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If \(y=x^{2}\) and \(d x / d t=3,\) then what is \(d y / d t\) when \(x=-1 ?\)

Short Answer

Expert verified
\(\frac{dy}{dt} = -6\) when \(x = -1\).

Step by step solution

01

Understand the Given Information

We are given the function \(y=x^2\) and the rate \(\frac{dx}{dt}=3\). Our task is to find \(\frac{dy}{dt}\) when \(x = -1\).
02

Apply Chain Rule for Differentiation

To find \(\frac{dy}{dt}\), we need to use the chain rule. The chain rule states that \(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}\).
03

Differentiate y with respect to x

Find the derivative of \(y = x^2\) with respect to \(x\) to get \(\frac{dy}{dx} = 2x\).
04

Substitute Derivatives and Given Values

Now, substitute \(x = -1\), \(\frac{dy}{dx} = 2x\), and \(\frac{dx}{dt} = 3\) into the chain rule formula: \[\frac{dy}{dt} = 2(-1) \times 3\].
05

Solve for dy/dt

Calculate the expression: \(\frac{dy}{dt} = 2(-1) \times 3 = -6\).

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Key Concepts

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

Chain Rule Simplified
The Chain Rule is a fundamental concept in calculus that helps us differentiate composite functions. More simply, it's about finding the rate of change of one variable concerning another.
When dealing with related rates, like in our problem, we can use the chain rule to connect the rates of change between different variables.
Here's how it works:
  • Identify the original function, such as the given function \( y = x^2 \).
  • Apply the chain rule formula: \( \frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} \).
This formula helps us understand the relationship between the rate of change of \( y \) with respect to time \( t \) (\( \frac{dy}{dt} \)), and the rate of change of \( x \) with respect to time \( t \) (\( \frac{dx}{dt} \)), leveraging the immediate transformation between \( y \) and \( x \) (\( \frac{dy}{dx} \)).
By using the chain rule, we effectively break down a complicated process into easier steps where each piece logically follows the next.
Understanding the Derivative
A derivative represents the rate of change of a variable. It is the foundation of differential calculus.
In our example, we need to find the derivative of \( y \) with respect to \( x \), which is the instantaneous rate of change of \( y \) for a small increment in \( x \).
  • The given function is \( y = x^2 \).
  • The derivative of this function, \( \frac{dy}{dx} \), measures how \( y \) changes as \( x \) changes, calculated by taking \( 2x \).
Derivatives help us understand how one quantity affects another. They are vital when calculating speed, growth rates, and other dynamic processes.
In our case, knowing the derivative allows us to find out how fast \( y \) is changing in relation to time \( t \). This is crucial for solving problems involving dynamic systems or moving objects.
Differentiation Steps Explained
Differentiating a function involves a series of steps that transform a given function into its derivative, allowing us to study how it changes over time.
Let's break down the differentiation steps for this problem:
  • **Given Function**: Start with the function \( y = x^2 \).
  • **Compute \( \frac{dy}{dx} \)**: Differentiate \( y \) with respect to \( x \). Here, the derivative of \( x^2 \) is \( 2x \).
  • **Apply Chain Rule**: Use the formula \( \frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} \). This allows us to link \( \frac{dy}{dt} \) with \( \frac{dx}{dt} \), given that \( \frac{dx}{dt} = 3 \).
  • **Substitute and Solve**: Plug in \( x = -1 \), yielding \( \frac{dy}{dt} = 2(-1) \times 3 \).
  • **Result**: Solve to find \( \frac{dy}{dt} = -6 \).
The process is structured to simplify complex functions, using both derivative calculations and substitutions to reach useful conclusions.
Following these steps ensures that you methodically address all aspects of a differentiation problem, giving you accurate and reliable results.

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Most popular questions from this chapter

Find \(d y / d t\) when \(x=1\) if \(y=x^{2}+7 x-5\) and \(d x / d t=1 / 3\)

Find the value of \((f \circ g)^{\prime}\) at the given value of \(x\). $$f(u)=\left(\frac{u-1}{u+1}\right)^{2}, \quad u=g(x)=\frac{1}{x^{2}}-1, \quad x=-1$$

(Continuation of Exercise \(65 .)\) The quotient $$ \frac{f(x+h)-f(x-h)}{2 h} $$ may have a limit as \(h \rightarrow 0\) when \(f\) has no derivative at \(x .\) As a case in point, take \(f(x)=|x|\) and calculate $$ \lim _{h \rightarrow 0} \frac{|0+h|-|0-h|}{2 h} $$ As you will see, the limit exists even though \(f(x)=|x|\) has no derivative at \(x=0 .\) Moral: Before using a centered difference quotient, be sure the derivative exists.

Give the position function \(s=f(t)\) of an object moving along the \(s\) -axis as a function of time \(t\). Graph \(f\) together with the velocity function \(v(t)=d s / d t=f^{\prime}(t)\) and the acceleration function \(a(t)=d^{2} s / d t^{2}=f^{\prime \prime}(t) .\) Comment on the object's behavior in relation to the signs and values of \(v\) and \(a .\) Include in your commentary such topics as the following: a. When is the object momentarily at rest? b. When does it move to the left (down) or to the right (up)? c. When does it change direction? d. When does it speed up and slow down? e. When is it moving fastest (highest speed)? Slowest? f. When is it farthest from the axis origin? \(s=60 t-4.9 t^{2}, \quad 0 \leq t \leq 12.5\) (a heavy object fired straight up from Earth's surface at \(60 \mathrm{m} / \mathrm{s}\) )

Use a CAS to perform the following steps a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point \(P\) satisfies the equation. b. Using implicit differentiation, find a formula for the derivative \(d y / d x\) and evaluate it at the given point \(P\). c. Use the slope found in part (b) to find an equation for the tangent line to the curve at \(P .\) Then plot the implicit curve and tangent line together on a single graph. $$x^{3}-x y+y^{3}=7, \quad P(2,1)$$

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