Solving PV & Tangent Line Equations in Calculus

  • Thread starter cscott
  • Start date
In summary, the problem is asking to find the rate at which the volume is decreasing at an instant where the volume is 480 cm^3, the pressure is 160 kPa, and the pressure is increasing at a rate of 15 kPa/min. To solve this, we use the equation PV=C (Boyle's Law) and plug in the known values to solve for the rate of change of volume. For the second part, we must find the equations of two lines that pass through the point (2,3) and are tangent to the parabola y=x^2+x. This can be done by setting the line equations equal to the parabola equation and solving for x and m. It is important to note
  • #1
cscott
782
1
PV = C (Boyle's Law)

At a certain instant, the volume is 480 cm^3, the pressure is 160 kPa, and the pressure is increasing at a rate of 15 kPa/min. At what rate is the volume decreasing at this instant?

----

Find the equations of both lines that pass through the point (2, 3) and are tangent to the parabola y = x^2 + x.



I don't really know where to start with these...
 
Physics news on Phys.org
  • #2
[tex]PV=C\Rightarrow\frac{d}{dt}(PV)=\frac{d}{dt}C\Rightarrow \frac{dP}{dt}V+P\frac{dV}{dt}=0[/tex]

then plug-in the known values of P,V, and dP/dt to solve for dV/dt.
 
  • #3
Any line through (2, 3) can be written as y= m(x- 2)+ 3 for some m.

If (x,y) is a point where that line intersects the parabola y= x2+ x, then we must have m(x-2)+ 3= x2+ x. If, in addition, the line is tangent to the parabola there, we must have
m= 2x+ 1. Solve those two equations for x and m.

I've edited this: before I had m= 2x- 1. Obviously, the derivative of x2+ x is 2x+ 1, not 2x- 1.
 
Last edited by a moderator:
  • #4
Thank you.
 

What is the purpose of solving PV & Tangent Line Equations in Calculus?

Solving PV & Tangent Line Equations in Calculus allows us to find the instantaneous rate of change, or slope, of a curve at a specific point. This is useful in many real-world applications, such as determining the velocity of an object at a given time.

What is the process for solving PV & Tangent Line Equations?

The first step is to find the derivative of the function, which represents the slope of the tangent line. Then, we substitute the x-value of the point we are interested in into the derivative to find the slope at that point. Finally, we use the slope and the point to write the equation of the tangent line using the point-slope form.

How do PV & Tangent Line Equations relate to the concept of limits in Calculus?

The process of finding the derivative and solving for the tangent line involves using the concept of limits. As the distance between two points on a curve gets closer and closer to 0, the slope of the secant line approaches the slope of the tangent line at that point. This is known as the limit of the slope, or the derivative, and is essential in solving PV & Tangent Line Equations.

Are there any common mistakes to avoid when solving PV & Tangent Line Equations?

One common mistake is mixing up the x and y values when substituting into the derivative. It is important to remember that the x-value represents the point we are interested in, and the y-value represents the corresponding output of the function. Another mistake is forgetting to include the units in the final answer when dealing with real-world applications.

How can understanding PV & Tangent Line Equations help in other areas of Calculus?

Solving PV & Tangent Line Equations is an essential skill in Calculus that can be applied to other topics, such as optimization and related rates. It also helps in understanding the behavior of functions and their graphs, which is crucial in topics like integration and curve sketching.

Similar threads

  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
4K
  • Calculus and Beyond Homework Help
Replies
2
Views
564
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
946
Replies
2
Views
877
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
  • Calculus and Beyond Homework Help
Replies
8
Views
348
Replies
2
Views
504
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Back
Top