Problem 45 - Calculus Single Variable

Question

Find the tangent line to the parametric curve \(x=\varphi_{1}(t), y=\varphi_{2}(t)\) at the point corresponding to the given value \(t_{0}\) of the parameter. $$ \varphi_{1}(t)=2^{t}+3, \varphi_{2}(t)=3^{t}+2 \quad t_{0}=1 $$

Step-by-Step Solution

Answer

The tangent line is given by \(y - 5 = \frac{3\ln(3)}{2\ln(2)} (x - 5)\).

Step 1: Evaluate the point at t_0

First, we need to find the coordinates of the point on the curve corresponding to the parameter value \(t_0 = 1\).\Calculate \(x(1) = \varphi_{1}(1) = 2^{1} + 3 = 5\).Calculate \(y(1) = \varphi_{2}(1) = 3^{1} + 2 = 5\).Hence, the point on the curve is \((5, 5)\).

Step 2: Find the derivatives

To find the tangent line, we need the slope of the line which is given by the derivative \(\frac{dy}{dx}\). \First, find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\):\(\frac{dx}{dt} = \frac{d}{dt}(2^{t} + 3) = 2^{t} \ln(2)\)\(\frac{dy}{dt} = \frac{d}{dt}(3^{t} + 2) = 3^{t} \ln(3)\)

Step 3: Calculate dy/dx at t_0

The slope \(\frac{dy}{dx}\) is given by \(\frac{dy/dt}{dx/dt}\).Substitute \(t = t_0 = 1\):\(\frac{dy}{dx} \bigg|_{t=1} = \frac{3^{1}\ln(3)}{2^{1}\ln(2)} = \frac{3\ln(3)}{2\ln(2)}\)

Step 4: Form the equation of the tangent line

The equation of a line with slope \(m\) passing through a point \((x_1, y_1)\) is given by:\(y - y_1 = m(x - x_1)\).Using \(m = \frac{3\ln(3)}{2\ln(2)}\) and the point \((5, 5)\):\(y - 5 = \frac{3\ln(3)}{2\ln(2)} (x - 5)\)

Step 5: Simplify if necessary

While optional, simplifying or arranging the equation can make it easier to understand or apply. In this case, expansion isn't strictly needed unless specified, so leave the equation as it is for a direct expression of the tangent line.

Key Concepts

Tangent Line

A tangent line to a curve at a specific point is a straight line that just "touches" the curve at that point. This line represents the instantaneous direction of the curve at that point.
It is like a shadow of the curve at that spot. In this exercise, we are working with parametric curves, which means both the x and y coordinates of the point are expressed in terms of a parameter, usually denoted by t.
Finding this tangent line involves determining the slope at a specific point on the curve, which will guide the line's direction.

• The key aspect of a tangent line is its slope at the point of contact.
• This slope is directly related to the derivatives of the parametric equations with respect to the chosen parameter.

The equation of the tangent line is usually expressed in point-slope form, which helps us quickly establish the relationship it has with the point on the curve and its slope.

Derivatives

Derivatives are at the heart of finding the tangent line. They allow us to determine the rate at which one quantity changes concerning another. When dealing with parametric equations, we often need to calculate two specific derivatives:

• \( \frac{dx}{dt} \), the rate of change of x with respect to t.
• \( \frac{dy}{dt} \), the rate of change of y with respect to t.

Once these are known, the slope of the tangent line, \( \frac{dy}{dx} \), relies on how dy relates to dx, given by the formula:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
This relationship helps us find the exact slope of the tangent line at any point corresponding to a parameter value, \( t_0 \). In our specific example, for \( t_0 = 1 \), these derivatives yield the slope of the tangent line.

Parametric Equations

Parametric equations allow us to express the coordinates (x, y) of a curve using a third variable, typically t. This method of representation captures the evolution of the curve in a way that individual functions for x and y in terms of each other might not solve easily.

For example, consider the equations given in the exercise:

• \( x = 2^t + 3 \)
• \( y = 3^t + 2 \)

These equations tell us how x and y change as t varies, making it easier to compute specific points, like (5, 5) when \( t_0 = 1 \), and their derivatives.
Parametric curves are particularly useful when dealing with motion scenarios or when the relation between x and y isn't straightforward or is multi-valued.

Slope of a Tangent

Understanding the slope of a tangent line is crucial for grasping how the line interacts with the curve at a particular point. This concept boils down to how steep the line is, which is quantitatively represented by the derivative \( \frac{dy}{dx} \).
To determine this slope, you find \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \), then substitute in the given parameter \( t_0 \) to evaluate these at a specific point.
For example, in our exercise, \( t_0 = 1 \):

• \(\frac{dx}{dt}= 2^{t} \ln(2)\)
• \(\frac{dy}{dt}= 3^{t} \ln(3)\)

The slope then becomes \(\frac{dy}{dx} = \frac{3 \ln(3)}{2 \ln(2)}\), which quantifies the tangent line's steepness at the point (5, 5). This value is unique to the point and gives a precise understanding of how the curve behaves at \( t_0 = 1 \).