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Chapter 7: Problem 75

For the following exercises, write the equation of the tangent line in Cartesian coordinates for the given parameter \(t\). \(x=e^{\sqrt{t}}, \quad y=1-\ln t^{2}, \quad t=1\)

Short Answer

Expert verified
The equation of the tangent line is \( y = -\frac{4}{e}x + 5 \).

Step by step solution

01

Find Derivatives

To find the equation of the tangent line, we first need the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). For \( x = e^{\sqrt{t}} \), use the chain rule to get: \[ \frac{dx}{dt} = \frac{d}{dt}(e^{\sqrt{t}}) = e^{\sqrt{t}} \cdot \frac{1}{2\sqrt{t}} \]For \( y = 1 - \ln(t^2) \), the derivative is: \[ \frac{dy}{dt} = \frac{d}{dt}(1 - \ln(t^2)) = -\frac{2}{t} \].
02

Calculate Slopes

To find the slope of the tangent line, we use the formula:\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \]Substitute the derivatives found:\[ \frac{dy}{dx} = \frac{-\frac{2}{t}}{e^{\sqrt{t}} \cdot \frac{1}{2\sqrt{t}}} = \frac{-4\sqrt{t}}{te^{\sqrt{t}}} \].
03

Evaluate Derivatives at \( t = 1 \)

Substitute \( t = 1 \) into the expressions for \( x \), \( y \), and \( \frac{dy}{dx} \):- For \( x \), \( x(1) = e^{\sqrt{1}} = e \).- For \( y \), \( y(1) = 1 - \ln(1^2) = 1 \).- For the slope: \[ \frac{dy}{dx} \bigg|_{t=1} = \frac{-4}{e} \].
04

Tangent Line Equation

With the point \((e, 1)\) and slope \( -\frac{4}{e} \), the tangent line can be written in the point-slope form:\[ y - y_1 = m(x - x_1) \]Substituting the known values gives:\[ y - 1 = -\frac{4}{e}(x - e) \].
05

Simplify the Equation

Rearrange the equation from the point-slope form: \[ y - 1 = -\frac{4}{e}x + 4 \]This simplifies to:\[ y = -\frac{4}{e}x + 5 \].

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Key Concepts

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

Parametric Equations
A parametric equation expresses the coordinates of the points making up a geometric object as functions of a parameter, often denoted as \( t \). This approach allows the representation of curves that might not be possible with a single Cartesian equation.
In this exercise, the parametric equations are given as:
  • \(x = e^{\sqrt{t}}\)
  • \(y = 1 - \ln t^2\)
Each function describes a different aspect of the curve, with \(t\) dictating the position on the curve. This setup is particularly useful for curves where one function fails to encompass the entire behavior of the point on the curve.
Derivative Calculation
The derivative represents the rate at which a function is changing and is key in finding tangent lines to curves at specific points. For parametric equations, we need to find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
For the problems given:
  • For \( x = e^{\sqrt{t}} \), using derivative properties for exponential functions gives \( \frac{dx}{dt} = e^{\sqrt{t}} \frac{1}{2\sqrt{t}} \).
  • For \( y = 1 - \ln(t^2) \), employing the properties of logarithmic derivatives results in \( \frac{dy}{dt} = -\frac{2}{t} \).
The derivatives enable us to explore how each function component changes with respect to the parameter \( t \), thus facilitating the calculation of the tangent line.
Chain Rule
The chain rule is a fundamental concept in calculus used to differentiate composite functions. It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
Let's take a look at how it's applied here:
  • When calculating \( \frac{dx}{dt} \) for \( x = e^{\sqrt{t}} \), the chain rule allows us to find the derivative effortlessly by considering \( \sqrt{t} \) as the inner function, yielding \( e^{\sqrt{t}} \cdot \frac{1}{2\sqrt{t}} \).
Applying the chain rule correctly is crucial in simplifying and accurately solving derivatives of complex functions.
Point-Slope Form
The point-slope form is a popular method for writing the equation of a line when you know a point on the line and its slope. It鈥檚 structurally expressed as:\[ y - y_1 = m(x - x_1) \]
  • \( m \) represents the slope.
  • \((x_1, y_1)\) represents the coordinates of the known point.
In the problem, after calculating the slope \(-\frac{4}{e}\) at \( t = 1 \) and identifying the point \((e, 1)\), we plug into the formula to derive the tangent equation:
\[ y - 1 = -\frac{4}{e}(x - e) \]
This specific form simplifies the process of quickly finding the equation of the tangent line, as it seamlessly incorporates both the slope and coordinates of the known point directly.

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Most popular questions from this chapter

For the following exercises, find the slope of the tangent line to the given polar curve at the point given by the value of \(\theta\) \(r=\ln \theta, \quad \theta=e\)

For the following exercises, sketch the graph of each conic. \(r=\frac{1}{1+\sin \theta}\)

For the following exercises, find the slope of the tangent line to the given polar curve at the point given by the value of \(\theta\) \(r=3 \cos \theta, \theta=\frac{\pi}{3}\)

Show that the curve \(r=\sin \theta \tan \theta\) (called a cissoid of Diocles) has the line \(x=1\) as a vertical asymptote.

Whispering galleries are rooms designed with elliptical ceilings. A person standing at one focus can whisper and be heard by a person standing at the other focus because all the sound waves that reach the ceiling are reflected to the other person. If a whispering gallery has a length of 120 feet and the foci are located 30 feet from the center, find the height of the ceiling at the center.
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