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Find \(\frac{d y}{d x}\) for the curve \(x^{3}+3 y+y^{2}=6\). What is the slope at \((2,-1)\) ? At what points is the slope zero?

Short Answer

Expert verified
The slope at \((2, -1)\) is \(-2\). The slope is zero at the points \((0, 1)\) and \((0, -3)\).

Step by step solution

01

Differentiate Both Sides of the Equation

Differentiate both sides of the equation with respect to \(x\). Use the chain rule when differentiating \(y\) (and \(y^2\)) treated as a function of \(x\). You get: \(3x^2 + 3\frac{dy}{dx} + 2y\frac{dy}{dx} = 0\).
02

Solve for \(\frac{dy}{dx}\)

Rearrange the differentiated equation to isolate \(\frac{dy}{dx}\) on one side: \(\frac{dy}{dx} = \frac{-3x^2}{3 + 2y}\).
03

Find the Slope at \((2,-1)\) by Substituting

To find the slope at the point \((2, -1)\), substitute \(x = 2\) and \(y = -1\) into the derivative equation from step 2: \(\frac{dy}{dx} = \frac{-3 \cdot (2)^2}{3 + 2\cdot (-1)} = -2\). So the slope at \((2, -1)\) is \(-2\).
04

Find the Points Where the Slope is Zero

To find the points where the slope is zero, set the derivative (from step 2) equal to zero and solve for \(x\). You'll get the equation \(0 = \frac{-3x^2}{3 + 2y}\). This only happens when the numerator \( -3x^2 = 0\) and solving for \(x\) gives \(x = 0\). Substitute \(x = 0\) back into the original equation to get the \(y\)-coordinate, which is \(y = 1\) or \(y = -3\). So the points where the slope is zero are \((0, 1)\) and \((0, -3)\).

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Key Concepts

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

Chain Rule
When we differentiate equations that involve more than one variable, like in the equation \(x^3 + 3y + y^2 = 6\), we often use a method called implicit differentiation. Implicit differentiation is particularly useful when we can't easily solve for \(y\) in terms of \(x\). This is where the chain rule comes into play.

The chain rule is a fundamental rule in calculus that allows us to differentiate composite functions. In this problem, when you differentiate terms like \(3y\) and \(y^2\), you treat \(y\) as a function of \(x\). The chain rule tells us how to differentiate these terms correctly:
  • For \(3y\), the derivative with respect to \(x\) is \(3\frac{dy}{dx}\).
  • For \(y^2\), the derivative is \(2y\frac{dy}{dx}\).
The chain rule is crucial here because it gives us the correct differentiation of terms involving \(y\), which can't be ignored, as they contribute to the overall rate of change of the curve.
Slope of a Curve
The slope of a curve at any given point reflects how steep the curve is at that particular point. The slope is essentially the same as the derivative \(\frac{dy}{dx}\).

In our example, once we differentiated the original equation using the chain rule, we obtained \(3x^2 + 3\frac{dy}{dx} + 2y\frac{dy}{dx} = 0\). Solving for \(\frac{dy}{dx}\) gave us \(\frac{dy}{dx} = \frac{-3x^2}{3 + 2y}\).
  • At the point \((2, -1)\), by substituting \(x = 2\) and \(y = -1\) into this derivative, we discovered the slope to be \(-2\).
  • This means the curve is decreasing at this point with a slope of \(-2\), indicating a fairly steep decline.
The slope effectively tells us the direction and steepness of the curve at the point, which is why it's such an important concept in calculus.
Equation of a Curve
Understanding the equation of a curve allows us to sketch its graph and optimize functions. In the problem \(x^3 + 3y + y^2 = 6\), this equation describes a relationship between \(x\) and \(y\) that forms a particular shape in a graph, which can be complex due to the presence of both \(x\) and \(y\) on the same side.

Using implicit differentiation and finding the slope at certain points—like we did at \((2, -1)\) and when finding where the slope is zero—helps us understand the nature of the curve's graph, including:
  • Using \(x = 0\) to find points like \((0, 1)\) and \((0, -3)\) where the slope is zero helps identify where the curve has a horizontal tangent.
  • These horizontal tangents can indicate minima or maxima points or changes in the direction of the curve.
These calculations are crucial for analyzing and understanding the behavior and shape of the curve dictated by the equation.

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Most popular questions from this chapter

Find the equation of the line tangent to the curve \(x^{3}+y^{3}-3 x^{2} y^{2}+1=0\) at the point \((1,1) .\)

In Barcelona there is a beautiful Spanish castle set \(1 / 4\) of a mile back from a straight road. A bicyclist rides by the castle at a velocity of \(15 \mathrm{mph}\). Assuming that the biker maintains this speed, how fast is the distance between the biker and the castle increasing 20 minutes later?

A conical container is used to hold oil. It is positioned upright with the tip of the cone at the bottom. Oil comes out a faucet at the base of the container. We know that the volume of a cone is \(V=\frac{1}{3} \pi r^{2} h\). (a) As oil leaves the cone the height, radius, and volume of oil in the container change with time. Find \(\frac{d V}{d t}\) in terms of \(r, h, \frac{d r}{d t}\), and \(\frac{d h}{d t}\). (b) Suppose the container has a height of 24 inches and a radius of 12 inches. Express the volume of oil in the container as a function of the height of oil in the container. (Hint: Use similar triangles to express the radius of the oil in terms of the height of the oil.) (c) Suppose oil is leaking out of the container at a rate of 5 cubic inches per hour. How fast is the height of the oil in the container decreasing when the height is 10 inches? When the height is 4 inches? Do the relative sizes of your answers make sense to you intuitively?

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Suppose we toss a rock into a pond causing a circular ripple. If the radius is increasing at a rate of 3 feet per second when the diameter is 4 , how fast is the area increasing?

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