Problem 4 Find an equation of the tangent ... [FREE SOLUTION]

Chapter 10: Problem 4

Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter. $$x=t-t^{-1}, \quad y=1+t^{2} ; \quad t=1$$

Short Answer

Expert verified

The equation of the tangent is $y = x + 2$.

Step by step solution

Find dx/dt and dy/dt

Determine the derivatives of the given parametric equations. First, find the derivative of the x equation with respect to the parameter $t$: $ \frac{dx}{dt} = 1 + \frac{1}{t^2} $. Then, find the derivative of the y equation: $ \frac{dy}{dt} = 2t $.

Compute the Slope of the Tangent

The slope of the tangent line for parametric equations is given by $ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} $. Substitute $ \frac{dy}{dt} $ and $ \frac{dx}{dt} $ into the equation: $ \frac{dy}{dx} = \frac{2t}{1 + \frac{1}{t^2}} $. Simplify: $ \frac{dy}{dx} = \frac{2t^3}{t^2 + 1} $.

Evaluate the Slope at t=1

Substitute $t = 1$ into the simplified $ \frac{dy}{dx} $: $ \frac{dy}{dx} = \frac{2(1)^3}{(1)^2 + 1} = \frac{2}{2} = 1 $. The slope of the tangent line at $t=1$ is 1.

Find the Point on the Curve

Substitute $t = 1$ into the original parametric equations to find the corresponding point. For $x$: $x = 1 - \frac{1}{1} = 0$. For $y$: $y = 1 + (1)^2 = 2$. The point is $(0, 2)$.

Write the Equation of the Tangent Line

Use the point-slope form $y - y_1 = m(x - x_1)$ with slope $m=1$ and point $(x_1, y_1) = (0, 2)$. Substitute the values: $y - 2 = 1(x - 0)$. Simplify to get the equation of the tangent line: $y = x + 2$.

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Key Concepts

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

Derivative

The concept of a derivative is fundamental in calculus. A derivative represents the rate at which a function is changing at any given point. In the context of parametric equations, we deal with derivatives with respect to the parameter, often denoted as $ t $. This involves finding $ \frac{dx}{dt} $ and $ \frac{dy}{dt} $ from the parametric equations $ x = t - t^{-1} $ and $ y = 1 + t^2 $.
$ \frac{dx}{dt} $ indicates how $ x $ changes as $ t $ changes, and $ \frac{dy}{dt} $ shows how $ y $ changes with $ t $.

For $ x = t - t^{-1} $, the derivative is $ \frac{dx}{dt} = 1 + \frac{1}{t^2} $. This subtracts the negative derivative of $ -t^{-1} $ from 1.
For $ y = 1 + t^2 $, the derivative is straightforward: $ \frac{dy}{dt} = 2t $, as the derivative of $ t^2 $ is $ 2t $ and the constant 1 disappears.

Through these derivatives, we understand the behaviors and trends of the curve as defined by the parametric equations.

Slope of Tangent Line

The slope of the tangent line is crucial for understanding how a curve behaves at a specific point. For parametric equations, this slope is not directly from $ x $ and $ y $, but is calculated as the ratio $ \frac{dy}{dx} $.
To find this, we use the formula $ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} $. Substitution leads us to $ \frac{dy}{dx} = \frac{2t}{1 + \frac{1}{t^2}} $.
This formula might initially seem complex, but it simplifies by clearing the fraction: $ \frac{dy}{dx} = \frac{2t^3}{t^2 + 1} $.

At a specific parameter value, say $ t = 1 $, this expression becomes $ \frac{dy}{dx} = \frac{2(1)^3}{(1)^2 + 1} $, which simplifies to $ 1 $.
This result tells us that the slope of the tangent at $ t = 1 $ is a simple 1, indicating a 45-degree angle with the x-axis.

This straightforward slope allows us to define the tangent line effectively.

Tangent Line Equation

Having found the slope, the next step is to derive the tangent line's equation. The tangent line to a curve at a point provides the best linear approximation of the curve near that point.
We use the point-slope form of the equation, which is $ y - y_1 = m(x - x_1) $, where $ m $ is the slope, and $(x_1, y_1)$ is the point of tangency.

From the previous calculations, the point of tangency at $ t = 1 $ is $ (0, 2) $, and the slope $ m $ is 1.
Substituting into the point-slope equation, we get $ y - 2 = 1(x - 0) $.
Simplifying this equation yields $ y = x + 2 $.

This equation $ y = x + 2 $ describes the tangent line. It intersects the curve at $(0, 2)$ with a slope of 1, aligning perfectly with the calculated values.

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91影视

Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter. $$x=t-t^{-1}, \quad y=1+t^{2} ; \quad t=1$$

Short Answer

Step by step solution

Find dx/dt and dy/dt

Compute the Slope of the Tangent

Evaluate the Slope at t=1

Find the Point on the Curve

Write the Equation of the Tangent Line

Key Concepts

Derivative

Slope of Tangent Line

Tangent Line Equation

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