Problem 5
Question
Suppose that \(z\) is a linear function of \(x\) and \(y\) with slope -5 in the \(x\) direction and slope 5 in the \(y\) direction. (a) A change of 0.2 in \(x\) and 0.5 in \(y\) produces what change in \(z ?\) change in \(z=\) __________. (b) If \(z=6\) when \(x=3\) and \(y=2\), what is the value of \(z\) when \(x=2.7\) and \(y=1.9 ?\) \(z=\) ___________.
Step-by-Step Solution
Verified Answer
(a) The change in \(z\) is 1.5
(b) The value of \(z\) when \(x=2.7\) and \(y=1.9\) is 7.
1Step 1: Calculate Δz based on Δx and Δy
For part (a), we need to find the change in \(z\) i.e \(\Delta z\) when \(\Delta x= 0.2\) and \(\Delta y=0.5\).
Multiply the change in x with its slope and change in y with its slope and calculate the change in z.
\[\Delta z = -5 \times 0.2 + 5 \times 0.5 \]
Now simplify the terms.
\[\Delta z = -1 + 2.5\]
2Step 2: Solve for Δz
So, by combining terms, we obtain the change in z.
\[\Delta z = 1.5\]
The change in \(z\) is 1.5 when there is a change of 0.2 in \(x\) and 0.5 in \(y\).
3Step 3: Solve for z using given values
For part (b), we are given that \(z = 6\) when \(x = 3\) and \(y = 2\). We can substitute these values into the equation and solve for \(C\).
\[6 = -5(3) + 5(2) + C\]
Now simplify the terms.
\[6 = -15 + 10 + C\]
4Step 4: Solve for C
To get the value of C, move the numeric terms to the other side of the equation.
\[C = 6 +15 -10\]
\[C = 11\]
So now we have the complete linear equation:
\[z = -5x + 5y + 11\]
5Step 5: Solve for z at x=2.7 and y=1.9
Now we need to find the value of z when \(x=2.7\) and \(y=1.9\). We can plug these values into the equation.
\[z = -5(2.7) + 5(1.9) + 11\]
Now simplify the terms.
\[z = -13.5 + 9.5 + 11\]
6Step 6: Solve for z
By combining terms, we obtain the final value for \(z\).
\[z = -13.5 + 9.5 + 11 = 7\]
So when \(x = 2.7\) and \(y = 1.9\), the value of \(z\) is 7.
Answer:
(a) The change in \(z\) is 1.5
(b) The value of \(z\) when \(x=2.7\) and \(y=1.9\) is 7.
Key Concepts
Partial DerivativesSlope of a FunctionMultivariable Calculus ApplicationsLinear Approximation
Partial Derivatives
Partial derivatives are fundamental tools in multivariable calculus. They represent the rate at which a function changes concerning one of its variables while keeping the other variables constant. Imagine slicing a multi-dimensional surface along one axis and examining the slope on just that slice.
For instance, if we have a function denoted as \( z(x, y) \), the partial derivative with respect to \( x \) is represented as \( \frac{\partial z}{\partial x} \). In the given exercise, the function has been described to have a slope of -5 in the \( x \)-direction. This implies that the partial derivative of \( z \) with respect to \( x \) is -5, capturing how \( z \) decreases as \( x \) increases. The concept of partial derivatives enables us to understand the individual effects of each variable on the outcome.
For instance, if we have a function denoted as \( z(x, y) \), the partial derivative with respect to \( x \) is represented as \( \frac{\partial z}{\partial x} \). In the given exercise, the function has been described to have a slope of -5 in the \( x \)-direction. This implies that the partial derivative of \( z \) with respect to \( x \) is -5, capturing how \( z \) decreases as \( x \) increases. The concept of partial derivatives enables us to understand the individual effects of each variable on the outcome.
Slope of a Function
The slope of a function in single-variable calculus is the rate of change of the function's value as its input changes. It's what you observe on a graph as the 'steepness' of the line. However, when dealing with functions of multiple variables, the 'slope' becomes a multi-dimensional concept.
In our exercise, we have a linear function of two variables, \( x \) and \( y \), with distinct slopes in the directions of each variable. The slopes given are the gradients along the axes - for every unit increase in \( x \), \( z \) changes by -5, and for every unit increase in \( y \), \( z \) changes by +5. These are intuitive measures akin to partial derivatives, enabling us to predict the function's behavior in response to varying each independent variable.
In our exercise, we have a linear function of two variables, \( x \) and \( y \), with distinct slopes in the directions of each variable. The slopes given are the gradients along the axes - for every unit increase in \( x \), \( z \) changes by -5, and for every unit increase in \( y \), \( z \) changes by +5. These are intuitive measures akin to partial derivatives, enabling us to predict the function's behavior in response to varying each independent variable.
Multivariable Calculus Applications
Multivariable calculus has wide-ranging applications in multiple fields such as physics, engineering, economics, and beyond. It helps us describe and analyze phenomena where several factors influence the outcome. In engineering, for example, it can model how different forces affect an object's motion.
In the context of our exercise, understanding how a change in two directions, \( x \) and \( y \), influences \( z \) could be akin to predicting weather changes based on temperature and humidity or optimizing a profit function based on different levels of product pricing and marketing spend. The ease with which multivariable calculus dissects complex relationships is what makes it so valuable in practical applications.
In the context of our exercise, understanding how a change in two directions, \( x \) and \( y \), influences \( z \) could be akin to predicting weather changes based on temperature and humidity or optimizing a profit function based on different levels of product pricing and marketing spend. The ease with which multivariable calculus dissects complex relationships is what makes it so valuable in practical applications.
Linear Approximation
Linear approximation is a technique to estimate the value of a function based on its behavior near a point. When a function is differentiable, it is closely captured by its tangent line or plane around that point. This is particularly useful when working with complex functions where exact values are hard to compute.
In our textbook solution, linear approximation allows us to estimate the change in \( z \) by simply adding the impacts of changes in \( x \) and \( y \), based on their slopes. This approach is applied when calculating \( z \) for a small change in \( x \) from 3 to 2.7 and in \( y \) from 2 to 1.9. The exercise shows the real-world utility of linear approximation in making quick and efficient predictions without the need for complex calculations.
In our textbook solution, linear approximation allows us to estimate the change in \( z \) by simply adding the impacts of changes in \( x \) and \( y \), based on their slopes. This approach is applied when calculating \( z \) for a small change in \( x \) from 3 to 2.7 and in \( y \) from 2 to 1.9. The exercise shows the real-world utility of linear approximation in making quick and efficient predictions without the need for complex calculations.
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