Question

In Exercises 74 and \(75,\) use the curve defined by the parametric equations \(x=t-\cos t, y=-1+\sin t\) Multiple Choice Which of the following is an equation of the tangent line to the curve at \(t=0 ?\) (A) \(y=x\) (B) \(y=-x\) (C) \(y=x+2\) (D) \(y=x-2 \quad(\) E) \(y=-x-2\)

Short answer

The equation of the tangent line to the curve at t=0 is \( y = x + 1 \), which does not match with any of the given multiple choices.

Step-by-step solution

Differentiate x and y with respect to t

Find the derivatives of the functions \( x = t - \cos t \) and \( y = -1 + \sin t \) with respect to t. This results in \( x' = 1+ \sin t \) and \( y' = \cos t \).

Evaluate the slope at t=0

Evaluate the function \( m = y' / x' \) at t=0, where m is the slope of the tangent. We have \( m = y'(0) / x'(0) = \cos 0 / (1+ \sin 0) = 1 \).

Formulate the equation of the tangent

The equation of the tangent line is given by point-slope form as \( y - y_1 = m(x - x_1) \). Substituting \( m=1 \), \( x_1 = 0 - cos(0) = 0 \) and \( y_1 = -1+ \sin 0 = -1 \), it results in \( y - (-1) = 1(x - 0) \) i.e. \( y = x + 1 \).

Match the equation with given options

The equation \( y = x+1 \) does not match with any of the available choices. Thus, there is an error in the multiple choice options provided as none of them is correct.

Key concepts

Differentiation

Differentiation is a fundamental operation in calculus that helps us understand how a function changes at any given point. It's like taking a snapshot of a function's velocity; how quickly the output of the function is changing with respect to its input. When we differentiate a function, we obtain its derivative, which tells us the rate of change or the slope of the tangent at every point along the curve.

In our exercise, we need to differentiate parametrically-defined functions. Here, both the coordinates x and y are described in terms of another variable, t. The derivatives, denoted as x' (dx/dt) and y' (dy/dt), are not direct rates of change of the functions themselves but are used to find the slope of the tangent to their curve at specific points.

Slope of Tangent

The slope of the tangent to a curve at any point is simply the slope of the line that just 'touches' the curve at that point without crossing it. It’s the best linear approximation to the curve at that point. In a parametric context, if you've differentiated x and y with respect to t to find x' and y', the slope of the tangent (m) is given by the ratio y'/x'.

In our exercise, after finding the derivatives, evaluating the slope at the point where t=0 gives us the specific angle at which the curve is moving at that instant. The slope helps us determine the direction and steepness of our tangent line.

Equation of Tangent Line

Once you know the slope of the tangent line, you can write its equation using the point-slope form: y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point the line passes through. This formulation allows us to construct the equation of a line given one point on the line and its slope.

Our problem at hand required us to plug in the slope we found at t=0, and the corresponding coordinates of the point on the curve at t=0 into the point-slope formula. A small hiccup with the provided choices reminds us to double-check our solution against the options presented.

Parametric Functions

Parametric functions are a neat way to represent curves by using a third variable, often called a parameter, to describe the x and y coordinates independently. This can make expressing complex curves and motions simpler, as each coordinate function only depends on the parameter, not directly on each other.

In our calculus problem, the functions for x and y are defined in terms of t, creating a set of parametric equations. They allow us to trace the path of a point moving along the curve, something particularly valuable in physics and engineering for modeling movement and change over time.

Calculus Problem Solving

Solving calculus problems often involves a step-by-step approach that lets you break down complicated problems into simpler parts. This methodical process includes understanding the problem, identifying what you need to find, breaking it down using known calculus rules (like differentiation), and carefully carrying out computations.

In the given problem, we followed a structured problem-solving approach. We differentiated, calculated the slope, then found the equation of the tangent line. Finally, it’s crucial to verify your answer, as sometimes even the textbook may have oversights, illustrated by the mismatch between our solution and the provided choices.