Problem 228
Question
Let \(z=x^{2} y,\) where \(x=t^{2}\) and \(y=t^{3} .\) Find \(\frac{d z}{d t}\).
Step-by-Step Solution
Verified Answer
The derivative \( \frac{dz}{dt} = 7t^6 \).
1Step 1: 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 \(z = (t^2)^2 (t^3)\).
2Step 2: Simplify the Expression for z
Simplify \(z = (t^2)^2 (t^3)\) to get \(z = t^4 \, t^3 = t^{7}\).
3Step 3: 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, \(\frac{dz}{dt} = 7 t^6\).
Key Concepts
Chain RuleDifferentiationSubstitution Method
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.
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:
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 \(\frac{dz}{dt} = 7 t^6\).
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 \(z = t^7\).
Other exercises in this chapter
Problem 226
Find \(\frac{d f}{d t}\) using the chain rule and direct substitution. \(\quad f(x, y)=x^{4}, \quad x=t, y=t\)
View solution Problem 227
Let \(w(x, y, z)=x^{2}+y^{2}+z^{2}\) \(x=\cos t, y=\sin t, \quad\) and \(z=e^{t} .\) Express \(w\) as a function of \(t\) and find \(\frac{d w}{d t}\) directly.
View solution Problem 229
Let \(u=e^{x} \sin y,\) where \(x=t^{2}\) and \(y=\pi t\). Find \(\frac{d u}{d t}\) when \(x=\ln 2\) and \(y=\frac{\pi}{4}\).
View solution Problem 230
Find \(\frac{d y}{d x}\) using partial derivatives. \(\sin (6 x)+\tan (8 y)+5=0\)
View solution