Finding the Equation of Variation

Oct 13, 2025 by recoveryshake.com 38 views

Introduction

In mathematics, variation is a concept that describes the relationship between two or more variables. When we say that one quantity varies jointly with two or more other quantities, it means that the first quantity is directly proportional to the product of the other quantities. In this article, we will explore the concept of joint variation and how to find the equation of variation when given specific values.

What is Joint Variation?

Joint variation is a type of variation where one quantity varies directly with the product of two or more other quantities. This means that as the product of the other quantities increases, the first quantity also increases at a constant rate. Mathematically, joint variation can be represented as:

s = k(tu)

where $$s$$ is the quantity that varies jointly with $$t$$ and $$u$$, $$k$$ is the constant of variation, and $$t$$ and $$u$$ are the quantities that vary jointly with $$s$$.

Finding the Equation of Variation

To find the equation of variation, we need to determine the constant of variation $$k$$. We are given that when $$t = -4$$ and $$u = -2$$, $$s = -40$$. We can use this information to find the value of $$k$$.

Step 1: Substitute the given values into the equation

We know that $$s = k(tu)$$. Substituting the given values, we get:

-40 = k(-4)(-2)

Step 2: Simplify the equation

Simplifying the equation, we get:

-40 = k(8)

Step 3: Solve for $$k$$

To solve for $$k$$, we can divide both sides of the equation by 8:

k = \frac{-40}{8}

k = -5

Step 4: Write the equation of variation

Now that we have found the value of $$k$$, we can write the equation of variation:

s = -5(tu)

Conclusion

In this article, we have explored the concept of joint variation and how to find the equation of variation when given specific values. We have used the given values to determine the constant of variation $$k$$ and have written the equation of variation in the form $$s = k(tu)$$. This equation can be used to predict the value of $$s$$ for different values of $$t$$ and $$u$$.

Example

Let's use the equation of variation to find the value of $$s$$ when $$t = 2$$ and $$u = 3$$.

s = -5(2)(3)

s = -30

This shows that when $$t = 2$$ and $$u = 3$$, the value of $$s$$ is $$-30$$.

Applications of Joint Variation

Joint variation has many applications in real-world problems. For example, in physics, the force of gravity varies jointly with the mass of two objects and the distance between them. In economics, the demand for a product varies jointly with the price of the product and the income of the consumer. In engineering, the stress on a material varies jointly with the force applied to the material and the cross-sectional area of the material.

Conclusion

In conclusion, joint variation is a powerful tool for modeling real-world problems. By understanding the concept of joint variation and how to find the equation of variation, we can make predictions and solve problems in a variety of fields.

Introduction

In our previous article, we explored the concept of joint variation and how to find the equation of variation when given specific values. In this article, we will answer some frequently asked questions about joint variation and provide additional examples to help solidify your understanding of the concept.

Q&A

Q: What is the difference between joint variation and inverse variation?

A: Joint variation is a type of variation where one quantity varies directly with the product of two or more other quantities. Inverse variation, on the other hand, is a type of variation where one quantity varies inversely with the product of two or more other quantities. In other words, as the product of the other quantities increases, the first quantity decreases at a constant rate.

Q: How do I determine whether a problem involves joint variation or inverse variation?

A: To determine whether a problem involves joint variation or inverse variation, look for the word "jointly" or "directly" in the problem statement. If the problem states that one quantity varies jointly with the product of two or more other quantities, then it involves joint variation. If the problem states that one quantity varies inversely with the product of two or more other quantities, then it involves inverse variation.

Q: Can I use the equation of variation to solve problems that involve multiple variables?

A: Yes, you can use the equation of variation to solve problems that involve multiple variables. However, you need to make sure that the equation of variation is in the correct form. For example, if the equation of variation is $$s = k(tu)$$, then you can use it to solve problems that involve the variables $$s$$, $$t$$, and $$u$$.

Q: How do I find the value of $$k$$ in the equation of variation?

A: To find the value of $$k$$ in the equation of variation, you need to substitute the given values into the equation and solve for $$k$$. For example, if the equation of variation is $$s = k(tu)$$ and you are given that $$s = 10$$, $$t = 2$$, and $$u = 3$$, then you can substitute these values into the equation and solve for $$k$$.

Q: Can I use the equation of variation to make predictions about real-world problems?

A: Yes, you can use the equation of variation to make predictions about real-world problems. For example, if you are given the equation of variation $$s = k(tu)$$ and you know the values of $$t$$ and $$u$$, then you can use the equation to predict the value of $$s$$.

Q: How do I apply the concept of joint variation to real-world problems?

