Question

In Exercises \(41-48,\) find the equation of the line tangent to the curve at the point defined by the given value of \(t\) . $$x=\cos t, \quad y=1+\sin t, \quad t=\pi / 2$$

Short answer

The equation of the line tangent to the curve at the point defined by \(t = \pi / 2\) is \(y = 2\).

Step-by-step solution

Find the Point of Intersection

Substitute the given value for \(t\) which is \(\pi / 2\) into the given expressions for \(x\) and \(y\) to find the coordinates (\(x\),\(y\)) of the point where the tangent line intersects the curve. This gives: \(x = \cos (\pi/2) = 0 \) and \(y = 1 + \sin (\pi/2) = 2\). Therefore, the point of tangency is (\(0,2\)).

Compute the slope of the tangent line at the given point

First find the derivative of \(y\) with respect to \(t\), \(dy/dt\) = \(\cos t\), and the derivative of \(x\) with respect to \(t\), \(dx/dt\) = \(-\sin t\). The slope of the tangent line is the ratio of these two derivatives, \(dy/dx = dy/dt / dx/dt = (\cos t) / (-\sin t)\). Substitute \(\pi / 2\) into this equation to get the slope \(m\) at the point \((0, 2)\): \(m = \cos (\pi / 2) / -\sin (\pi / 2) = 0 / -1 = 0\).

Write out the equation of the tangent line

The formula for the tangent line to a curve at a specific point is in the point-slope form, \(y - y_1 = m (x - x_1)\), where \((x_1 , y_1)\) is the point of tangency and \(m\) is the slope of the tangent line. Substituting \((0, 2)\) for the point and \(0\) for the slope gives: \(y - 2 = 0 (x - 0)\). So the equation of the tangent line is \(y = 2\).

Key concepts

Point of Tangency

The point of tangency is the specific point where a tangent line touches a curve. It's a critical concept for understanding the nature of the tangent line because the tangent line only intersects the curve at this one exact point—assuming that the curve does not curve back on itself locally.

In our example, the equations provided for the curve are parametric, with the parameter being \( t \). The point of tangency is found by substituting the given value of \( t \) into the parametric equations. Here, with \( t = \frac{\pi}{2} \), we substitute into \( x = \text{cos}(t) \) and \( y = 1 + \text{sin}(t) \) yielding the point \( (0,2) \). This point is where the tangent will 'kiss' the curve exactly at one single point.

Derivative as Slope of Tangent

Understanding the derivative as the slope of the tangent line is essential in calculus. The derivative basically represents the rate of change of a function at a certain point, which can be visually interpreted as the slope of the tangent line to the curve at that point.

To find the slope of the tangent line for parametric equations, like \( x=\text{cos}(t) \) and \( y=1+\text{sin}(t) \), we use derivatives with respect to the parameter \( t \). The slope \( m \) of the tangent line is given by \( m = \frac{dy/dt}{dx/dt} \). In this exercise, we computed \( dy/dt = \text{cos}(t) \) and \( dx/dt = -\text{sin}(t) \). At \( t = \frac{\text{pi}}{2} \), this calculation gives us a slope of 0, indicating that the tangent line is horizontal at the point of tangency.

Parametric Equations Derivatives

When dealing with parametric equations, such as the ones given for the curve in this example, the derivatives are not computed directly with respect to \( x \) or \( y \), but rather with respect to the parameter \( t \).

To find the slope of a tangent for parametric equations, we compute the derivatives of both \( x \) and \( y \) with respect to \( t \): \( dx/dt \) and \( dy/dt \), which represent the rates of change of \( x \) and \( y \) with respect to the parameter. Dividing the derivative of \( y \) by the derivative of \( x \) yields the slope of the tangent line at a particular value of \( t \). This interplay of derivatives is crucial for understanding curves described parametrically.

Point-Slope Form Equation

The point-slope form is an efficient way to write the equation of a line when you know a point on the line (the point of tangency) and its slope (the derivative at that point). The general form is \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is the point on the line and \( m \) is the slope.

In the example, because our slope \( m \) is 0, the equation simplifies significantly. Substituting \( (0, 2) \) for the point of tangency and \( 0 \) for the slope into the point-slope form, the equation of the tangent line becomes \( y - 2 = 0 \), which simplifies to \( y = 2 \). This signifies a horizontal line that crosses the y-axis at 2, congruous with the horizontal tangent indicated by a slope of 0.