Question

Tangent lines Find an equation of the line tangent to the curve at the point corresponding to the given value of \(t\). $$x=e^{t}, y=\ln (t+1) ; t=0$$

Short answer

The equation of the tangent line at the given point is \(y = x - 1\).

Step-by-step solution

Find \(\frac{dy}{dx}\) in terms of \(t\) using Chain Rule

To find the derivative, \(\frac{dy}{dx}\), we'll first need to compute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). Our equations are: $$x = e^t$$ $$y = \ln(t+1)$$ Differentiating both equations with respect to \(t\), we get: $$\frac{dx}{dt} = e^t$$ $$\frac{dy}{dt} = \frac{1}{t+1}$$ Now, using the Chain Rule, we can find \(\frac{dy}{dx}\): $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{1}{(t+1)e^t}$$

Determine the point of tangency corresponding to \(t=0\)

To find the point of tangency, we need to plug the given value of \(t=0\) into the x and y parametric equations: $$x = e^0 = 1$$ $$y = \ln(0+1) = 0$$ So the point of tangency is \((1,0)\).

Calculate the slope of the tangent line at \(t=0\)

To find the slope of the tangent line at the point \((1,0)\), we need to evaluate the derivative \(\frac{dy}{dx}\) at \(t=0\). Plugging \(t=0\) in the expression we have found in Step 1, we get: $$\frac{dy}{dx}(0) = \frac{1}{(0+1)e^0} = \frac{1}{1} = 1$$ So the slope of the tangent line at the point \((1,0)\) is \(1\).

Write the equation of the tangent line using the point-slope form

Now that we have the point \((1,0)\) and the slope \(1\), we can use the point-slope form to write the equation of the tangent line. The point-slope form is given by: $$y - y_1 = m(x - x_1)$$ Here, \((x_1, y_1) = (1,0)\) is the point and \(m = 1\) is the slope. Plugging these values in, we get: $$y - 0 = 1(x - 1)$$ Simplifying the equation, we have: $$y = x - 1$$ So, the equation of the line tangent to the curve at the point corresponding to \(t=0\) is $$y = x - 1$$.

Key concepts

Parametric Equations

In mathematics, parametric equations provide a way to express the coordinates of a curve using a parameter, usually denoted as \(t\). This is particularly useful in scenarios where a function cannot be easily expressed in the form \(y=f(x)\). Instead, you define both \(x\) and \(y\) in terms of \(t\). This method offers flexibility and is often used to describe the positions of a point in motion.
For instance, consider the equations \(x = e^t\) and \(y = \ln(t+1)\). Here, \(t\) serves as the parameter that drives the change in both \(x\) and \(y\), enabling you to trace the path of the curve by varying \(t\).

The primary advantage of parametric equations is how they decouple the \(x\) and \(y\) coordinates, allowing for more complex shapes and motions to be articulated in a concise manner.
Whenever you're tackling problems involving parametric equations, always focus on:

• Identifying the parameter and how it affects the equations.
• Understanding the range of \(t\) to determine the curve's extent.
• Computing derivatives using \(t\) to find slopes and other motion characteristics.

Chain Rule

The chain rule is a pivotal calculus tool used for differentiating composite functions. It becomes crucial when you're dealing with parametric equations or any situation where one variable depends on another. In simple terms, the chain rule allows you to find the derivative of a function by breaking it down into its components.
When dealing with parametric curves where \(x\) and \(y\) are defined by \(t\), you can express the slope \(\frac{dy}{dx}\) using the chain rule. This is done by calculating \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\), which are the derivatives of \(y\) and \(x\) with respect to \(t\). Then, applying the chain rule, you find:
\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\]
This formula essentially measures how \(y\) changes with respect to \(x\) by considering the rates of change with respect to \(t\).

Mastering the chain rule involves:

• Identifying composite functions and their derivatives.
• Practicing the application of \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\) in parametric cases.
• Understanding how changes in \(t\) impact both \(x\) and \(y\).

Understanding the chain rule equips you with the capability to solve a broader range of mathematical and real-world problems involving rates of change and motion.

Point-Slope Form

The point-slope form is a straightforward way to describe the equation of a line when you know a point on the line and its slope. This form is particularly useful when defining tangent lines, which are lines that just touch a curve at a single point without crossing it.
The formula for the point-slope form is:\[y - y_1 = m(x - x_1)\]
Here, \((x_1, y_1)\) is the known point on the line, and \(m\) is the slope. Whenever you find the slope of a tangent line and a point on the curve, you can immediately use this form to construct the equation of the tangent line.
In practice, use the point-slope form when you:

• Have calculated the slope of a line, particularly a tangent line, as the derivative at a point.
• Know the coordinates of at least one point on the line.
• Need to quickly write down the line's equation to further analyze the curve or solution.

This form’s practicality lies in its simplicity, allowing for an efficient method to derive a line equation that reveals insights about the behavior of functions at specific points.