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Chapter 3: Problem 3

Why are both the \(x\) -coordinate and the \(y\) -coordinate generally needed to find the slope of the tangent line at a point for an implicitly defined function?

Short Answer

Expert verified
Both the \(x\)-coordinate and the \(y\)-coordinate are needed because the derivative (slope) of an implicitly defined function depends on both variables. Since we cannot express \(y\) explicitly as a function of \(x\), we must use both coordinates to find the slope of the tangent line at a specific point, ensuring accurate calculation and representation of the function's behavior.

Step by step solution

01

Understanding Implicitly Defined Functions

An implicitly defined function is a relation between variables where the dependent variable, \(y\), cannot be explicitly expressed as a function of the independent variable, \(x\). In other words, we can't isolate \(y\) on one side of the equation. As a result, it's challenging to use traditional methods such as finding the derivative, which is needed to calculate the slope of the tangent line.
02

Differentiating Implicit Functions

In order to find the slope of the tangent line for an implicitly defined function, we need to differentiate the function with respect to \(x\). However, since we can't express \(y\) explicitly as a function of \(x\), we need to use the chain rule. To differentiate the implicit function, we differentiate both sides of the equation with respect to \(x\) and use the chain rule \((\frac{dy}{dx})\) for terms containing \(y\).
03

Finding the Slope of the Tangent Line

After differentiating the implicit function, we'll have an expression that includes both \(x\) and \(y\) terms. To find the slope of the tangent line at a specific point \((x_0, y_0)\), we need to substitute both the \(x\)-coordinate and the \(y\)-coordinate into the expression, since the derivative (slope) depends on both variables.
04

Why Both Coordinates are Needed

Both the \(x\)-coordinate and the \(y\)-coordinate are needed to find the slope of the tangent line at a point for an implicitly defined function because the derivative (slope) depends on both variables. Since we can't express \(y\) explicitly as a function of \(x\), we must use both coordinates to find the slope of the tangent line at a specific point. Otherwise, we won't be able to fully describe the function's behavior and calculate the slope accurately.

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Most popular questions from this chapter

Earth's atmospheric pressure decreases with altitude from a sea level pressure of 1000 millibars (a unit of pressure used by meteorologists). Letting \(z\) be the height above Earth's surface (sea level) in \(\mathrm{km}\), the atmospheric pressure is modeled by \(p(z)=1000 e^{-z / 10}.\) a. Compute the pressure at the summit of Mt. Everest which has an elevation of roughly \(10 \mathrm{km}\). Compare the pressure on Mt. Everest to the pressure at sea level. b. Compute the average change in pressure in the first \(5 \mathrm{km}\) above Earth's surface. c. Compute the rate of change of the pressure at an elevation of \(5 \mathrm{km}\). d. Does \(p^{\prime}(z)\) increase or decrease with \(z\) ? Explain. e. What is the meaning of \(\lim _{z \rightarrow \infty} p(z)=0 ?\)

The flow of a small stream is monitored for 90 days between May 1 and August 1. The total water that flows past a gauging station is given by $$V(t)=\left\\{\begin{array}{ll}\frac{4}{5} t^{2} & \text { if } 0 \leq t<45 \\\\-\frac{4}{5}\left(t^{2}-180 t+4050\right) & \text { if } 45 \leq t<90, \end{array}\right.$$ where \(V\) is measured in cubic feet and \(t\) is measured in days, with \(t=0\) corresponding to May 1. a. Graph the volume function. b. Find the flow rate function \(V^{\prime}(t)\) and graph it. What are the units of the flow rate? c. Describe the flow of the stream over the 3 -month period. Specifically, when is the flow rate a maximum?

General logarithmic and exponential derivatives Compute the following derivatives. Use logarithmic differentiation where appropriate. $$\frac{d}{d x}(2 x)^{2 x}$$.

Work carefully Proceed with caution when using implicit differentiation to find points at which a curve has a specified slope. For the following curves, find the points on the curve (if they exist) at which the tangent line is horizontal or vertical. Once you have found possible points, make sure they actually lie on the curve. Confirm your results with a graph. $$x^{2}(y-2)-e^{y}=0$$

Work carefully Proceed with caution when using implicit differentiation to find points at which a curve has a specified slope. For the following curves, find the points on the curve (if they exist) at which the tangent line is horizontal or vertical. Once you have found possible points, make sure they actually lie on the curve. Confirm your results with a graph. $$x\left(1-y^{2}\right)+y^{3}=0$$
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