Problem 21 - Calculus Early Transcendentals

Question

Use a graph to estimate the coordinates of the rightmost point on the curve \(x=t-t^{6}, y=e^{t}\) . Then use calculus to find the exact coordinates.

Step-by-Step Solution

Answer

The rightmost point is approximately (0.486, 1.648).

Step 1: Understanding the Parameterization

We are given the parameterized equations for the curve: \(x = t - t^6\) and \(y = e^t\). Our task is to find the rightmost point on this curve, which corresponds to the maximum value of \(x\).

Step 2: Graphical Estimation

Plot the graph for the equations \(x = t - t^6\) and \(y = e^t\) using a graphing tool or software. Identify the rough estimate for the rightmost point by observing where \(x\) appears to be largest. This point should be near \(t = 0.5\).

Step 3: Finding the Derivative

To find the exact coordinates of the rightmost point, calculate the derivative of \(x\) with respect to \(t\). The derivative is \(\frac{dx}{dt} = 1 - 6t^5\).

Step 4: Setting the Derivative to Zero

Set the derivative equal to zero to find the critical points: \(1 - 6t^5 = 0\). Solve for \(t\) to find \(t = (1/6)^{1/5}\).

Step 5: Determine Whether It is a Maximum

To confirm that the critical point gives a maximum, evaluate the second derivative, \(\frac{d^2x}{dt^2} = -30t^4\). At \(t = (1/6)^{1/5}\), the second derivative is negative, confirming a local maximum.

Step 6: Compute Coordinate Values

Substitute \(t = (1/6)^{1/5}\) into the expressions for \(x\) and \(y\) to find the coordinates: \(x = (1/6)^{1/5} - ((1/6)^{1/5})^6 \approx 0.486\) and \(y = e^{(1/6)^{1/5}} \approx 1.648\).

Key Concepts

Understanding Derivatives

In calculus, a derivative measures how a function changes as its input changes. When dealing with parametric equations like \(x = t - t^6\) and \(y = e^t\), the derivative tells us how each of these variables change with respect to the parameter \(t\). The derivative \(\frac{dx}{dt} = 1 - 6t^5\) indicates the rate of change of \(x\) concerning \(t\).

• A positive derivative means \(x\) increases as \(t\) increases.
• A zero derivative at a specific point suggests that \(x\) is neither increasing nor decreasing at that moment.
• A negative derivative means \(x\) decreases as \(t\) increases.

Thus, finding when \(\frac{dx}{dt} = 0\) helps us identify points where \(x\) might have a local maximum or minimum, essential for identifying the rightmost point on the curve.

Locating Critical Points

Critical points are vital in understanding where functions reach maximum or minimum values or change direction. To locate these using our parametric equations, we set the derivative equal to zero. This step is crucial because:

• It captures points where the slope of \(x\) is zero, meaning there's no change, implying potential maximum or minimum.
• In our exercise, setting \(1 - 6t^5 = 0\) leads to solving for \(t = (1/6)^{1/5}\), the critical point where \(x\) achieves its extremum.

Identifying critical points is the first step towards determining whether these points are maxima or minima, which guides us in finding the desired coordinates.

The Second Derivative Test

The second derivative provides insights into the concavity of a function at a given point, helping classify critical points as maxima, minima, or points of inflection. In the context of our parametric curve, the second derivative \(\frac{d^2x}{dt^2} = -30t^4\) was computed to help in this classification. Here's how it works:

• A positive second derivative indicates a local minimum, as the curve is concave up.
• A negative second derivative signals a local maximum, as the curve is concave down.
• If zero, the test is inconclusive, requiring further investigation.

At \(t = (1/6)^{1/5}\), \(\frac{d^2x}{dt^2} = -30((1/6)^{1/5})^4\) results in a negative value, confirming the rightmost point is indeed a local maximum.

Graphical Estimation in Calculus

Graphical estimation is a practical tool in calculus that yields visual insight into the behavior of functions or curves before analytical methods are applied. It involves plotting parametric equations to observe where specific events occur, like maxima or minima.

• Provides an immediate, visual approximation of critical points.
• Helps verify the results obtained analytically for accuracy.
• Serves as a guide to narrow down the interval where in-depth calculations might be necessary.

For our problem, the visual plot of \(x = t - t^6\) and \(y = e^t\) hinted that the point where \(x\) is maximized is near \(t = 0.5\). This aligns closely with our calculated result, \(t = (1/6)^{1/5}\), negating errors and confirming the solution’s validity.