Question

Find the equations of the tangent and normal lines to the graph of the function at the given point. \(f(t)=e^{t}+3\) at \(t=0\).

Short answer

Tangent line: \(y = x + 4\); Normal line: \(y = -x + 4\).

Step-by-step solution

Identify the Point

First, we need to find the coordinates of the point on the graph of \(f(t) = e^{t} + 3\) at \(t=0\). Substitute \(t=0\) into the function: \(f(0) = e^{0} + 3 = 1 + 3 = 4\). Thus, the point is \((0, 4)\).

Compute the Derivative

Calculate the derivative \(f'(t)\) to find the slope of the tangent line. The function is \(f(t) = e^{t} + 3\), so the derivative is \(f'(t) = e^{t}\).

Evaluate the Derivative at the Given Point

Find the slope of the tangent line at \(t=0\) by evaluating \(f'(t)\) at \(t=0\). Substitute \(t=0\) into the derivative: \(f'(0) = e^{0} = 1\).

Equation of the Tangent Line

Use the point-slope form of a line, \(y - y_1 = m(x - x_1)\), with \((x_1, y_1) = (0, 4)\) and \(m = 1\). The equation becomes \(y - 4 = 1(x - 0)\) or simply \(y = x + 4\).

Equation of the Normal Line

The slope of the normal line is the negative reciprocal of the tangent line's slope. Since the tangent slope is 1, the normal slope is \(-1\). Using the point-slope form again with slope \(-1\), the equation is \(y - 4 = -1(x - 0)\) or \(y = -x + 4\).

Key concepts

Derivative

Let's begin by understanding the concept of a derivative. In calculus, the derivative of a function is a measure of how a function changes as its input changes. It's like asking "What is the rate of change or the slope of the function at a certain point?" Here's a bit more detail:- For any curve described by a function, the derivative helps us find the slope of the tangent line to the curve at a given point.- In mathematical notation, the derivative of a function \( f(t) \) is often written as \( f'(t) \), \( \frac{df}{dt} \), or \( \dot{f} \). - To find the derivative, we use rules like the Power Rule, Product Rule, and Chain Rule depending on the function.In the context of our exercise, we found the derivative of \( f(t) = e^t + 3 \), which is \( f'(t) = e^t \). This tells us how steep the curve is at any point \( t \). For \( t=0 \), the derivative evaluates to 1, indicating a slope of 1 at this point.

Tangent Line

The tangent line is a straight line that touches a curve at a single point without crossing it at that point. Imagine drawing a line that just slightly touches the curve at a point — that's your tangent line!Here's why the tangent line is important:- It reflects the "instantaneous" rate of change of the function at that point.- Tangent lines are used to approximate the curve at a very small interval around the point.To find the tangent line's equation, we:- Calculate the slope using the derivative of the function.- Use the point-slope equation \( y - y_1 = m(x - x_1) \), where \( m \) is the slope found from the derivative, and \( (x_1, y_1) \) is the point.In the exercise, using the point \((0, 4)\) and slope 1, the equation of the tangent line is \( y = x + 4 \). This line just gently kisses the curve at the point (0, 4) and moves on.

Normal Line

A normal line is perpendicular to the tangent line at the point of tangency. It provides a nice contrasting perspective to the way a tangent represents rate of change.Key features of normal lines include:- They are always at a 90-degree angle to the tangent line at the point of contact.- The slope of a normal line is the negative reciprocal of the tangent line's slope.For finding its equation:- Determine the normal slope. If the tangent line's slope is \( m \), the normal slope will be \(-\frac{1}{m}\).- Use the point-slope form \( y - y_1 = m(x - x_1) \) with the calculated normal slope.In our example, since the tangent slope at \( t=0 \) is 1, the normal slope becomes -1. Then the equation of the normal line using point \( (0, 4) \) is \( y = -x + 4 \), which cuts through the curve perpendicularly at (0, 4).