Problem 31 - Applied Calculus

Question

Find the equation of the tangent line to \(y=e^{-2 t}\) at \(t=0 .\) Check by sketching the graphs of \(y=e^{-2 t}\) and the tangent line on the same axes.

Step-by-Step Solution

Answer

The equation of the tangent line is \( y = -2t + 1 \).

Step 1: Find the derivative of the function

To find the slope of the tangent line, we need the derivative of the function. Differentiate the function \( y = e^{-2t} \) with respect to \( t \). The derivative is \( \frac{dy}{dt} = -2e^{-2t} \).

Step 2: Evaluate the derivative at the given point

To find the slope of the tangent line at \( t = 0 \), substitute \( t = 0 \) into the derivative. This gives: \( \frac{dy}{dt} \bigg|_{t=0} = -2e^{0} = -2 \).

Step 3: Find the point on the function

Evaluate the original function at \( t = 0 \) to find the y-value of the point where the tangent line touches the function. Substitute \( t = 0 \) into \( y = e^{-2t} \) to get: \( y = e^{0} = 1 \). Thus, the point is \((0, 1)\).

Step 4: Write the equation of the tangent line

Use the point-slope form of the line equation: \( y - y_1 = m(t - t_1) \), where \( m \) is the slope and \( (t_1, y_1) \) is the point. Substituting \( m = -2 \), \( t_1 = 0 \), and \( y_1 = 1 \), we get: \( y - 1 = -2(t - 0) \) or \( y = -2t + 1 \).

Step 5: Sketch the graphs

Sketch the graph of \( y = e^{-2t} \), which is a decaying exponential curve. Then sketch the tangent line \( y = -2t + 1 \), which has a slope of -2, starting at the point \((0, 1)\). The line should touch the curve exactly at this point.

Key Concepts

Derivative

The concept of a derivative is crucial in understanding how functions change. The derivative represents the slope of the tangent line at any given point on the graph of a function.
In mathematical terms, the derivative of a function at a certain point tells us how the function value changes as its input changes. It's like asking: "How is my position changing as I move forward?"
For the exponential function given, the original expression is \(y = e^{-2t}\).
To find the derivative, you apply the chain rule of calculus, which involves taking the derivative of each part of the function.

• First, recognize that the outer function is an exponential function, \(e^u\), and its derivative is \(e^u\).
• Then, notice the inner function is \(-2t\), and its derivative is \(-2\).

By combining these, the derivative is \(\frac{dy}{dt} = -2e^{-2t}\).
This derivative tells us not just how steep the curve is but also whether the function is increasing or decreasing at that point.

Exponential Function

An exponential function is a mathematical function in the form \(f(t) = a \cdot e^{bt}\), where \(e\) is Euler's number, a fundamental constant approximately equal to 2.71828. Exponential functions are widespread in nature; they often describe growth processes or decay, which can be rapid and significant.

For the function \(y = e^{-2t}\), we specifically have:

• \(a = 1\), meaning it starts at the value of 1 when \(t = 0\).
• The exponent \(-2t\) indicates the function is decreasing, or decaying, over time. This is because of the negative sign in front of \(2t\).

When graphed, the curve of \(y = e^{-2t}\) will start at 1 when \(t = 0\), then decrease steeply at first, flattening out as \(t\) increases. This curve represents how quickly values decrease in systems characterized by this type of exponentially decreasing behavior.

Point-Slope Form

The point-slope form is a method used to find the equation of a line when you know a point on the line and the slope. It is especially useful for finding tangent lines to curves. The formula is presented as:
\[ y - y_1 = m(t - t_1) \]where:

• \(y_1\) is the y-coordinate of the known point.
• \(m\) is the slope of the line.
• \(t_1\) is the t-coordinate of the known point.

In our problem, at the point \((0, 1)\), the slope \(m\) was derived from the derivative, giving us \(m = -2\). Substituting these into the point-slope form results in:
\( y - 1 = -2(t - 0)\), which simplifies to \(y = -2t + 1\).
This equation represents the tangent line, touching the curve of the function at exactly one point on the graph. It's a straight line that matches the slope and the position at that instant, providing insight into how the function is changing right at that spot.