1. Related Rates problem

A point moves along the curve y= 2x^2 - 1 in such a way that the y value is decreasing at the rate of 2 units per second. At what rate is x changing when x = -3/2 ?

Not really looking to solve just an explanation on the process of how to get to the solution. Thanks!!

2. Originally Posted by Hibijibi
A point moves along the curve y= 2x^2 - 1 in such a way that the y value is decreasing at the rate of 2 units per second. At what rate is x changing when x = -3/2 ?

Not really looking to solve just an explanation on the process of how to get to the solution. Thanks!!

Well this is pretty general so here it goes

$\displaystyle y=f(x)$ taking the derivative we get

$\displaystyle \frac{dy}{dx}=f'(x)$ abusing notation we multiply by dx to get

$\displaystyle dy=f'(x)dx$ plug in what we know and solve for dx.

Hint: member dx and dy represent the rates of change of x and y.

Good luck.

Originally Posted by Hibijibi
A point moves along the curve y= 2x^2 - 1 in such a way that the y value is decreasing at the rate of 2 units per second. At what rate is x changing when x = -3/2 ?

Not really looking to solve just an explanation on the process of how to get to the solution. Thanks!!
what you do is that you take the derivative of your function, your new function is
$\displaystyle \frac{dy}{dx}=4x$, then
$\displaystyle dy=4xdx$
you know what $\displaystyle dy$ is, it is -2, the rate of change of y with respect to x. x is -3/2, and dx is what you are looking for, the rate of x.

4. Originally Posted by Mr. Engineer
what you do is that you take the derivative of your function, your new function is
$\displaystyle \frac{dy}{dx}=4xdx$
you know what $\displaystyle \frac{dy}{dx}$ is, it is -2, the rate of change of y with respect to x. x is -3/2, and dx is what you are looking for, the rate of x.
Ah, thank you! Very clear answer. Also I got 1/3, and how do I know if it's increasing or decreasing?

5. no prob

no problem

6. Based on what how would I know whether that value is an increasing value or decreasing?

7. Originally Posted by Hibijibi
Ah, thank you! Very clear answer. Also I got 1/3, and how do I know if it's increasing or decreasing?
since you said in your problem the value of the rate of y was -2, the graph of the function would be decreasing.
you could try the first derivative test to see at which intervals the function would be increasing or decreasing, but from the values you have, as x gets bigger, y gets smaller

8. A spherical balloon filled with helium at the rate of 2 in^2 / sec. Find the rate of change of the radius at the moment when the radius is 1 inch. Any advice? (I'm working on the final exam review sheet my professor gave me so lots of questions ^_^)

9. 2nd

using properties from parabolas, the parabola opens up and its vertex is in the center. at -3/2, you have a decreasing slope, until you hit the vertex at (0,-1) where the slop is zero, then it becomes positive. just another way to look at it

Originally Posted by Hibijibi
A spherical balloon filled with helium at the rate of 2 in^2 / sec. Find the rate of change of the radius at the moment when the radius is 1 inch. Any advice? (I'm working on the final exam review sheet my professor gave me so lots of questions ^_^)
for this and other problems with related rates, make sure you know the formulas. for a sphere $\displaystyle V=\frac{4}{3}\pi r^3$
when you differentiate the function you get
$\displaystyle \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$
you know that $\displaystyle \frac{dV}{dt}$ is 2in^2/sec, volume with respect to time. radius is 1, just solve for dr/dt, the rate of change of the radius with respect to time. with these problems, its just taking your derivative, realize what is changing and plug in what you know
make sure you use the thank button

11. A maximization problem. A square is to be cut from each corner of a piece of paper which is 8 cm by 10 cm, and the sides are to be folded up to create an open box. What should the side of the square be for maximum volume? What's the basic process for maximizing problems?

Originally Posted by Hibijibi
A maximization problem. A square is to be cut from each corner of a piece of paper which is 8 cm by 10 cm, and the sides are to be folded up to create an open box. What should the side of the square be for maximum volume? What's the basic process for maximizing problems?
find a way to relate length width and height...if u draw a picture you will see that the length is 10-x and the width is 8-x for the base of the box. the height is simply x, so V=(10-x)(8-x)(x). you can multiply, then differentiate the function. you set it equal to zero and you will get values for x. you must use the 1st or sometimes 2nd derivative test to see whether the values you got are maximums or minimums.

13. good night

i wish i can help you longer, but i have to get up early tomorrow, good night and good luck on your exam

14. Originally Posted by Mr. Engineer
i wish i can help you longer, but i have to get up early tomorrow, good night and good luck on your exam
My Exam still isn't for another 2 days, and thank you. Also I think it would be 10-2x and 8-2x correct? Since each corner on both sides you're taking out an x?