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Chapter 12: Problem 75

Tangent lines Find an equation of the line tangent to the curve at the point corresponding to the given value of \(t\). $$x=\cos t+t \sin t, y=\sin t-t \cos t ; t=\pi / 4$$

Short Answer

Expert verified
Answer: The equation of the tangent line at \(t=\frac{\pi}{4}\) is \(y = x - \sqrt{2}\).

Step by step solution

01

Find the point on the curve corresponding to \(t=\frac{\pi}{4}\)

To find the point on the curve corresponding to \(t=\frac{\pi}{4}\), we need to plug \(\frac{\pi}{4}\) into the expressions for \(x(t)\) and \(y(t)\): $$x(t) = \cos t + t \sin t$$ $$x(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) + \frac{\pi}{4}\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}+\frac{\pi\sqrt{2}}{8}$$ $$y(t) = \sin t - t \cos t$$ $$y(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) - \frac{\pi}{4}\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}-\frac{\pi\sqrt{2}}{8}$$ So the point on the curve corresponding to \(t=\frac{\pi}{4}\) is: $$(\frac{\sqrt{2}}{2}+\frac{\pi\sqrt{2}}{8}, \frac{\sqrt{2}}{2}-\frac{\pi\sqrt{2}}{8})$$
02

Find the slope of the tangent line

To find the slope of the tangent line, we need to compute the derivatives of \(x(t)\) and \(y(t)\) with respect to \(t\): $$\frac{dx}{dt} = -\sin t + \sin t + t \cos t = t\cos t$$ $$\frac{dy}{dt}=\cos t - \cos t + t \sin t=t\sin t$$ Now we can find the slope of the tangent line at \(t=\frac{\pi}{4}\): $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{t\sin t}{t\cos t} = \frac{\sin t}{\cos t}$$ Substituting \(t=\frac{\pi}{4}\) into the expression for the slope: $$\frac{dy}{dx}=\frac{\sin(\frac{\pi}{4})}{\cos(\frac{\pi}{4})} = 1$$ So the slope of the tangent line at \(t=\frac{\pi}{4}\) is \(1\).
03

Find the equation of the tangent line

Now that we have the point on the curve and the slope of the tangent line, we can use the point-slope form of a line to find the equation of the tangent line: $$y-y_1 = m(x-x_1)$$ Plugging in the point \((\frac{\sqrt{2}}{2}+\frac{\pi\sqrt{2}}{8}, \frac{\sqrt{2}}{2}-\frac{\pi\sqrt{2}}{8})\) and the slope \(m=1\): $$y-\left(\frac{\sqrt{2}}{2}-\frac{\pi\sqrt{2}}{8}\right) = 1\left(x-\left(\frac{\sqrt{2}}{2}+\frac{\pi\sqrt{2}}{8}\right)\right)$$ Simplify the expression: $$y = x - \sqrt{2}$$ Thus, the equation of the tangent line to the curve at \(t=\frac{\pi}{4}\) is: $$y = x - \sqrt{2}$$

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Key Concepts

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

Differentiation
Differentiation is the process of finding the derivative of a function, which tells us how the function changes at any given point. In this problem, we are dealing with parametric equations, so we find derivatives of both the x and y components with respect to the parameter \(t\). This helps us understand the rate of change in each direction.
For differentiation, we apply the basic rules:
  • The derivative of \(\cos t\) is \(-\sin t\).
  • The derivative of \(\sin t\) is \(\cos t\).
  • The derivative of \(t\) is \(1\), and using the product rule for terms like \(t \sin t\).
This results in the derivatives \(\frac{dx}{dt} = t \cos t\) and \(\frac{dy}{dt} = t \sin t\). Differentiation is crucial in finding the slope of the tangent line, which involves these derived functions.
Parametric Equations
Parametric equations allow us to describe curves in terms of a third variable, usually \(t\). This approach gives flexibility in representing complex curves:
  • The x-component is given by \(x = \cos t + t \sin t\).
  • The y-component is \(y = \sin t - t \cos t\).
Unlike traditional equations involving x and y directly, parametric equations provide a way to explore a wide variety of curves by letting \(t\) run through a range of values. In the context of our problem, \(t = \frac{\pi}{4}\) gives us specific points and changes, allowing us to find the tangent line's slope and equation at that point.
Slope of Tangent Line
The slope of the tangent line describes the steepness and direction at any point on the curve. It tells us how the y-component of the curve changes with the x-component.
To find it, we divide the derivative of \(y\) by the derivative of \(x\):
  • The formula is \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).
For our specific point \(t = \frac{\pi}{4}\), the derivatives simplify to \(\sin t/\cos t\), resulting in a slope of 1. This means the tangent line is at a 45-degree angle, equally rising as it runs along the x-axis.
Point-Slope Form
The point-slope form gives us an easy way to write the equation of a line when we know:
  • A specific point \((x_1, y_1)\) on the line.
  • The slope \(m\) of the line.
The form is defined by the equation: \(y - y_1 = m(x - x_1)\). Using the point \((\frac{\sqrt{2}}{2} + \frac{\pi\sqrt{2}}{8}, \frac{\sqrt{2}}{2} - \frac{\pi\sqrt{2}}{8})\) and a slope of 1, we substitute these into the formula and simplify:
Ultimately, we get the equation \(y = x - \sqrt{2}\), describing the tangent line. The point-slope form makes it straightforward to find this equation, ensuring students understand how to connect slopes and points seamlessly.

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Most popular questions from this chapter

Let \(C\) be the curve \(x=f(t)\), \(y=g(t),\) for \(a \leq t \leq b,\) where \(f^{\prime}\) and \(g^{\prime}\) are continuous on \([a, b]\) and C does not intersect itself, except possibly at its endpoints. If \(g\) is nonnegative on \([a, b],\) then the area of the surface obtained by revolving C about the \(x\)-axis is $$S=\int_{a}^{b} 2 \pi g(t) \sqrt{f^{\prime}(t)^{2}+g^{\prime}(t)^{2}} d t$$. Likewise, if \(f\) is nonnegative on \([a, b],\) then the area of the surface obtained by revolving C about the \(y\)-axis is $$S=\int_{a}^{b} 2 \pi f(t) \sqrt{f^{\prime}(t)^{2}+g^{\prime}(t)^{2}} d t$$ (These results can be derived in a manner similar to the derivations given in Section 6.6 for surfaces of revolution generated by the curve \(y=f(x)\).) Consider the curve \(x=3 \cos t, y=3 \sin t+4,\) for \(0 \leq t \leq 2 \pi\) a. Describe the curve. b. If the curve is revolved about the \(x\) -axis, describe the shape of the surface of revolution and find the area of the surface.

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