Problem 49

Question

Find the equation of the tangent line to $y=x^{2}-2 x+2$ at the point $(1,1)$.

Step-by-Step Solution

Verified

Answer

The equation of the tangent line is $y = 1$.

1Step 1: Determine the derivative

To find the equation of the tangent line, we first need the derivative of the function. The function given is $y = x^{2} - 2x + 2$. We find the derivative using basic differentiation rules: $y' = 2x - 2$. This derivative gives us the slope $m$ of the tangent line at any point $x$ on the curve.

2Step 2: Evaluate the derivative at the given point

Substitute $x = 1$ into the derivative to find the slope of the tangent line at that point. $y' = 2(1) - 2 = 0$. Thus, the slope of the tangent line at $(1, 1)$ is $0$.

3Step 3: Write the tangent line equation

The general equation of a line is $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept. With the point $(1, 1)$ and the slope $m = 0$, the line equation is $y = 0 \cdot x + c$. Substituting the point $(1, 1)$ into this equation, we find $1 = c$. Therefore, the equation of the tangent line is $y = 1$, which is a horizontal line passing through $(1, 1)$.

Key Concepts

DifferentiationSlope of the TangentDerivative Evaluation

Differentiation

Differentiation is a fundamental concept in calculus. It helps us understand how a function changes at any given point. The process of differentiation involves finding the derivative of a function. A derivative gives the rate at which one quantity changes with respect to another. For functions like polynomials, using basic rules can easily find the derivative.

Power rule: If you have a function $f(x) = x^n$, then its derivative is $f'(x) = nx^{n-1}$.
Sum rule: The derivative of a sum of functions is the sum of the derivatives.
Constant rule: The derivative of a constant is always zero.

In our exercise, we differentiated $y = x^{2} - 2x + 2$. Applying the rules, we find the derivative $y' = 2x - 2$. This expression tells us how the function $y$ changes as $x$ changes.

Slope of the Tangent

The slope of the tangent line at a particular point on a curve is crucial. It indicates how steep the curve is at that exact location. Mathematically, the slope at a point is the value of the derivative evaluated at that point. In our problem, the function $y = x^{2} - 2x + 2$ was differentiated to get $y' = 2x - 2$.
To find the slope of the tangent line at the point $(1,1)$, we plug $x = 1$ into the derivative:

$y' = 2(1) - 2 = 0$.

This result, $0$, tells us that the slope of the tangent at $(1,1)$ is zero, indicating a horizontal line. A zero slope suggests no vertical change in the line at that point.

Derivative Evaluation

Evaluating the derivative at a specific point allows us to determine important properties like the tangent slope. This process involves substituting the particular value of $x$ into the derivative to see how the original function behaves.
In our example, we had $y' = 2x - 2$ and checked it for $x = 1$. We found the slope of the tangent was zero:

$y' = 2(1) - 2 = 0$.

This evaluation confirmed that the tangent at $(1, 1)$ is horizontal. Knowing how to evaluate derivatives quickly allows you to find tangent lines efficiently. It helps apply calculus concepts to solve real-world problems, like cutting-edge technologies and engineering solutions.

Problem 49

Other exercises in this chapter

Problem 49

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Problem 49

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Problem 49

Find $\Delta y$ for the given values of $x_{1}$ and $x_{2}($ see Example 7). $$ y=\frac{3}{x+1}, x_{1}=2.34, x_{2}=2.31 $$

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Problem 50

$$ \underline{\phantom{xxx}} , \text { find the indicated derivative. } $$ $$ D_{x}\left(\ln x^{2}\right)^{2 x+3} $$

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