Question

Consider the following parametric curves. a. Determine \(d y / d x\) in terms of \(t\) and evaluate it at the given value of \(t.\) b. Make a sketch of the curve showing the tangent line at the point corresponding to the given value of \(t.\) $$x=2+4 t, y=4-8 t ; t=2$$

Short answer

Question: Find the slope of the tangent line to the curve defined by the parametric equations \(x = 2 + 4t\) and \(y = 4 - 8t\) at \(t = 2\). Also, provide a sketch of the curve and tangent line. Answer: The slope of the tangent line to the curve at \(t = 2\) is -2. The curve is a straight line with a constant slope of -2, passing through the point (10, -12). The tangent line is a straight line with the same slope as the curve itself, touching the curve at the point (10, -12).

Step-by-step solution

Find dx/dt and dy/dt

To find \(dy/dx\) in terms of t, we need to first find the derivatives of x and y with respect to t. These are \(dx/dt\) and \(dy/dt\). Differentiate the given parametric equations for x and y with respect to t: $$\frac{dx}{dt} = \frac{d(2+4t)}{dt} = 4$$ $$\frac{dy}{dt} = \frac{d(4-8t)}{dt} = -8$$

Find dy/dx in terms of t using the Chain Rule

Now that we have \(dx/dt\) and \(dy/dt\), we can use the Chain Rule to find \(dy/dx\): $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-8}{4} = -2$$ This result tells us that the slope at any point on the curve is constant; it does not depend on the parameter t.

Evaluate dy/dx at t=2

Since dy/dx is constant, its value at \(t=2\) is the same as anywhere else on the curve: $$\frac{dy}{dx}\bigg|_{t=2} = -2$$

Find the point on the curve corresponding to t=2

Substitute the value of t=2 into the parametric equations to find the coordinates of the point at which the tangent line will be drawn: $$x = 2 + 4(2) = 10$$ $$y = 4 - 8(2) = -12$$ The point is \((10, -12)\).

Sketch the curve and tangent line

Now that we have the slope of the tangent line and the point on the curve corresponding to t=2, we can sketch the curve and tangent line. The curve is a straight line with a slope of -2, passing through the point (10, -12). To draw the tangent line, use the point-slope form of a line equation: $$y - (-12) = -2(x - 10)$$ So the tangent line is also a straight line with the same slope as the curve itself, touching the curve at the point (10, -12). Keep in mind that the final sketch should visually show (a) the curve derived from the parametric equations, (b) the point on the curve corresponding to t=2, and (c) the tangent line to the curve at that point.

Key concepts

dy/dx in terms of t

Understanding the rate of change of one variable with respect to another is a foundational concept in calculus. When dealing with parametric equations, we have the expressions for both x and y in terms of a third parameter t. To find the slope of the curve represented by these parametric equations, or in calculus terms, to determine \( \frac{dy}{dx} \) in terms of t, we must utilize the derivatives of these equations.

The first step is to differentiate both the x-component and y-component with respect to the parameter t. After obtaining \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), the slope of the tangent line at any point on the curve can be found by dividing \(\frac{dy}{dt}\) by \(\frac{dx}{dt}\). This is effectively the Chain Rule in action, linking the rate of change from y to t, and then from t to x, resulting in the rate of change from y to x.

In the given exercise, the procedure uncovers that the slope, \(\frac{dy}{dx} = -2\), is constant regardless of the value of t. This constant slope simplifies the process of sketching the tangent line at any given point for this specific curve.

Chain Rule

The Chain Rule is an essential tool in calculus for differentiation, especially when variables are linked in a composite manner. Imagine having a set of gears, with each gear representing a different function or rate of change. The Chain Rule helps us understand how rotating one gear (changing one variable) affects the motion (rate of change) of another linked gear.

In the context of parametric curves, we often need to find the derivative of y with respect to x, even when both variables are functions of t. The Chain Rule provides a straightforward method to compute this derivative by multiplying the derivative of y with respect to t (\(\frac{dy}{dt}\)) by the reciprocal of the derivative of x with respect to t (\(\frac{dt}{dx}\)). Symbolically, this is represented as \(\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}\).

To put it simply, the Chain Rule allows us to 'chain' together derivative relationships across multiple variables, thereby arriving at the relationship between variables that aren't directly related through a shared parameter. This bridge is crucial in solving our given problem of finding the slope of the tangent line in parametric equations.

tangent lines in parametric equations

Tangent lines hold a special place in calculus, representing the slope of a curve at a particular point and serving as a local linear approximation of the curve at that point. For curves expressed in parametric form, finding the tangent line involves a few specific steps, linking together several calculus concepts.

After determining the slope of the tangent line \(\frac{dy}{dx}\) at a particular value of t, the next step is to identify the point on the curve that corresponds to this t-value. This is done by substituting the value of t back into the original parametric equations for x and y. These coordinates give us a specific point on the curve where the tangent touches.

When sketching, you would draw the curve based on a set of points generated from the parametric equations. You then plot the point of tangency, and use the calculated slope to draw the tangent line at that point. The tangent line should look like it's just brushing the curve at that point and not cutting across it. In our textbook exercise, the curve turned out to be a straight line itself, with the tangent line coinciding with the curve at the evaluated point. This unique scenario brings out an interesting aspect of tangency in parametric curves; sometimes, the curve can be its own tangent, simplifying the drawing process but also providing a clear visual representation of the constant slope concept.