Problem 20 - Calculus Single Variable

Question

Variables \(x\) and \(y\) are functions of a parameter \(t\) and are related by a given equation. A point \(P_{0}\) on the graph of that equation is also given, as is one of the following two values: $$ v_{0}=\left.\frac{d x}{d t}\right|_{P_{0}} \quad \text { or } \quad s_{0}=\left.\frac{d y}{d t}\right|_{P_{0}} $$ Find the other value. Also, referring to the Insight that follows Example \(2,\) find the tangent line to the graph of the given equation at \(P_{0}\). $$ 2 x-\sqrt{x}-\sqrt{y}=3, \quad P_{0}=(4,9), \quad s_{0}=2 $$

Step-by-Step Solution

Answer

The missing value is \( v_0 = \frac{8}{21} \), and the tangent line is \( y = \frac{21}{4}x - 3 \).

Step 1: Differentiate the Given Equation

Differentiate the equation \( 2x - \sqrt{x} - \sqrt{y} = 3 \) with respect to the parameter \( t \). We obtain: \[ 2 \frac{dx}{dt} - \frac{1}{2\sqrt{x}} \frac{dx}{dt} - \frac{1}{2\sqrt{y}} \frac{dy}{dt} = 0. \]

Step 2: Substitute Known Values

Substitute the known values of \( P_0 = (4, 9) \) and \( s_0 = \left. \frac{dy}{dt} \right|_{P_0} = 2 \) into the differentiated equation: \[ 2v_0 - \frac{1}{2\cdot 2}v_0 - \frac{1}{2\cdot 3} \cdot 2 = 0. \]

Step 3: Simplify the Equation to Solve for Velocity \( v_0 \)

Simplify the equation: \[ 2v_0 - \frac{1}{4}v_0 - \frac{1}{3} \cdot 2 = 0. \] Simplifying further, we get \( \frac{7}{4}v_0 - \frac{2}{3} = 0 \).

Step 4: Solve for \( v_0 \)

Rearrange the equation to solve for \( v_0 \): \[ \frac{7}{4}v_0 = \frac{2}{3}. \] Therefore, \[ v_0 = \frac{2}{3} \times \frac{4}{7} = \frac{8}{21}. \]

Step 5: Find the Slope of the Tangent Line

The slope of the tangent line at \( P_0 \) is given by \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2}{\frac{8}{21}} = \frac{21}{4}. \)

Step 6: Write the Equation of the Tangent Line

Using the point-slope form, the equation of the tangent line is \( y - 9 = \frac{21}{4}(x - 4) \). Simplifying, the tangent line equation becomes \( y = \frac{21}{4}x - \frac{21}{4} \cdot 4 + 9 = \frac{21}{4}x - 12 + 9 \), which simplifies to \( y = \frac{21}{4}x - 3 \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus focusing on how a function changes when its variables change. In this problem, we are working with functions defined by parametric equations, where both \( x \) and \( y \) are functions of a third variable, \( t \). To differentiate our given equation, \( 2x - \sqrt{x} - \sqrt{y} = 3 \), we take the derivative of each term with respect to \( t \).

While differentiating:

• The term \( 2x \) becomes \( 2 \frac{dx}{dt} \), because the derivative of a constant times a variable function is the constant times the derivative of the function.
• The derivative of \( \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \), applying the chain rule and differentiating with respect to \( t \) gives us \( \frac{1}{2\sqrt{x}} \frac{dx}{dt} \).
• Similarly, for \( \sqrt{y} \), the process is analogous, yielding \( \frac{1}{2\sqrt{y}} \frac{dy}{dt} \).

After substituting these into the equation, we solved for the unknown derivative \( \frac{dx}{dt} \). Differentiation is the tool we use to find rates of change, represented here as \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).

Tangent Line

A tangent line is a straight line that just "touches" a curve at a particular point without crossing it. The slope of this tangent line gives us an immediate measure of the curve's steepness at that position.

To find the tangent line at \( P_0 = (4, 9) \), we first find the slope of the tangent line, which is determined by the derivatives \( \frac{dy}{dx} \), using our results for \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). Specifically, the slope at \( P_0 \) is:

• \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \).

Using this slope, we can construct the equation of the tangent line using the point-slope form of a line: \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is our point \( P_0 \) and \( m \) is the slope. This neatly gives us the full equation for the tangent line, matching the problem's requirements.

Parametric Equations

Parametric equations allow us to describe curves not just as \( y \) as a function of \( x \), but in terms of a third variable, \( t \), which acts as a parameter. This is especially useful when dealing with complex curves that cannot be easily expressed as simple functions in Cartesian coordinates. In this exercise, both \( x \) and \( y \) are dependent on \( t \).

With parametric equations, we express each variable in relation to \( t \):

• \( x = f(t) \)
• \( y = g(t) \)

This approach provides a powerful way to handle situations where changes over time or another parameter are involved, giving a clearer picture of motion or growth. For instance, in physics, parametric equations often describe trajectories, allowing us to examine how objects move along curved paths.

In solving the exercise, parametric differentiation helps us track changes in both \( x \) and \( y \) as \( t \) changes. The derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) reveal how fast each variable is changing with respect to the parameter \( t \). Understanding and using parametric equations can provide a deeper insight into behavior and relationships of dynamic systems.