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For Activities 1 through 6 a. Write the mathematical notation for the partial rate-ofchange function needed to answer the question posed. b. Write the units of measure for that rate-of-change function. \(p(k, c)\) dollars is the price of a diamond that weighs \(k\) karats and is of color code \(c\). How quickly is the price of a diamond changing with respect to weight?

Short Answer

Expert verified
The partial derivative is \( \frac{\partial p}{\partial k} \) with units dollars per karat.

Step by step solution

01

Identify the Variables

Here we first identify the independent variables on which the function depends. The function given is \( p(k, c) \), where \( k \) represents weight in karats and \( c \) is the color code of the diamond.
02

Determine the Variable of Change

Since we are asked to determine how quickly the price changes with respect to weight, we focus on the variable \( k \), keeping the color code \( c \) constant.
03

Write the Partial Derivative

The partial rate-of-change of the price with respect to the weight is represented by the partial derivative \( \frac{\partial p}{\partial k} \). This notation indicates the rate at which the price changes as the weight changes, while color code stays constant.
04

Determine Units of Measure

In this context, the units of \( \frac{\partial p}{\partial k} \) would be in dollars per karat, as \( p \) is measured in dollars and \( k \) is measured in karats.

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Key Concepts

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

Rate of Change
Understanding the concept of rate of change is crucial when dealing with changes in measurable quantities. In mathematics, the rate of change tells us how one quantity changes in relation to another quantity. Consider the price of a diamond, which can depend on various factors such as weight and color. When we talk about how quickly the price changes with respect to weight, we are referring to its rate of change regarding weight. This is particularly useful to know if you're interested in identifying how much more expensive a diamond gets as it gains weight.

In our particular example, we are concerned with the price of the diamond as its weight (measured in karats) changes. Hence, the rate of change here is represented by the partial derivative, which helps isolate the effect of varying weight while keeping other variables constant. This makes it easier to understand the impact of each variable separately.
Units of Measure
Units of measure are essential in providing context and understanding for any rate of change. They tell us how we are measuring each variable involved. In our diamond price example, the function given is in dollars, expressed as \(p(k, c)\), and the weight is in karats, represented by \(k\).

When we calculate the partial derivative \( \frac{\partial p}{\partial k} \), we are essentially determining how many dollars the price rises for each additional karat of weight. Thus, the units for the rate of change, \( \frac{\partial p}{\partial k} \), become dollars per karat. This gives a clear picture of how significant an increase in weight affects the price.

Understanding these units of measure helps us interpret the rate of change more meaningfully, enabling better decision-making whether in purchasing or appraising diamonds.
Mathematical Notation
Mathematical notation is a standardized way of representing mathematical concepts with symbols and formulas, making it easier to read and understand complex information. For partial derivatives, the notation \( \frac{\partial p}{\partial k} \) is used to denote the partial derivative of a function \(p\) with respect to the variable \(k\).

This notation is particularly advantageous because:
  • It clearly communicates that the change is being observed in one variable while the other is held constant.
  • It prevents confusion between different rates of change, especially when multiple variables are in play, like our diamond's weight and color.

By employing consistent mathematical notation, we can efficiently convey complex concepts and ensure precise communication in mathematics and related disciplines.

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Most popular questions from this chapter

Sunflower Pigment A process to extract pectin and pigment from sunflower heads involves washing the sunflower heads in heated water. It has been shown that the percentage of pigment that can be removed from a sunflower head by washing for 20 minutes can be modeled as $$ p(r, t)=306.761-9.6544 t+1.9836 r $$ \(+0.07368 t^{2}-0.02958 r^{2}\) percent where \(r\) milliliters of \(t^{\circ} \mathrm{C}\) water is used for each gram of sunflower heads. (Source: X. Q. Shi et al., "Optimizing Water Washing Process for Sunflower Heads Before Pectin Extraction," Journal of Food Science, vol. \(61,\) no. \(3(1996),\) pp. \(608-612)\) a. Sketch the \(53 \%\) contour curve for \(20 \leq r \leq 45\) and \(85 \leq t \leq 88\) b. Draw the line(s) tangent to the \(53 \%\) contour curve when the temperature is \(86.5^{\circ} \mathrm{C}\). c. Write a formula for the rate of change of temperature with respect to a change in the amount of water used when \(53 \%\) of the pigment is removed. d. Use the formula from part \(c\) to calculate the slope(s) of the tangent line(s) drawn in part \(b\).

For Activities 17 through \(22,\) write the first and second partial derivatives. $$ f(x, y)=2 x y+8 x^{2} y^{3}+5 e^{2 y}+10 $$

Future Value The value \(F(P, r)\) of an investment of \(P\) dollars after 2 years in an account with annual percentage yield \(100 r \%\) is given by the function \(F(P, r)=P(1+r)^{2}\) dollars. a. Write the first partial derivatives of \(F\). b. Write each of the second partial derivative formulas and interpret them for \(P=10,000\) and \(r=0.09 .\)

For Activities 9 through \(16,\) write formulas for the indicated partial derivatives for each of the multivariable functions. \(f(x, y)=3 x^{2}+5 x y+2 y^{3}\) a. \(\frac{\partial f}{\partial x}\) b. \(\frac{\partial f}{\partial y}\) c. \(\left.\frac{\partial f}{\partial x}\right|_{y=7}\)

Draw the contour curves \(f(L, K)=50, f(L, K)=55,\) and \(f(L, K)=60\) for the function \(f(L, K)=38.9 L^{0.5} K^{0.5}\) for \(0.25 \leq L \leq 1\).

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