Problem 75 - Calculus Single Variable

Question

Plot the given parametric curve \(x=\varphi_{1}\) \((t), y=\varphi_{2}(t)\) in a viewing rectangle that contains the given point \(P_{0}\). Find the equation of the tangent line at \(P_{0}\). Add the tangent line to your plot. $$ \begin{array}{l} \varphi_{1}(t)=\left(1+t^{2}\right) /\left(1+3 t^{2}\right), \varphi_{2}(t)=4 t /\left(1+3 t^{2}\right. \\ P_{0}=(1 / 2,1)(t=1) \end{array} $$

Step-by-Step Solution

Answer

The tangent line at \(P_0 = (\frac{1}{2}, 1)\) is \(y = -2x + 2\).

Step 1: Understand the Parametric Equations

The parametric equations given are \(x = \varphi_{1}(t) = \frac{1 + t^2}{1 + 3t^2}\) and \(y = \varphi_{2}(t) = \frac{4t}{1 + 3t^2}\). These equations determine the coordinates of a point \((x,y)\) on the curve for any given parameter \(t\).

Step 2: Substitute to Find Specific Point

We need to verify the given point \(P_0\) where \(t = 1\). Substitute \(t = 1\) into the parametric equations: \[ x(1) = \frac{1 + 1^2}{1 + 3(1)^2} = \frac{2}{4} = \frac{1}{2}\] \[ y(1) = \frac{4(1)}{1 + 3(1)^2} = \frac{4}{4} = 1\] This confirms \(P_0 = \left(\frac{1}{2}, 1\right)\).

Step 3: Find the Tangent Line

The tangent to the parametric curve can use the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\): \[ \frac{dx}{dt} = \frac{d}{dt}\left(\frac{1+t^2}{1+3t^2}\right)\] Using the quotient rule we get: \[ \frac{dx}{dt} = \frac{2t(1+3t^2)-3t(1+t^2)}{(1+3t^2)^2} = \frac{-t + 3t^3}{(1+3t^2)^2}\] Find \(\frac{dy}{dt}\): \[ \frac{dy}{dt} = \frac{d}{dt}\left(\frac{4t}{1+3t^2}\right)\] \[ \Rightarrow \frac{dy}{dt} = \frac{4(1+3t^2) - 4t(6t)}{(1+3t^2)^2} = \frac{4-8t^2}{(1+3t^2)^2}\]

Step 4: Calculate the Slope at t=1

Use the derivatives to find the slope \(m\) of the tangent line at \(t = 1\) using \(m = \frac{dy/dt}{dx/dt}\): \[ \frac{dy}{dt}\bigg|_{t=1} = \frac{4 - 8(1)^2}{(1+3(1)^2)^2} = \frac{-4}{16} = -\frac{1}{4} \] \[ \frac{dx}{dt}\bigg|_{t=1} = \frac{3(1)^3 - 1}{16} = \frac{2}{16} = \frac{1}{8} \] Thus, \( m = \frac{-\frac{1}{4}}{\frac{1}{8}} = -2\).

Step 5: Equation of the Tangent Line

The equation of the tangent line is \(y - y_0 = m(x - x_0)\) where \((x_0, y_0) = (\frac{1}{2}, 1)\) and \(m = -2\): \[ y - 1 = -2(x - \frac{1}{2}) \] \[ y = -2x + 1 + 1 \] \[ y = -2x + 2 \]

Step 6: Plot the Curve and Tangent

Plot the parametric equations \(\varphi_{1}(t)\) and \(\varphi_{2}(t)\) for a suitable range of \(t\) around the given point. On the same graph, also plot the tangent line \(y = -2x + 2\) and verify that it touches the curve at the point \(P_0\).

Key Concepts

Tangent Line

A tangent line is a straight line that touches a curve at a single point without crossing it. This line represents the instantaneous direction of the curve at that point. Imagine holding a pencil and touching just the edge of a curved path—it gently kisses the curve only at that point. The slope of the tangent line gives us a snapshot of how steep the curve is at that specific location. In our exercise, the tangent line is calculated at point \( P_0 \), which is \( \left( \frac{1}{2}, 1 \right) \). By finding the tangent line, we understand the direction and rate of change of the curve at exactly that point.

Parametric Equations

Parametric equations offer a great way of defining curves by expressing their \(x\) and \(y\) coordinates as continuous functions of a third variable, commonly denoted as \(t\). This way of defining curves is particularly useful in scenarios where curves loop or reverse direction. In the given exercise, we have \( \varphi_1(t) = \frac{1+t^2}{1+3t^2} \) and \( \varphi_2(t) = \frac{4t}{1+3t^2} \). Here, \(t\) is the parameter that varies, thereby tracing out a curve as it moves through its range. By substituting different values of \(t\), we can find various points on the parametric curve, providing us with a comprehensive view of its shape and behavior.

Derivatives of Parametric Equations

To find the slope at a specific point on a parametric curve, we need to find the derivatives of the parametric equations with respect to the parameter \(t\). These derivatives give us essential information about how each coordinate (\(x\) and \(y\)) changes as \(t\) changes.

• \( \frac{dx}{dt} \) indicates the rate at which the \(x\) coordinate changes.
• \( \frac{dy}{dt} \) shows the rate for the \(y\) coordinate.

For our exercise, the derivatives were calculated as:

• \( \frac{dx}{dt} = \frac{-t + 3t^3}{(1+3t^2)^2} \)
• \( \frac{dy}{dt} = \frac{4-8t^2}{(1+3t^2)^2} \)

Using these derivatives, we calculated the slope of the tangent by dividing \( \frac{dy}{dt} \) by \( \frac{dx}{dt} \), revealing the steepness and the immediate direction of the curve at the point of interest.

Quotient Rule

The quotient rule is a formula used to find the derivative of a function that is the quotient of two differentiable functions. It's very handy when you have a "top" function divided by a "bottom" function, written as \( \frac{u(t)}{v(t)} \). The rule states that:\[ \left( \frac{u}{v} \right)^{\prime} = \frac{v \cdot u' - u \cdot v'}{v^2} \]We use the quotient rule in our exercise to differentiate both \( \varphi_1(t) \) and \( \varphi_2(t) \), as both are expressed as quotients of polynomials:

• For \( \varphi_1(t) = \frac{1+t^2}{1+3t^2} \) and \( \varphi_2(t) = \frac{4t}{1+3t^2} \), we apply the quotient rule to find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \), respectively.

The quotient rule simplifies the process of differentiation, making it easier to work with complex expressions and find accurate derivatives essential for calculating the slope of the tangent.