Problem 51
Question
In Exercises \(50-53,\) verify that point \(P\) is on the graph of function \(F,\) and calculate the tangent line to the graph of \(F\) at \(P\) $$ F(x)=\int_{1}^{x} 1 / t d t \quad P=(e, 1) $$
Step-by-Step Solution
Verified Answer
Yes, point P is on the graph, and the tangent line is \(y = \frac{1}{e}x\).
1Step 1: Confirm that Point P is on the Graph
To verify that point \(P = (e, 1)\) is on the graph of \(F(x)\), we calculate \(F(e)\).The function \(F(x)\) is defined as \(F(x) = \int_{1}^{x} \frac{1}{t} \, dt\).Evaluating the integral, we get:\[F(x) = \ln |x| - \ln |1| = \ln x\]Therefore, \(F(e) = \ln e = 1\).Since \(F(e) = 1\), point \(P = (e, 1)\) is indeed on the graph of \(F(x)\).
2Step 2: Differentiate F to Find the Slope
To find the equation of the tangent line, first determine the slope at the point \(P\). The slope of the tangent line is the derivative of \(F(x)\) at \(x = e\).Given \(F(x) = \ln x\), the derivative is:\[F'(x) = \frac{d}{dx} (\ln x) = \frac{1}{x}\]Therefore, the slope of the tangent line at \(x = e\) is \(F'(e) = \frac{1}{e}\).
3Step 3: Use the Point-Slope Form to Write Equation
With the slope calculated, use the point-slope form of a line equation to find the tangent line.The point-slope form is:\[y - y_1 = m(x - x_1)\]Here, \((x_1, y_1) = (e, 1)\) and the slope \(m = \frac{1}{e}\).Substituting in these values gives:\[y - 1 = \frac{1}{e}(x - e)\]Rearrange to find the equation of the tangent line:\[y = \frac{1}{e}x - 1 + 1 = \frac{1}{e}x\]So the tangent line at \(P = (e, 1)\) is \(y = \frac{1}{e}x\).
Key Concepts
Integral CalculusDerivativesPoint-Slope Form
Integral Calculus
Integral calculus is a vital branch of mathematics that deals with the concept of integration. This process helps in finding the overall accumulation of quantities, such as areas under curves and the total change in functions. A function's integral is often represented by \(\int\) followed by the function being integrated. For instance, in the exercise provided, the function \(F(x) = \int_{1}^{x} \frac{1}{t} \, dt\) represents the integral of \(\frac{1}{t}\) from 1 to \(x\). This integral calculates the area under the curve \(\frac{1}{x}\) from 1 to \(x\).
An essential aspect of integral calculus is understanding the relationship between a function and its integral. In our case, this relationship is explored using the fundamental theorem of calculus. The evaluated integral \(\ln x\) shows how the original function behaves, particularly in locating point \(P\). By confirming this point lies on the function's graph, we verify our understanding of integral calculations and their applications in graphical representations.
An essential aspect of integral calculus is understanding the relationship between a function and its integral. In our case, this relationship is explored using the fundamental theorem of calculus. The evaluated integral \(\ln x\) shows how the original function behaves, particularly in locating point \(P\). By confirming this point lies on the function's graph, we verify our understanding of integral calculations and their applications in graphical representations.
Derivatives
In calculus, derivatives are crucial for understanding how functions change. A derivative measures a function's instantaneous rate of change at a particular point or, visually, the slope of the tangent to the function's graph. For the function \(F(x) \) in our exercise, which simplifies to \(\ln x\), finding the derivative lets us determine the slope of the tangent line at a specific point.
Calculating the derivative, \(F'(x) = \frac{1}{x}\), allows us to see how changes in \(x\) affect \(F(x)\). This specific derivative implies that at any point \(x\), the derivative or slope is \(\frac{1}{x}\). Hence, for our point \(P = (e, 1)\), \(F'(e) = \frac{1}{e}\), confirming the slope at this point. Understanding derivatives is essential not just in finding tangent lines but also in optimizing functions and analyzing their behavior.
Calculating the derivative, \(F'(x) = \frac{1}{x}\), allows us to see how changes in \(x\) affect \(F(x)\). This specific derivative implies that at any point \(x\), the derivative or slope is \(\frac{1}{x}\). Hence, for our point \(P = (e, 1)\), \(F'(e) = \frac{1}{e}\), confirming the slope at this point. Understanding derivatives is essential not just in finding tangent lines but also in optimizing functions and analyzing their behavior.
Point-Slope Form
The point-slope form is a useful formula for writing the equation of a line given a point and a slope. It is defined as \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a point on the line, and \(m\) is the slope. In the exercise, we use this form to write the tangent line at point \(P\) on the graph of \(F(x)\).
Given \((x_1, y_1) = (e, 1)\) and the previously calculated slope \(m = \frac{1}{e}\), we substitute these values into the point-slope form to find the equation of our tangent line. Solving gives us \(y = \frac{1}{e}x\), an equation that represents the tangent line at point \(P\).
Given \((x_1, y_1) = (e, 1)\) and the previously calculated slope \(m = \frac{1}{e}\), we substitute these values into the point-slope form to find the equation of our tangent line. Solving gives us \(y = \frac{1}{e}x\), an equation that represents the tangent line at point \(P\).
- This form is simple yet powerful.
- It facilitates determining the line's equation quickly when a point and slope are known.
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