60–62. {Use of Tech} Multiple tangent lines Complete the following steps. a. Find equations of all lines tangent to the curve at the given value of x.x+y³−y=1; x=1

Ch. 3 - Derivatives

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 3 - Derivatives

문제 3.8.60a

3장, 문제 3.8.60a

60–62. {Use of Tech} Multiple tangent lines Complete the following steps. <IMAGE>
a. Find equations of all lines tangent to the curve at the given value of x.
x+y³−y=1; x=1

검증된 단계별 안내

First, we need to find the derivative of the given equation with respect to x. The equation is x + y³ - y = 1. To find the derivative, we will use implicit differentiation.

Differentiate both sides of the equation with respect to x. Remember that y is a function of x, so when differentiating terms involving y, apply the chain rule. The derivative of x is 1, the derivative of y³ is 3y²(dy/dx), and the derivative of y is (dy/dx).

After differentiating, we get: 1 + 3y²(dy/dx) - (dy/dx) = 0. Now, solve for dy/dx to find the slope of the tangent line. Combine like terms to get: dy/dx(3y² - 1) = -1.

Solve for dy/dx by dividing both sides by (3y² - 1): dy/dx = -1 / (3y² - 1). This expression gives us the slope of the tangent line at any point (x, y) on the curve.

Substitute x = 1 into the original equation to find the corresponding y-values. Solve 1 + y³ - y = 1 to find the y-values. Use these y-values in the derivative expression to find the slopes of the tangent lines. Finally, use the point-slope form of a line, y - y₁ = m(x - x₁), to write the equations of the tangent lines.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.

영상 길이:

10m

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Implicit Differentiation

Implicit differentiation is a technique used to differentiate equations where the dependent and independent variables are not explicitly separated. In this case, the equation x + y³ - y = 1 involves both x and y, requiring us to differentiate both sides with respect to x while treating y as a function of x. This method allows us to find dy/dx, which is essential for determining the slope of the tangent line.

Tangent Line Equation

The equation of a tangent line at a given point on a curve can be expressed using the point-slope form: y - y₀ = m(x - x₀), where (x₀, y₀) is the point of tangency and m is the slope at that point. Once we find the slope using implicit differentiation, we can substitute the coordinates of the point where x = 1 into this formula to derive the equation of the tangent line.

Finding Points on the Curve

To find the equations of the tangent lines, we first need to determine the corresponding y-values for the given x-value of 1. This involves substituting x = 1 into the original equation x + y³ - y = 1 and solving for y. The solutions will provide the points on the curve where the tangent lines will be calculated, which is crucial for completing the problem.

60–62. {Use of Tech} Multiple tangent lines Complete the following steps. <IMAGE>a. Find equations of all lines tangent to the curve at the given value of x.x+y³−y=1; x=1