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Chapter 3: Problem 17

Distance Let \(x\) and \(y\) be differentiable functions of \(t,\) and let \(s=\sqrt{x^{2}+y^{2}}\) be the distance between the points \((x, 0)\) and \((0, y)\) in the \(x y\) -plane. a. How is \(d s / d t\) related to \(d x / d t\) if \(y\) is constant? b. How is \(d s / d t\) related to \(d x / d t\) and \(d y / d t\) if neither \(x\) nor \(y\) is constant? c. How is \(d x / d t\) related to \(d y / d t\) if \(s\) is constant?

Short Answer

Expert verified
a) \(\frac{ds}{dt} = \frac{x}{\sqrt{x^2 + c^2}} \frac{dx}{dt}\) when \(y\) is constant. b) \(\frac{ds}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{\sqrt{x^2 + y^2}}\). c) \(\frac{dx}{dt} = -\frac{y}{x} \frac{dy}{dt}\) when \(s\) is constant.

Step by step solution

01

Given Information and Definitions

We have the formula for the distance: \( s = \sqrt{x^2 + y^2} \). Here, \(s\) is the distance between points \((x, 0)\) and \((0, y)\) in the xy-plane. The exercise involves finding the derivatives \(\frac{ds}{dt}, \frac{dx}{dt},\) and \(\frac{dy}{dt}\) under different conditions on \(x\) and \(y\).
02

Part A - Derivative Relationship with Constant y

When \(y\) is constant, \(y = c\), the derivative \(\frac{dy}{dt} = 0\). We substitute \(s = \sqrt{x^2 + c^2}\). Differentiate \(s\) with respect to \(t\) using the chain rule: \[ \frac{ds}{dt} = \frac{1}{2}(x^2 + c^2)^{-1/2} \cdot (2x \cdot \frac{dx}{dt}) = \frac{x}{\sqrt{x^2 + c^2}} \cdot \frac{dx}{dt} \].This shows that \(\frac{ds}{dt} = \frac{x}{\sqrt{x^2+c^2}}\frac{dx}{dt}\).
03

Part B - Derivative Relationships with Variable x and y

When neither \(x\) nor \(y\) is constant, we differentiate \(s = \sqrt{x^2 + y^2}\) using implicit differentiation:\[ \frac{ds}{dt} = \frac{1}{2}(x^2 + y^2)^{-1/2} \cdot (2x\frac{dx}{dt} + 2y\frac{dy}{dt}) = \frac{x\frac{dx}{dt} + y\frac{dy}{dt}}{\sqrt{x^2 + y^2}} \].Thus, \(\frac{ds}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{s}\).
04

Part C - Derivative Relationship with Constant s

If \(s\) is constant, \(\frac{ds}{dt} = 0\). Using the result from the previous step, we set it to zero:\[ \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{s} = 0 \rightarrow x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \].Solving for \(\frac{dx}{dt}\), we obtain: \[ x \frac{dx}{dt} = -y \frac{dy}{dt} \rightarrow \frac{dx}{dt} = -\frac{y}{x} \frac{dy}{dt} \].

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Key Concepts

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

Understanding Derivatives in Calculus
A derivative in calculus measures how a function changes as its input changes. It answers the question: "How does the output of a function change as we slightly tweak its input?" For differentiable functions of time, such as position over time, derivatives offer insights into velocity or rate of change.

In practical terms:
  • The derivative of a function \(x(t)\) with respect to time \(t\) is written as \(\frac{dx}{dt}\), which represents the rate of change of \(x\) at any point \(t\).
  • If we have \(y(t)\) as well, \(\frac{dy}{dt}\) expresses the same concept for \(y\).
Derivatives are powerful tools, especially when dealing with equations like distance where variables are dependent on time. They allow us to understand how one variable impacts another, offering a deeper insight into dynamic systems.
Exploring Implicit Differentiation
In many real-world situations, not all functions are presented in a straightforward "one-variable in terms of another" format. Often, we deal with relationships where variables are entwined - this is where implicit differentiation shines.

For instance, the equation \( s = \sqrt{x^2 + y^2} \) represents the distance between two points as a function of \(x\) and \(y\). In this context:
  • Implicit differentiation allows us to find how \(s\) changes with time \(t\), even when \(x\) and \(y\) both depend on \(t\).
  • This technique treats every variable as a function of time, applying the chain rule to differentiate.
A key benefit of implicit differentiation is its ability to handle scenarios where distribution of derivative impact is complex, capturing changes across multiple interdependent variables seamlessly.
The Distance Formula in the XY-plane
Understanding distance in the context of the xy-plane provides a foundational application of calculus in geometry. The distance formula \( s = \sqrt{x^2 + y^2} \) captures the straight-line distance between the origin \((0, 0)\) and a point \((x, y)\).

Why it matters:
  • For a constant \(y\) (part a), as \(y\) does not change, its derivative \(\frac{dy}{dt} = 0\), simplifying the relationship between \(\frac{ds}{dt}\) and \(\frac{dx}{dt}\).
  • When both \(x\) and \(y\) vary (part b), the formula captures the combined effect of each variable鈥檚 motion on the overall distance.
  • In scenarios where \(s\) stays constant (part c), understanding \(x\) and \(y\)'s relationship helps balance their rates of change to keep \(s\) unchanged.
This formula forms the backbone of many physics and engineering problems where motion and force are calculated across two-dimensional planes.

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Most popular questions from this chapter

You will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS: a. Plot the function \(y=f(x)\) together with its derivative over the given interval. Explain why you know that \(f\) is one-to-one over the interval. b. Solve the equation \(y=f(x)\) for \(x\) as a function of \(y,\) and name the resulting inverse function \(g\). c. Find an equation for the tangent line to \(f\) at the specified $$ \text { point }\left(x_{0}, f\left(x_{0}\right)\right) $$ d. Find an equation for the tangent line to \(g\) at the point \(\left(f\left(x_{0}\right), x_{0}\right)\) located symmetrically across the \(45^{\circ}\) line \(y=x\) (which is the graph of the identity function). Use Theorem 3 to find the slope of this tangent line. e. Plot the functions \(f\) and \(g\), the identity, the two tangent lines, and the line segment joining the points \(\left(x_{0}, f\left(x_{0}\right)\right)\) and \(\left(f\left(x_{0}\right), x_{0}\right) .\) Discuss the symmetries you see across the main diagonal (the line \(y=x\) ). $$y=e^{x}, \quad-3 \leq x \leq 5, \quad x_{0}=1$$

Find the derivative of \(y\) with respect to the given independent variable. $$y=t^{1-e}$$

In Exercises \(69-74,\) use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval I. Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\). c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can that satisfies \(|x-a|<\delta \Rightarrow|f(x)-L(x)|<\varepsilon\) for \(\varepsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see whether your \(\delta\) -estimate holds true. $$f(x)=x 2^{x}, \quad[0,2], \quad a=1$$

What is special about the functions $$f(x)=\sin ^{-1} \frac{1}{\sqrt{x^{2}+1}} \text { and } g(x)=\tan ^{-1} \frac{1}{x} ?$$ Explain.

Find \(d y / d x\) $$x^{y}=y^{x}$$
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