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Notes: Direct and Inverse Variations #2 (ppt)

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Page 1: Notes: Direct and Inverse Variations #2 (ppt)
Page 2: Notes: Direct and Inverse Variations #2 (ppt)
Page 3: Notes: Direct and Inverse Variations #2 (ppt)

The general equation for DIRECT VARIATION is y =kx with k≠0.

k is called the constant of variation.

We will do an example together.

Page 4: Notes: Direct and Inverse Variations #2 (ppt)

If y varies directly as x, and y=24 and x=3 find: (a) the constant of variation

(b) Find y when x=2

(a) Find the constant of variationy=kx Write the general equation

24=k⋅3 Substitute

k=8

Page 5: Notes: Direct and Inverse Variations #2 (ppt)

(b) Find y when x=2

First we find the constant of variation, which was k=8

Now we substitute into y=kx.

y=kxy=8⋅2

y=16

Page 6: Notes: Direct and Inverse Variations #2 (ppt)

Another method of solving direct variation problems is to use proportions.

If y1 =kx1, then k=y1x1

and

If y2 =kx2, then k=y2x2

Therefore...

Page 7: Notes: Direct and Inverse Variations #2 (ppt)

y1x1

=y2x2

So lets look at a problem that can by solved by either of these two methods.

Page 8: Notes: Direct and Inverse Variations #2 (ppt)

If y varies directly as x and y=6 when x=5, then find y when x=15.

Proportion Method:65 = y

15 Let x1 =5, y1 =6, x2 =15, y2 =y

5y=90y=18

Page 9: Notes: Direct and Inverse Variations #2 (ppt)

Now lets solve using the equation.

y=kx6=k⋅5k=6

5

y=kx

y=65⋅15

y=18

Either method gives the correct answer, choose the easiest for you.

Page 10: Notes: Direct and Inverse Variations #2 (ppt)

Now you do one on your own.

y varies directly as x, and x=8 when y=9. Find y when x=12.

Answer: 13.5

Page 11: Notes: Direct and Inverse Variations #2 (ppt)

What does the graph y=kx look like?A straight line with a y-intercept of 0.

5

-5

-10 10

f x( ) = 3⋅x

Page 12: Notes: Direct and Inverse Variations #2 (ppt)

Looking at the graph, what is the slope of the line?

Answer: 3

Looking at the equation, what is the constant of variation?

Answer: 3The constant of variation and the slope are the same!!!!

Page 13: Notes: Direct and Inverse Variations #2 (ppt)

We will apply what we know and try this problem.

According to Hook’s Law, the force F required to stretch a spring x units beyond its natural length varies directly as x. A force of 30 pounds stretches a certain spring 5 inches. Find how far the spring is stretched by a 50 pound weight.

Page 14: Notes: Direct and Inverse Variations #2 (ppt)

F1x1

=F2x2

Set up a proportion

305 =50

xSubstitute

30x=250x=81

3 inches

Page 15: Notes: Direct and Inverse Variations #2 (ppt)

Now try this problem.

Use Hook’s Law to find how many pounds of force are needed to stretch a spring 15 inches if it takes 18 pounds to stretch it 13.5 inches.

Answer: 20 pounds

Page 16: Notes: Direct and Inverse Variations #2 (ppt)

Inverse Variation

y varies inversely as x if k≠0such that xy=k or y=k

x

Just as with direct variation, a proportion can be set up solve problems of indirect variation.

Page 17: Notes: Direct and Inverse Variations #2 (ppt)

x1y2

=x2y1

A general form of the proportion

Lets do an example that can be solved by using the equation and the proportion.

Page 18: Notes: Direct and Inverse Variations #2 (ppt)

Find y when x=15, if y varies inversely as x and x=10 when y=12

Solve by equation:xy=k

10⋅12=k120=k

xy=k15⋅y=120

y=8

Page 19: Notes: Direct and Inverse Variations #2 (ppt)

Solve by proportion:x1y2

=x2y1

1512=10

y

15y=120

y=8

Page 20: Notes: Direct and Inverse Variations #2 (ppt)

Solve this problem using either method.

Find x when y=27, if y varies inversely as x and x=9 when y=45.

Answer: 15

Page 21: Notes: Direct and Inverse Variations #2 (ppt)

Lets apply what we have learned.

The pressure P of a compressed gas is inversely proportional to its volume V according to Boyle’s Law. A pressure of 40 pounds per square inch is created by 600 cubic inches of a certain gas. Find the pressure when the gas is compressed to 200 cubic inches.

Page 22: Notes: Direct and Inverse Variations #2 (ppt)

Step #1: Set up a proportion.

x1y2

=x2y1

40200= x

600

200x=24000

x=120 pounds/ in2

Page 23: Notes: Direct and Inverse Variations #2 (ppt)

Now try this one on your own.

A pressure of 20 pounds per inch squared is exerted by 400 inches cubed of a certain gas. Use Boyle’s Law to find the pressure of the gas when it is compressed to a volume of 100 inches cubed.

Answer: 80 pounds/ in2

Page 24: Notes: Direct and Inverse Variations #2 (ppt)

What does the graph of xy=k look like? Let k=5 and graph.

6

4

2

-2

-4

-6

-10 -5 5 10

f x( ) = 5

x

Page 25: Notes: Direct and Inverse Variations #2 (ppt)

This is a graph of a hyperbola.

Notice: That in the graph, as the x values increase the y values decrease.

alsoAs the x values decrease the y values increase.