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Chapter 16: Problem 14

Find the equation of the tangent to \(y=\frac{1}{x^{2}}\) where \(x=-1 .\)

Short Answer

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Question: Find the equation of the tangent to the curve \(y = \frac{1}{x^2}\) at the point where \(x = -1\). Answer: The equation of the tangent to the curve at the given point is \(y = 2x + 3\).

Step by step solution

01

Find the Point on the Curve

We are given the \(x\) value as \(x= -1\). We now find the \(y\) value by plugging this into the equation of the curve. This will give us the point \((x,y)\) where the tangent line touches the curve: $$ y = \frac{1}{x^2} = \frac{1}{(-1)^2} = 1 $$ So, the tangent touches the curve at the point \((-1, 1)\).
02

Differentiate the Equation of the Curve

Differentiate \(y = \frac{1}{x^2}\) with respect to \(x\) to obtain the slope function of the curve: $$ y' = \frac{d}{dx} \left( \frac{1}{x^2} \right) = -\frac{2}{x^3} $$
03

Find the Slope of the Tangent at the Given Point

Substitute the given \(x\) value (\(x=-1\)) into the derivative to find the slope of the tangent at the point: $$ m = y'(-1) = -\frac{2}{(-1)^3} = 2 $$ The slope of the tangent at the point \((-1,1)\) is \(2\).
04

Write the Equation of the Tangent Line

Now we can use the point-slope form of a linear equation (\(y-y_1=m(x-x_1)\)) to write the tangent equation using the point \((-1,1)\) and slope \(2\): $$ y-1 = 2(x-(-1)) $$ Simplify the equation to obtain the final tangent equation: $$ y = 2x+3 $$

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Key Concepts

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

Derivatives
In calculus, derivatives signify the rate at which a function changes at any given point. Essentially, when you find the derivative of a function, you are determining how the y-value (output) changes as the x-value (input) changes.
For any function, its derivative is a new function that provides the slope or steepness of the tangent line at any given point along the original function's curve.
  • For example, the derivative of the function \( y = \frac{1}{x^2} \) is \( y' = -\frac{2}{x^3} \), representing how quickly this function is changing at any value of \( x \).
  • This derivative is essential for calculating the slope of the tangent line at any point.
Tangent Line
The tangent line is a straight line that touches a curve at only one point. This line represents the localized linear approximation of the curve at the point of contact.
Tangent lines are crucial as they provide a simple representation of the curve near a tiny section around the point of tangency.
  • It's identified by a slope equal to the derivative of the function at the given point.
  • In our example, the tangent line to the curve \( y = \frac{1}{x^2} \) at \( x = -1 \) touches the curve at only the point \( (-1,1) \).
Calculating the tangent line at a given point involves finding the slope using the derivative and using this slope along with a specific point to form the line's equation.
Differentiation
Differentiation is the process of finding the derivative of a function. It involves applying specific rules to discover a slope function based on the original equation of the curve. This process allows us to determine how a function behaves over time or across different values compared to the initial function.
For a function \( y = \frac{1}{x^2} \), the differentiation of this function results in \( y' = -\frac{2}{x^3} \).
  • Here, "-" indicates the function decreases as x increases and vice versa.
  • The process of differentiation helps us examine changes in functions at exact points, facilitating the sketching of tangent lines.
Remember, different rules like the power rule, product rule, or chain rule might apply depending on the complexity of the function you're dealing with.
Point-Slope Form
The point-slope form provides an essential equation to express a line on a graph based on its slope and a particular point it passes through. This approach is especially handy when you're working on finding an equation for a tangent line.
In the point-slope form, the equation is expressed as:
  • \( y - y_1 = m(x-x_1) \),
  • where \((x_1, y_1)\) is the specific point and \(m\) is the slope.
In the example provided, using the point \((-1,1)\) and the slope \(m=2\), the point-slope form becomes:
  • \( y - 1 = 2(x + 1) \).
This equation reflects how the tangent line behaves near the curve and gives a simple linear equation representing that specific section of the curve.

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Most popular questions from this chapter

Find the equation of the tangent to \(y=x \mathrm{e}^{x}\) where \(x=1\).

Differentiate each of the following functions: (a) \(\left(x^{3}+2\right)^{6}\) (b) \(\sqrt{\sin x}\) (c) \(\left(\mathrm{e}^{x}+1\right)^{7}\) (d) \((\cos 2 x)^{5}\) (e) \(\ln (x+1)\)

Find \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) given \(\ln (x+y)=k, k\) constant.

State the range of values of \(x\) for which each of the following functions is (i) concave up (ii) concave down: (a) \(y=\frac{x^{3}}{6}-\frac{5 x^{2}}{2}+3 x-9\) (b) \(y=3+x+\frac{x^{2}}{2}-\frac{x^{4}}{12}\) (c) \(y=\mathrm{e}^{x}-100 x-100\) (d) \(y=(x-1)^{4}\) (e) \(y=\frac{x^{2}}{2} \ln x-\frac{x^{2}}{4}\)

Find \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) given (a) \(x(t)=t^{2}+3, y(t)=2 t^{2}+t+1\) (b) \(x(t)=t^{2}, y(t)=t^{3}+k, k\) constant (c) \(x(t)=\frac{1}{t}, y(t)=\sin t\) (d) \(x(t)=2 \mathrm{e}^{t}, y(t)=t \mathrm{e}^{t}\) (e) \(x(t)=\sqrt{t, y(t)}=\sqrt{2 t+1}\)
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