91Ó°ÊÓ

Parametric equations express coordinates of points on a curve as functions of a parameter, often denoted as \( t \). To find the rate of change of \( y \) with respect to \( x \) when both \( x \) and \( y \) depend on \( t \), we utilize the derivatives \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \). The chain rule in calculus tells us that to determine \( \frac{dy}{dx} \), we need to derive the quotient \( \frac{dy/dt}{dx/dt} \).

Here's a step-by-step breakdown:

• Differentiation of \( x(t) = e^{t/10} \cos t \) and \( y(t) = e^{t/10} \sin t \) using the product rule gives us the necessary components of \( dx/dt \) and \( dy/dt \).
• Combine these components to form a single fraction \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \).
• This quotient provides the slope of the tangent at any point on the parametric curve.

By evaluating this derivative at specific values of \( t \), such as \( t = \pi/2 \) in our example, we find the exact slope of the curve at a particular point. This slope plays a crucial role in identifying and sketching tangent lines to the curve.

In calculus, tangent and normal lines give valuable insights into the behavior of curves at specific points. Once we have the derivative from the parametric equations, we can move forward to find equations of these lines.

• The tangent line has a slope equal to \( \frac{dy}{dx} \) evaluated at the given point. Its equation follows the point-slope form: \( y - y_1 = m(x - x_1) \).
• The normal line is perpendicular to the tangent, thus its slope is the negative reciprocal. If the tangent slope is \( m \), the normal line slope is \( -1/m \).

At the point where \( t = \pi/2 \), we found the tangent slope to be \(-\frac{1}{10}\), leading us to write the equation of the tangent line as \( y = -\frac{1}{10}x + e^{\pi/20} \).
Meanwhile, the normal line at the same point has a slope of \( 10 \) and a resulting equation of \( y = 10x + e^{\pi/20} \). Understanding these elements is key to studying the behavior of the parametric curve.

Sketching a parametric curve involves plotting points and understanding their formation as \( t \) varies. Unlike traditional \( y = f(x) \) graphs, parametric curves can loop, spiral, or take complex forms based on the equations.

• The values of \( x(t) \) and \( y(t) \) define the path of the curve in the coordinate plane as \( t \) changes.
• Focus on key points, like the ones calculated at specific \( t \) values, to achieve a well-defined curve.
• Include the tangent and normal lines to give viewers a sense of directionality and orientation at critical points.

In our exercise, sketching would begin with plotting the point at \( (0, e^{\pi/20}) \), corresponding to \( t = \pi/2 \). Then, the tangent and normal lines can be drawn to visualize how the curve behaves at this point. This creates a comprehensive view for analyzing the parametric equation.

Mastering calculus problems, especially those involving parametric equations, requires a systematic approach. Start by understanding what each piece signifies in a parametric context.

• Begin with appropriate differentiation techniques to find slopes or rates of change like \( \frac{dy}{dx} \).
• Clearly identify the points of interest by evaluating the parametric equations at given parameter values.
• Use the slopes found to work out tangent and normal line equations systematically.
• Finally, confidently sketch the resulting curve, tangent, and normal lines, providing a visual understanding of the mathematical functions at hand.

Applying these techniques not only solves individual problems but also builds a foundation for tackling more complex calculus issues. Understanding each step deepens your overall mathematical proficiency and enhances problem-solving skills.