Problem 33 - Calculus: Early Transcendental Functions

Question

Find an equation of the tangent line at the given value of \(x\) $$y=\int_{0}^{x} \sin \sqrt{t^{2}+\pi^{2}} d t, x=0$$

Step-by-Step Solution

Answer

The equation of the tangent line to the function at \(x=0\) is \(y=0\).

Step 1: Use the Fundamental Theorem of Calculus to find the derivative

The Fundamental Theorem of Calculus allows us to transform the integral into a function that can be differentiated. If we have \(y=\int_{a}^{x} f(t) dt\), the derivative of \(y\) with respect to \(x\) is just \(f(x)\). Apply this rule to the given integral, we get \(y'(x)=\sin\sqrt{x^{2}+\pi^{2}}\).

Step 2: Evaluate the derivative at the given x to find the slope of the tangent line

Substitute \(x=0\) into the derivative function. This gives \(y'(0)=\sin\sqrt{0^{2}+\pi^{2}}\), which simplifies to \(y'(0)=\sin\pi=0\). This value is the slope of the tangent line at \(x=0\).

Step 3: Use the slope and the point to write the equation of the tangent line

The equation of the tangent line follows the slope-intercept form \(y=mx+b\), where \(m\) is the slope, \(b\) is the y-intercept. Since we know the slope is 0 and the line is tangent to the curve at a point \((x, y)=(0, y(0))\), we can solve for \(b\) and write the equation for the tangent line. Since the integral evaluated at \(x=0\) gives \(y(0)=0\), we can see \(b=y(0)=0\). This simplifies the equation of the tangent line to \(y=0\).

Key Concepts

Tangent Line

A tangent line is a straight line that touches a curve at a single point and has the same slope as the curve at that point. This characteristic makes the tangent line a best linear approximation of the curve at the given point. It does not cross the curve at this point, meaning it only "touches" the curve.

• To find the equation of the tangent line, we need the slope of the tangent and a point through which the line passes.
• The slope of the tangent line is given by the derivative of the curve evaluated at the point of tangency.
• Once the slope is known, the point-slope form of a line can be used to find its equation.

For the given problem, after finding the derivative using the Fundamental Theorem of Calculus, we evaluated it to find that the slope at the point \((0, 0)\) is 0. Therefore, the tangent line is simply the horizontal line \(y = 0\). This tangent line mirrors the flat nature of the curve's slope at \(x = 0\).

Derivative

The concept of a derivative represents the rate of change of a function as one of its variables changes. It is essentially the slope of the tangent line to the curve at any given point.When using differentiation, especially to solve for tangent lines, keep in mind:

• The derivative of a function \(f(x)\) at a point \(x\) is given by \(f'(x)\).
• If the derivative at a point is 0, this means that the curve is flat (no slope) at that point.
• The Fundamental Theorem of Calculus relates differentiation and integration, allowing one to differentiate an integral function directly.

In the exercise, the derivative is found using the Fundamental Theorem of Calculus, resulting in \(y'(x) = \sin\sqrt{x^2 + \pi^2}\). Evaluating at \(x = 0\), we find the slope is 0, confirming a flat tangent line at that point.

Integral Function

An integral function provides the accumulated area under a curve and is essentially the inverse operation of differentiation. This makes it an essential tool in analyzing the behavior of functions over intervals.

• An integral from \(a\) to \(x\) can be used to define a new function. Here, the integral gives the net area under the curve from \(t = 0\) to \(t = x\).
• The resulting function is often used to find antiderivatives or solutions to accumulation problems.

In the problem, we start with the integral function \(y = \int_{0}^{x} \sin \sqrt{t^2 + \pi^2} \, dt\), which describes the area under the curve of \(\sin \sqrt{t^2 + \pi^2}\) from \(0\) to \(x\). The Fundamental Theorem of Calculus then helps derive this function, enabling us to determine tangents and other rate of change information. By evaluating at \(x = 0\), an important property emerges: the integral function value is directly \(y(0) = 0\), illustrating the direct relationship between the area function and its derivative.