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Let \(z=f(x, y) .\) Suppose \(\frac{\partial}{\partial y}(1,2)=-5 .\) Interpret this partial derivative as an instantaneous rate of change.

Short Answer

Expert verified
The partial derivative \(-5\) indicates that \(z\) decreases by 5 units for every unit increase in \(y\) at \((1, 2)\), keeping \(x\) constant.

Step by step solution

01

Understanding the Partial Derivative

The partial derivative \(\frac{\partial z}{\partial y}(x, y)\) represents how the function \(z = f(x, y)\) changes instantaneously with respect to \(y\), while keeping \(x\) constant. Here, we evaluate this at the point \((x, y) = (1, 2)\).
02

Given Information

We are given that \(\frac{\partial}{\partial y}(1,2) = -5\). This means that the partial derivative of \(z\) with respect to \(y\) at the point \((1, 2)\) is \(-5\).
03

Interpretation of the Partial Derivative

The value of \(-5\) tells us that at the point \((1, 2)\), for an infinitesimally small increase in \(y\), while \(x\) is kept constant, \(z\) decreases by 5 times that small increase. This reflects how \(z\) changes instantaneously at the given point as \(y\) changes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Instantaneous Rate of Change
The concept of the instantaneous rate of change is central to calculus and is used to describe how a quantity varies at a specific point. When dealing with a function of two variables, like our function \( z = f(x, y) \), the partial derivative \( \frac{\partial z}{\partial y} \) measures how fast \( z \) changes as \( y \) changes slightly, while keeping \( x \) constant.
In simpler terms, if you've ever looked at a speedometer and wondered what it tells you at a single moment, that's an instantaneous rate—how fast you are going right now, not an average over a trip. Similarly, here, \( -5 \) represents that if \( y \) changes ever so slightly from \( 2 \) by a little increase, \( z \) will change (specifically decrease) by 5 times that tiny change.
It highlights how responsive our function \( f(x, y) \) is to changes in \( y \) at the specific point we are examining.
Function of Two Variables
When considering a function of two variables, say \( z = f(x, y) \), we visualize it as a surface over a plane. This can be challenging, but it is an extension of functions with one variable.
Each pair of \( (x, y) \) returns a unique \( z \), forming a surface; imagine a landscape, with hills and valleys, that changes as you move around. As both \( x \) and \( y \) vary, the output \( z \) moves up or down over this surface.
In real-world applications, functions of two variables can represent a variety of phenomena: heat over an area, stock prices influenced by time and economic conditions, or even the topography of a geographic region.
  • This multi-dimensional approach gives us greater control and insight into complex systems.
  • They enable predictions and optimizations by understanding how changes in one variable can affect outcomes.
Interpretation of Derivatives
Derivatives offer a powerful tool to interpret how functions behave. When dealing with a partial derivative like \( \frac{\partial z}{\partial y} \), it tells us the specific rate at which the function \( z \) changes as just one input, \( y \), varies while others remain constant.
Understanding this is crucial, as it allows us to isolate the effect of one variable in multifaceted problems. This is why derivatives are often seen in physics, engineering, and economics, where systems may have multiple inputs affecting outputs.
In our case, the derivative value of \( -5 \) at the point \( (1, 2) \) is interpreted as: if \( y \) increases slightly at the point where \( x = 1 \) and \( y = 2 \), the function \( z \) will decrease rapidly at 5 times the rate of that change in \( y \).
  • This immediate responsiveness is crucial to modeling real-world scenarios accurately, allowing for both predictions and adjustments.
  • Such interpretations assist in understanding system sensitivity, crucial for optimization and control.

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Most popular questions from this chapter

Integrating first with respect to one of \(x\) or \(y\) is difficult, while integrating first with respect to the other is not. Find \(\iint_{D} f(x, y) d A\) the easier way. \(f(x, y)=1, D\) is the region in the first quadrant bounded by the curves \(x=y-2, x=y^{3}, x=0,\) and \(x=1\).

The Production Function Suppose a production function is in the form of a Cobb-Douglas production function \(P=a x^{b} y^{1-b}\) and a cost function is \(C=p_{1} x+p_{2} y,\) where \(x\) is the number of items made at a cost of \(p_{1}\) and \(y\) is the number of items made at a cost of \(p_{2}\). Find the cost as a function of \(P,\) that is, find \(C=f(P)\). Hint: Solve for \(x\) in the production function, substitute this \(x\) in the cost function. The cost function then becomes a function of \(y .\) Differentiate with respect to \(y,\) set equal to zero, and solve for \(y .\) Use this to find \(x,\) substitute into the cost equation, and finally obtain $$ C=\left(\frac{1}{b}-1\right)^{b}\left(\frac{p_{1}}{p_{2}}\right)^{b}\left[\frac{b}{1-b} p_{2}+1\right] \frac{P}{a} $$.

Find the volume under the surface of the given function and over the indicated region. \(f(x, y)=1, D\) is the region in the first quadrant bounded by the curves \(y^{2}=x^{3}\) and \(y=x\).

Engineering Production Function An engineering production function in the natural gas transmission industry was given by Cullen \(^{20}\) as $$ Q=Q(H, d, L)=0.33 \frac{H^{0.27} d^{1.8}}{L^{0.36}} $$ where \(Q\) is the output in cubic feet of natural gas, \(H\) is the station horsepower, \(d\) is the inside diameter of the transmission line in inches, and \(L\) is the length of the pipeline. Find the three first-order partial derivatives of \(Q(H, d, L)\).

Agriculture Paris \(^{24}\) showed that an approximate relationship between yield \(y\) of corn and the amounts of nitrogen \(N\) and phosphorus \(P\) fertilizer used is given by $$ \begin{aligned} y=-0.075+0.584 N+0.664 P-& 0.158 N^{2} \\ &-0.18 P^{2}+0.081 P N\end{aligned} $$ where \(y, N,\) and \(P\) are in appropriate units. Find the number of units of nitrogen and phosphorus that will maximize the corn yield according to this model. Use your grapher to find the critical point.

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