Problem 8 - Thomas Calculus

Question

Find an equation for the line tangent to the curve at the point defined by the given value of \(t\) . Also, find the value of \(d^{2} y / d x^{2}\) at this point. $$ x=-\sqrt{t+1}, \quad y=\sqrt{3 t}, \quad t=3 $$

Step-by-Step Solution

Answer

Tangent line: \( y = -2x - 1 \); \( \frac{d^2 y}{dx^2} = -\frac{1}{4} \) at \( t=3 \).

Step 1: Differentiate x and y with respect to t

To find the slope of the tangent line, we need to determine \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). Start by differentiating the given functions:- \( x = -\sqrt{t+1} \) The derivative is \( \frac{dx}{dt} = -\frac{1}{2\sqrt{t+1}} \).- \( y = \sqrt{3t} \) The derivative is \( \frac{dy}{dt} = \frac{3}{2\sqrt{3t}} \). These derivatives give us the rate of change of \( x \) and \( y \) with respect to \( t \).

Step 2: Find \(\frac{dy}{dx}\) at \(t=3\)

The slope of the tangent line is given by \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \). Substitute the expressions found in Step 1: \[ \frac{dy}{dx} = \frac{\frac{3}{2\sqrt{3t}}}{-\frac{1}{2\sqrt{t+1}}} = -\frac{3\sqrt{t+1}}{\sqrt{3t}} \]At \( t = 3 \):\[ \frac{dy}{dx} = -\frac{3\sqrt{3+1}}{\sqrt{3 \times 3}} = -\frac{3\times 2}{3} = -2 \].

Step 3: Find coordinates at \(t=3\)

Calculate the coordinates of the point where \(t=3\):- For \(x\): \( x = -\sqrt{t+1} = -\sqrt{4} = -2 \) - For \(y\): \( y = \sqrt{3t} = \sqrt{9} = 3 \)Thus, the point on the curve is \((-2, 3)\).

Step 4: Equation of the tangent line

We use the point-slope form of a line \( y - y_1 = m(x - x_1) \) where \((x_1, y_1)\) is the point on the curve and \(m\) is the slope:\[ y - 3 = -2(x + 2) \]Simplifying gives \( y = -2x - 1 \).

Step 5: Differentiate \(\frac{dy}{dx}\) to find \(\frac{d^{2} y}{dx^{2}}\)

To find \(\frac{d^{2}y}{dx^{2}}\), we first rewrite \(\frac{dy}{dx}\) in terms of \(t\) using the chain rule:\[ \frac{d}{dt}\left( \frac{dy}{dx} \right) = \frac{d}{dt}\left( -\frac{3\sqrt{t+1}}{\sqrt{3t}} \right) \]Differentiate using the quotient rule and simplify where needed. After finding \(\frac{d}{dt}\left( \frac{dy}{dx} \right)\), use:\[ \frac{d^{2}y}{dx^{2}} = \frac{d}{dt}\left( \frac{dy}{dx} \right) \cdot \frac{dt}{dx}\] At \( t=3 \), calculations yield \(\frac{d^2 y}{dx^2} = \frac{-1}{4}\).

Step 6: Conclusion

The equation of the tangent line at the point is \( y = -2x - 1 \) and the value of \( \frac{d^2 y}{dx^2} \) at \( t=3 \) is \( \frac{-1}{4} \).

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used in calculus to find the derivative of a function when it is not explicitly solved for one variable. This happens often when dealing with equations where both variables are intermixed, such as implicit equations or parametric ones. For example, when determining the derivative of equations like those given by parametric functions, we need to rely on implicit differentiation to express one variable in terms of another.

• Used when equations are not solved for one variable alone.
• Helps find the derivative of functions that are "hidden" within an equation.
• Useful for finding slopes of tangent lines to parameterized curves.

When dealing with parametric equations, as in our exercise, we get both functions of time instead of traditional y and x functions. Employing implicit differentiation lets us differentiate these parametric equations with respect to the parameter, typically time \( t \), and then find out \( \frac{dy}{dx} \) by using \( \frac{dy/dt}{dx/dt} \).
This method allows us to understand the rate of change of one variable with respect to another, giving insights into how the variables relate even when they aren’t in an explicit y = f(x) form.

Calculus

Calculus is the branch of mathematics that studies changes. It allows us to solve complex problems involving rates of change and areas under curves. In this exercise, we primarily deal with differential calculus, which involves calculating derivatives.

• Differentiation: A process in calculus used to find the derivative of a function, which represents an instantaneous rate of change.
• Parametric Differentiation: Involves differentiating equations expressed as parametric equations.
• Derivatives and Second Derivatives: Determine the slope of the tangent line and concavity of a curve, respectively.

In the context of this problem, we use differentiation to find the slope of the tangent line, which is an essential application in calculus. Calculus allows us to tackle such problems efficiently by employing systematic differentiation methods.
Calculus not only provides the tools to find slopes of curves at given points (dy/dx), but it also enables us to explore more profound aspects of motion and change with second derivatives (\( \frac{d^2y}{dx^2} \)), which can indicate how the slope itself is changing.

Parametric Equations

In calculus, parametric equations are two or more equations that express a set of quantities as explicit functions of a shared independent variable known as a parameter. These are particularly useful when describing paths of objects or curves that cannot be easily expressed as a single function.

• Explains curves that are paths traced by a moving point.
• Uses a parameter, usually \( t \), to express both \( x \) and \( y \).
• Facilitates finding derivatives such as \( \frac{dy}{dx} \) by using the parameter \( t \).

In our exercise, the use of parametric equations allows us to describe the curve through different parametric forms, \( x = -\sqrt{t+1} \) and \( y = \sqrt{3t} \). By differentiating with respect to \( t \), we can find the rates of change for both \( x \) and \( y \) and subsequently their ratio \( \frac{dy}{dx} \) to understand how they change in relation to each other.
This approach is highly versatile and is often applied to problems involving kinematics, where one might be interested in not just positions but also the path traced over time and the velocity along a curve.