A: To apply the concept of joint variation to real-world problems, you need to identify the variables involved in the problem and determine whether they vary jointly or inversely. Then, you can use the equation of variation to make predictions or solve the problem.

Examples

Example 1: Joint Variation in Physics

In physics, the force of gravity varies jointly with the mass of two objects and the distance between them. If the force of gravity is 10 N when the mass of the first object is 5 kg, the mass of the second object is 3 kg, and the distance between them is 2 m, then what is the force of gravity when the mass of the first object is 10 kg, the mass of the second object is 6 kg, and the distance between them is 3 m?

Solution

We can use the equation of variation $$F = k(m_1m_2)$$ to solve this problem. We are given that $$F = 10$$ N, $$m_1 = 5$$ kg, $$m_2 = 3$$ kg, and $$d = 2$$ m. We can substitute these values into the equation and solve for $$k$$:

10 = k(5)(3)

k = \frac{10}{15}

k = \frac{2}{3}

Now that we have found the value of $$k$$, we can use the equation of variation to make predictions about the force of gravity:

F = \frac{2}{3}(10)(6)

100 = k(10)(20) </span></p> <p class='katex-block'><span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>k</mi><mo>=</mo><mfrac><mn>100</mn><mn>200</mn></mfrac></mrow><annotation encoding="application/x-tex">k = \frac{100}{200} </annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.6944em;"></span><span class="mord mathnormal" style="margin-right:0.03148em;">k</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:2.0074em;vertical-align:-0.686em;"></span><span class="mord"><span class="mopen nulldelimiter"></span><span class="mfrac"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:1.3214em;"><span style="top:-2.314em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">200</span></span></span><span style="top:-3.23em;"><span class="pstrut" style="height:3em;"></span><span class="frac-line" style="border-bottom-width:0.04em;"></span></span><span style="top:-3.677em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">100</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.686em;"><span></span></span></span></span></span><span class="mclose nulldelimiter"></span></span></span></span></span></span></p> <p class='katex-block'><span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>k</mi><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac></mrow><annotation encoding="application/x-tex">k = \frac{1}{2} </annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.6944em;"></span><span class="mord mathnormal" style="margin-right:0.03148em;">k</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:2.0074em;vertical-align:-0.686em;"></span><span class="mord"><span class="mopen nulldelimiter"></span><span class="mfrac"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:1.3214em;"><span style="top:-2.314em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">2</span></span></span><span style="top:-3.23em;"><span class="pstrut" style="height:3em;"></span><span class="frac-line" style="border-bottom-width:0.04em;"></span></span><span style="top:-3.677em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">1</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.686em;"><span></span></span></span></span></span><span class="mclose nulldelimiter"></span></span></span></span></span></span></p> <p>Now that we have found the value of $$k$$, we can use the equation of variation to make predictions about the demand for the product:</p> <p class='katex-block'><span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>D</mi><mo>=</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mo stretchy="false">(</mo><mn>15</mn><mo stretchy="false">)</mo><mo stretchy="false">(</mo><mn>30</mn><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">D = \frac{1}{2}(15)(30) </annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.6833em;"></span><span class="mord mathnormal" style="margin-right:0.02778em;">D</span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:2.0074em;vertical-align:-0.686em;"></span><span class="mord"><span class="mopen nulldelimiter"></span><span class="mfrac"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:1.3214em;"><span style="top:-2.314em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">2</span></span></span><span style="top:-3.23em;"><span class="pstrut" style="height:3em;"></span><span class="frac-line" style="border-bottom-width:0.04em;"></span></span><span style="top:-3.677em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">1</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height:0.686em;"><span></span></span></span></span></span><span class="mclose nulldelimiter"></span></span><span class="mopen">(</span><span class="mord">15</span><span class="mclose">)</span><span class="mopen">(</span><span class="mord">30</span><span class="mclose">)</span></span></span></span></span></p> <p class='katex-block'><span class="katex-error" title="ParseError: KaTeX parse error: Can&#x27;t use function &#x27;$&#x27; in math mode at position 8: D = 225$̲$ units This s…" style="color:#cc0000">D = 225$$ units This shows that when the price of the product is $15 and the income of the consumer is $30, the demand for the product is 225 units. ## Conclusion In conclusion, joint variation is a powerful tool for modeling real-world problems. By understanding the concept of joint variation and how to find the equation of variation, you can make predictions and solve problems in a variety of fields. We hope that this Q&amp;A article has helped to clarify any questions you may have had about joint variation and has provided additional examples to help solidify your understanding of the concept.</span></p>