Problem 64 - Calculus Early Transcendentals

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 \(t.\) $$x=\sqrt{t}, y=2 t ; t=4$$

Step-by-Step Solution

Answer

The slope of the tangent line at the point corresponding to \(t=4\) is 8.

Step 1: Find \(dx/dt\) and \(dy/dt\)

Differentiate each equation \(x=\sqrt{t}\) and \(y=2t\) with respect to \(t\). $$\frac{dx}{dt}=\frac{d}{dt}\sqrt{t}=\frac{1}{2\sqrt{t}}$$ $$\frac{dy}{dt}=\frac{d}{dt}(2t)=2$$

Step 2: Find \(dy/dx\)

Use the Chain Rule to find the derivative \(dy/dx\) by dividing \(dy/dt\) by \(dx/dt\). $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2}{\frac{1}{2\sqrt{t}}}=(2)\cdot(2\sqrt{t})=4\sqrt{t}$$

Step 3: Evaluate \(dy/dx\) at \(t=4\)

Plug the given value of \(t=4\) into the derivative. $$\frac{dy}{dx}\Big|_{t=4}=4\sqrt{4}=4\cdot2=8$$

Step 4: Find the point corresponding to \(t=4\)

Plug the given value of \(t=4\) into the equations for \(x\) and \(y\). $$x\Big|_{t=4}=\sqrt{4}=2$$ $$y\Big|_{t=4}=2\cdot4=8$$ So, the point corresponding to \(t=4\) is \((2,8)\).

Step 5: Sketch the curve and tangent line

Sketch the curve defined by the parametric equations \(x=\sqrt{t}\) and \(y=2t\). Then, draw the tangent line at the point \((2,8)\) with a slope of \(8\). To make a simple sketch, label the axes, plot the point corresponding to \(t=4\), and draw the tangent line at the point. The precise sketching can be obtained through graphing software or calculator.

Key Concepts

Differential Calculus

Differential calculus is a subfield of calculus concerned with the study of rates at which quantities change. It provides an arithmetical procedure to compute the rate of change of a function in correspondence to changes in its variables. When dealing with parametric curves, such as in this exercise, the rate of change is not directly between the variables x and y, but instead it is considered in terms of a third parameter, t.

In our case, where a curve is described parametrically by equations in the form of 'x' and 'y' as functions of 't', differential calculus allows us to find the relationship between these functions. By differentiating the functions for x and y separately with respect to 't', we obtain derivatives that can describe the slope of the curve at any point when evaluated at particular values of 't'. This is exactly what is done in Step 1 and Step 3 of the solution provided, leading to a concrete slope value at t = 4.

Tangent Lines

Tangent lines are straight lines that lightly touch a curve at a single point, without crossing through it at that instant. They represent the instantaneous direction of the curve and the slope of the tangent line at any given point is equal to the derivative of the function at that point.

In parametric equations, the process of finding a tangent line's slope is nuanced because each variable depends on the parameter 't'. We calculate the derivatives with respect to 't' for both x (dx/dt) and y (dy/dt), then use these to find the slope dy/dx. This slope tells us the gradient of the tangent at any given point on the curve, as seen in Step 2 of the solution.

How to visualize the tangent

To sketch the tangent at a specific value of 't', you first calculate the corresponding x and y values to locate the point on the curve. Then, using the found slope (in this example, evaluated at t = 4), you draw a line passing through the point with this slope. A correct tangent line at t = 4 has been accomplished in Step 5 of the solution.

Chain Rule

The Chain Rule is a critical tool in calculus used to compute the derivative of the composition of two or more functions. It states that the derivative of a composite function equals the derivative of the outside function evaluated at the inside function multiplied by the derivative of the inside function.

With parametric equations where 'x' and 'y' are both given as functions of 't', the Chain Rule becomes essential for finding \(dy/dx\) — the rate of change of 'y' with respect to 'x'. We essentially treat \(dy/dt\) as the 'outside function' and \(dx/dt\) as the 'inside function', deploying the Chain Rule to divide \(dy/dt\) by \(dx/dt\) to find \(dy/dx\). In our example, Step 2 shows the application of the Chain Rule, providing a clear mechanism to get the slope of the tangent line to the curve at any point by only knowing the values of \(dy/dt\) and \(dx/dt\), and without the need for an explicit equation relating 'x' and 'y'.