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Chapter 10: Problem 64

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

Short Answer

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

Step by step solution

01

Find \(dx/dt\) and \(dy/dt\)

Differentiate each equation \(x=\sqrt{t}\) and \(y=2t\) with respect to \(t\). $$\frac{dx}{dt}=\frac{d}{dt}\sqrt{t}=\frac{1}{2\sqrt{t}}$$ $$\frac{dy}{dt}=\frac{d}{dt}(2t)=2$$
02

Find \(dy/dx\)

Use the Chain Rule to find the derivative \(dy/dx\) by dividing \(dy/dt\) by \(dx/dt\). $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2}{\frac{1}{2\sqrt{t}}}=(2)\cdot(2\sqrt{t})=4\sqrt{t}$$
03

Evaluate \(dy/dx\) at \(t=4\)

Plug the given value of \(t=4\) into the derivative. $$\frac{dy}{dx}\Big|_{t=4}=4\sqrt{4}=4\cdot2=8$$
04

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 \((2,8)\).
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 \(dy/dx\) 鈥� 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'.

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Most popular questions from this chapter

Consider positive real numbers \(x\) and \(y\). Notice that \(4^{3}<3^{4},\) while \(3^{2}>2^{3},\) and \(4^{2}=2^{4} .\) Describe the regions in the first quadrant of the \(x y\) -plane in which \(x^{y}>y^{x}\) and \(x^{y}

Graph the following spirals. Indicate the direction in which the spiral is generated as \(\theta\) increases, where \(\theta>0 .\) Let \(a=1\) and \(a=-1\). Logarithmic spiral: \(r=e^{a \theta}\)

Graph the following conic sections, labeling the vertices, foci, directrices, and asymptotes (if they exist). Use a graphing utility to check your work. $$r=\frac{6}{3+2 \sin \theta}$$

An epitrochoid is the path of a point on a circle of radius \(b\) as it rolls on the outside of a circle of radius \(a\). It is described by the equations $$\begin{array}{l}x=(a+b) \cos t-c \cos \left(\frac{(a+b) t}{b}\right) \\\y=(a+b) \sin t-c \sin \left(\frac{(a+b) t}{b}\right)\end{array}$$ Use a graphing utility to explore the dependence of the curve on the parameters \(a, b,\) and \(c.\)

Sector of a hyperbola Let \(H\) be the right branch of the hyperbola \(x^{2}-y^{2}=1\) and let \(\ell\) be the line \(y=m(x-2)\) that passes through the point (2,0) with slope \(m,\) where \(-\infty < m < \infty\). Let \(R\) be the region in the first quadrant bounded by \(H\) and \(\ell\) (see figure). Let \(A(m)\) be the area of \(R .\) Note that for some values of \(m\) \(A(m)\) is not defined. a. Find the \(x\) -coordinates of the intersection points between \(H\) and \(\ell\) as functions of \(m ;\) call them \(u(m)\) and \(v(m),\) where \(v(m) > u(m) > 1 .\) For what values of \(m\) are there two intersection points? b. Evaluate \(\lim _{m \rightarrow 1^{+}} u(m)\) and \(\lim _{m \rightarrow 1^{+}} v(m)\) c. Evaluate \(\lim _{m \rightarrow \infty} u(m)\) and \(\lim _{m \rightarrow \infty} v(m)\) d. Evaluate and interpret \(\lim _{m \rightarrow \infty} A(m)\)
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