/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 Carry out the following steps. ... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 30

Carry out the following steps. a. Verify that the given point lies on the curve. b. Determine an equation of the line tangent to the curve at the given point. $$\begin{array}{l} \left(x^{2}+y^{2}\right)^{2}=\frac{25}{4} x y^{2} \\ (1,2) \end{array}$$

Short Answer

Expert verified
a. After plugging in the given point (1,2) into the equation and showing that both sides are equal, we verified that the point lies on the curve. b. Using implicit differentiation and solving for the slope at the given point, we found that the slope of the tangent line at (1,2) is -2. Therefore, the equation of the tangent line at this point is \(y = -2x + 4\).

Step by step solution

01

Verify the point lies on the curve

To verify if the point (1,2) lies on the curve, we will plug the x and y values into the equation: \((x^2+y^2)^2=\frac{25}{4}xy^2\). Substitute the values: \((1^2 + 2^2)^2 = \frac{25}{4}(1)(2^2)\) Simplify the equation: \((1 + 4)^2 = \frac{25}{4}(4)\) After solving, we get: \(5^2 = 25\) Since the left side equals the right side, the point (1,2) lies on the curve.
02

Find the derivative using implicit differentiation

Now we need to find the derivative of the equation \((x^2+y^2)^2=\frac{25}{4}xy^2\) with respect to x using implicit differentiation. Differentiating both sides of the equation with respect to x gives: \(2(x^2+y^2)(2x+2y\frac{dy}{dx})=\frac{25}{4}(y^2+2xy\frac{dy}{dx})\) Now we will solve the above equation for \(\frac{dy}{dx}\).
03

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

Solve for \(\frac{dy}{dx}\) using the implicit differentiation result: \(4(x^2+y^2)(x+y\frac{dy}{dx})=\frac{25}{4}(4y^2+8xy\frac{dy}{dx})\) Rearrange the terms to collect \(\frac{dy}{dx}\) on one side: \(4(x^2+y^2)(y\frac{dy}{dx}-2(x+y\frac{dy}{dx}))=25(2y^2-4xy\frac{dy}{dx})\) Factor out \(\frac{dy}{dx}\): \(\frac{dy}{dx}(4(x^2+y^2)(y-2)-25(8x))=25(4y^2)\) Now divide both sides by the terms in the parenthesis: \(\frac{dy}{dx}=\frac{25(4y^2)}{4(x^2+y^2)(y-2)-25(8x)}\) Now we will substitute the coordinates (1,2) into the above equation to find the slope of the tangent at that point.
04

Find the slope of the tangent at (1,2)

Substitute the coordinates (1,2) into the equation: \(\frac{dy}{dx}=\frac{25(4(2)^2)}{4((1)^2+(2)^2)(2-2)-25(8(1))}\) Simplify the equation: \(\frac{dy}{dx}=\frac{25(16)}{-25(8)}\) After solving, we get: \(\frac{dy}{dx} = -2\)
05

Find the equation of the tangent line

We found that the slope of the tangent at the point (1,2) is -2. Now we will use the slope-point form of the line \((y-y_1) = m(x-x_1)\) to find the equation of the tangent line. Substitute the coordinates (1,2) and the slope -2 into the equation: \((y-2)=-2(x-1)\) Rearrange and simplify the equation to get the tangent line in the form \(y=mx+b\): \(y=-2x+2+2\) The equation of the tangent line is: \(y=-2x+4\).

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Key Concepts

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

Tangent Line Equation
When studying curves, whether they are simple lines or complex equations, finding the tangent line at a specific point is a fundamental concept in calculus. The tangent line equation is the linear approximation of a curve at a particular point. It essentially represents the slope of the curve (the derivative) at that exact location. The equation of a tangent line can be found using a standard formula, known as the point-slope form of a linear equation:
\[y - y_1 = m(x - x_1)\]
where \(m\) denotes the slope of the tangent line and \( (x_1, y_1) \) are the coordinates of the point of tangency. This equation is particularly valuable because it not only allows us to visualize the slope at a certain point but also serves many practical applications in physics, engineering, and other sciences where the rate of change is critical.
Derivative Calculation
The primary step in working out the equation of a tangent line is to calculate the derivative of the function, which represents the slope. Derivatives play a crucial role in describing the rate at which one quantity changes with respect to another. In the explicit functions, derivatives can be calculated directly. However, for implicit functions 鈥� those involving both \(x\) and \(y\) intermixed 鈥� finding the derivative requires implicit differentiation, a process where both sides of the equation are differentiated with respect to \(x\), treating \(y\) as an implicit function of \(x\). This can lead to having the term \(dy/dx\) in the resulting equation, and through algebraic manipulation and solving, we can isolate \(dy/dx\) to find the slope at any given point on the curve. The calculation of \(dy/dx\) is precise and essential because it gives us the exact rate of change of \(y\) with respect to \(x\), which is synonymous with the slope of the tangent line on a curve.
Implicit Differentiation Application
Applying implicit differentiation can seem complex at first, but with practice, it becomes a powerful tool to handle equations where \(y\) is not isolated. This technique enables us to differentiate equations that describe curves not easily represented by a function in the format \(y=f(x)\). When you are given a point that lies on the curve, implicit differentiation allows you to calculate the derivative at that point even without solving for \(y\) explicitly 鈥� a common scenario in problems involving geometric shapes, economics, and other disciplines.

In the example of our exercise, we used implicit differentiation to find the slope of the tangent line at a given point on an implicitly defined curve. After differentiation, we substituted the coordinates of the given point to solve for the slope \(m\), allowing us to write the tangent line equation. This application highlights the practical benefits of implicit differentiation in calculus, especially when dealing with more complex equations where determining the slope directly would be difficult or impossible otherwise.

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Most popular questions from this chapter

A lighthouse stands 500 m off a straight shore and the focused beam of its light revolves four times each minute. As shown in the figure, \(P\) is the point on shore closest to the lighthouse and \(Q\) is a point on the shore 200 m from \(P\). What is the speed of the beam along the shore when it strikes the point \(Q ?\) Describe how the speed of the beam along the shore varies with the distance between \(P\) and \(Q\). Neglect the height of the lighthouse.

Find the following higher-order derivatives. $$\frac{d^{2}}{d x^{2}}\left(\log _{10} x\right)$$

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