function like the one in the original exercise, where a variable is defined in terms of two others (\[z = x^{2}y\] with \(x = t^{2}\) and \(y = t^{3}\)), the chain rule helps us find how \(z\) changes with respect to \(t\).
  • The essence of the chain rule is to find the derivative of each component and multiply them appropriately. This means we derive \(z\) with respect to \(x\) and \(y\), then \(x\) and \(y\) with respect to \(t\), and combine these results.
The concept might seem complex, but it's simply about linking together the derivatives of each dependent relationship in the composite function. Understanding this can greatly simplify differentiation in multivariable calculus.
Differentiation
Differentiation is the process of finding the derivative, or the rate of change, of a function. This is a cornerstone of calculus, and it helps us understand how one quantity changes in relation to another.
For the function \(z = t^7\) from the exercise, differentiating with respect to \(t\) involves applying the power rule:
  • The power rule states that if you have \(t^n\), the derivative with respect to \(t\) is \(n t^{n-1}\).
  • In this example, since \(z = t^7\), differentiating gives \(\frac{dz}{dt} = 7 t^6\).
Differentiation tells us how small changes in \(t\) will affect \(z\), which is crucial for understanding behavior of functions in all areas of science and engineering.
Substitution Method
The substitution method is a strategy for simplifying complex expressions or equations by substituting variables with their known relationships. It's particularly useful in differentiation when a function is given in terms of several variables. For instance, in the given exercise:
  • Original expressions are \(x = t^2\) and \(y = t^3\).
  • Substitute these into \(z = x^2 y\): replacing \(x\) and \(y\) gives us \(z = (t^2)^2 (t^3)\).
  • Simplifying the expression yields \(z = t^7\).
This substitution significantly simplifies the process of differentiation. Often, by condensing expressions into powers of a single variable like \(t\), complex problems become straightforward to solve. Using substitution to address multivariable problems is a powerful method for tackling integrals and derivatives efficiently.

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Most popular questions from this chapter

Find \(\frac{d y}{d x}\) using partial derivatives. \(x^{2} y^{3}+\cos y=0\)

For the following exercises, determine the extreme values and the saddle points. Use a CAS to graph the function. $$[T]f(x, y)=y e^{x}-e^{y}$$

Maximize \(f(x, y)=\sqrt{6-x^{2}-y^{2}}, x+y-2=0\).

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Find \(\frac{d z}{d t}\) by the chain rule where \(z=\cosh ^{2}(x y), x=\frac{1}{2} t,\) and \(y=e^{t}\)
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