Word problem using derivatives - struggling with it

In summary, the question is asking how fast the pressure is increasing when the volume is decreasing at a rate of 10cm^3/minute, given an initial pressure of 100g/cm^2 and initial volume of 20 cm^3. The equation for this problem is PV=c, also known as Boyle's law. Using the product rule, you can find the derivative of PV and solve for dP/dt. Then, substitute the given values to find the rate of pressure increase.
  • #1
meredith
16
0

Homework Statement



the equation is PV = c; p = pressure, v = volume, c=constant. (also known as Boyle's law)
the question: if volume is decreasing at a rate of 10cm^3/minute, how fast is the pressure increasing when the pressure is 100g/cm^2 and volume is 20 cm^3


Homework Equations



none


The Attempt at a Solution


dv/dt = -10cm^3/min
dp/dt = ?
equation: PV=c
derivative:
dp/dt x dv/dt = 0
dp/dt = -dv/dt

but i know what I am doing isn't right. can anyone help me? THANKS!
 
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  • #2
meredith said:

Homework Statement



the equation is PV = c; p = pressure, v = volume, c=constant. (also known as Boyle's law)
the question: if volume is decreasing at a rate of 10cm^3/minute, how fast is the pressure increasing when the pressure is 100g/cm^2 and volume is 20 cm^3


Homework Equations



none


The Attempt at a Solution


dv/dt = -10cm^3/min
dp/dt = ?
equation: PV=c
derivative:
dp/dt x dv/dt = 0
The above isn't right. It is not true that d/dt(PV) = dP/dt * dV/dt. You need to use the product rule.
meredith said:
dp/dt = -dv/dt

but i know what I am doing isn't right. can anyone help me? THANKS!

After you differentiate PV, solve algebraically for dP/dt, and then substitute the values you have.
 

FAQ: Word problem using derivatives - struggling with it

1. What is a derivative in the context of a word problem?

A derivative is a mathematical concept that represents the rate of change of a function at a specific point. In the context of a word problem, it can be thought of as the instantaneous rate of change of a quantity with respect to another quantity.

2. How do I identify when to use derivatives in a word problem?

Derivatives are commonly used in word problems involving rates of change, optimization, and related rates. Look for key words or phrases such as "rate," "change," "maximize," or "minimize" to determine if a derivative is needed in the problem.

3. What are some common techniques for solving word problems using derivatives?

Some common techniques for solving word problems using derivatives include setting up an equation, finding the derivative of the equation, and using algebraic manipulation or calculus rules to solve for the desired quantity. It is also helpful to draw a diagram or graph to visualize the problem.

4. How can I check my answer in a word problem using derivatives?

To check your answer, you can use basic algebraic techniques to plug your solution back into the original equation and see if it satisfies the given conditions. You can also use a graphing calculator or online tool to graph the original equation and your solution and see if they intersect at the desired point.

5. What are some common mistakes to avoid when solving word problems using derivatives?

Some common mistakes to avoid when solving word problems using derivatives include forgetting to label units, not clearly defining variables, and using the wrong derivative formula or calculus rule. It is important to carefully read the problem and double check your work to avoid errors.

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