Problem 56 - Calculus Early Transcendentals

Question

Consider the following surfaces specified in the form \(z=f(x, y)\) and the curve \(C\) in the \(x y\) -plane given parametrically in the form \(x=g(t), y=h(t)\). a. In each case, find \(z^{\prime}(t)\). b. Imagine that you are walking on the surface directly above the curve \(C\) in the direction of increasing t. Find the values of \(t\) for which you are walking uphill (that is, \(z\) is increasing). $$z=2 x^{2}+y^{2}+1, C: x=1+\cos t, y=\sin t ; 0 \leq t \leq 2 \pi$$

Step-by-Step Solution

Answer

Answer: \(z\) is increasing for the values of \(t\) in the interval \((\pi, 2\pi)\).

Step 1: Find the partial derivatives of z

For the function \(z=2x^2+y^2+1\), we can find the partial derivatives with respect to x and y: $$\frac{\partial z}{\partial x} = 4x$$ $$\frac{\partial z}{\partial y} = 2y$$

Step 2: Find the derivatives of x(t) and y(t)

For the parametric curve C: \(x=1+\cos t\) and \(y=\sin t\), we need to find the derivatives of x with respect to t and y with respect to t: $$\frac{dx}{dt} = -\sin t$$ $$\frac{dy}{dt} = \cos t$$

Step 3: Apply the chain rule to obtain z'(t)

Now, we can find the derivative of z with respect to t using the chain rule: $$z'(t) = \frac{\partial z}{\partial x}\frac{dx}{dt} +\frac{\partial z}{\partial y}\frac{dy}{dt}$$ Substitute the derivatives we found in steps 1 and 2: $$z'(t) = (4x)(-\sin t) + (2y)(\cos t)$$ Now we substitute the expressions for x(t) and y(t) into the equation: $$z'(t)= (4(1+\cos t))(-\sin t) + (2(\sin t))(\cos t)$$

Step 4: Simplify z'(t)

To simplify, we can expand the brackets and combine terms: $$z'(t) = -4\sin t - 4\cos t\sin t + 2\sin t \cos t$$

Step 5: Determine when z'(t) is positive

We want to find the values of t for which we are walking uphill, meaning \(z'(t) > 0\). To do this, we will analyze the inequality: $$ -4\sin t - 4\cos t\sin t + 2\sin t \cos t > 0 $$ Now we can factor out \(\sin t\): $$\sin t(-4-4\cos t+2\cos t) > 0$$ Simplify the expression: $$ \sin t(-4-2\cos t) > 0$$ Now we need to find the values of t in the interval \(0 \leq t \leq 2\pi\) for which the above inequality is true. These values of t will represent when we are walking uphill.

Step 6: Solve the inequality

To solve this inequality, we need to find the intervals for which the expression \((-4-2\cos t)\) has the same sign as \(\sin t\): 1. The first case is when both are positive: \(-4-2\cos t > 0\) and \(\sin t > 0\) 2. The second case is when both are negative: \(-4-2\cos t < 0\) and \(\sin t < 0\) Now we need to solve each of these cases in the interval \(0 \leq t \leq 2\pi\). For the first case, we combine the two conditions: \(-4-2\cos t > 0\) and \(0 < t < \pi\). If \(-4-2\cos t > 0\), then \(\cos t < -2\). However, since \(-1\leq \cos t \leq 1\) for all values of t, there is no solution for this case. For the second case, we combine the two conditions: \(-4-2\cos t < 0\) and \(\pi < t < 2\pi\). If \(-4-2\cos t < 0\), then \(\cos t > -2\). Since \(\cos t\) is always greater than -1, we have solutions for all values of t in the interval \((\pi, 2\pi)\). #Conclusion# The values of t for which we are walking uphill (z is increasing) are in the interval \((\pi, 2\pi)\).

Key Concepts

Parametric Curves

In mathematical modeling, a parametric curve is a curve that's represented by parameters.
This means instead of directly relating the two axes, we use a third variable,often denoted as \(t\), to express the \(x\) and \(y\) coordinates.

For example, given a curve \(C\) where \(x = g(t)\) and \(y = h(t)\), the parameter \(t\) could represent time or any other variable through which the motion along the curve is described.
This approach allows us to easily express more complex motions, like those found in circular paths or ellipses.

The power of parametric form is that it allows for a broader range of curves and makes differentiation and integration under constrained contexts easier. Some of the key benefits include:

• Describing curves that aren’t functions in the traditional sense, like circles.
• Handling motion and dynamic scenarios where forces only act in certain dimensions at specific instances.

Chain Rule

The chain rule is a fundamental tool in calculus, employed for finding the derivative of composite functions.
In the context of partial derivatives, the chain rule helps us connect changes in a function's outputwith respect to changes in its input via an intermediate variable.

For functions expressed in parametric form, certain changes are not direct between dependent and independent variables.
That’s when the chain rule comes into play.

For the function \(z = f(x, y)\), where \(x = g(t)\) and \(y = h(t)\), we usethe chain rule to express \(\frac{dz}{dt}\), which equates to:

• \( \frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt} \)

This expression combines partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\),with the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), respectively.

Curve Analysis

Curve analysis involves studying the characteristics and behavior of curves. It often includes finding velocities, locating points of interest,and understanding the direction of motion.
When walking along a parametric curve, we're primarily interested in the manner the path develops over time.

This involves:

• Understanding the relationship between \(x\) and \(y\) coordinates, as defined parametrically.
• Using calculus techniques, such as differentiation, to find the rate of change—in other words, the slope or velocity along the curve.
• Locating specific lengths or features such as turning points or maximum and minimum coordinates.

By performing these analyses, we gain insight into not only the geometric properties of the curve but also the dynamics of any object following that path.

Surface Gradient

In multivariable calculus, the gradient is a vector field pointing in the direction of greatest increase of a function.
For a surface described by \(z = f(x, y)\), the gradient \(abla z\) consists of the partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\).

This gradient vector

• Indicates the direction and rate of maximum increase of the function \(z\).
• Helps in visualizing the slope of a surface at any given point.

The concept of the gradient is essential when analyzing a surface relative to a curve, especially when the curve dictates the movement of an object—and determining the conditions under which one might experience uphill or downhill motion.
The gradient not only exposes the changes in the surface itself but also aids in determining the optimal path to takebased on our specific trajectory or guidance needs.