Question

y varies jointly with p and q. When p = 5 and q = 2, y = 80. Find y when p = 4 and q = 5.
a, 20
b. 120
c. 160
d. 240

Answer 1

If y varies jointly with p and q, that means we can write
\[y = kpq\] where \(k\) is a proportionality constant. Plug in your known initial values of \(p,q,y\) and solve for \(k\). Update the equation to use that value, and evaluate it to find \(y\) with the final values of \(p,q\).

Direct variation of thing1 with thing2: thing1 = k * thing2
Indirect variation of thing1 with thing2: thing1 = k / thing2
Joint variation of thing1 with thing2 and thing3: thing1 = k * thing2 * thing3

and of course you can combine them:
joint variation of thing1 with thing2 and thing3 coupled with inverse variation with the square of thing4 : thing1 = k * thing2 * thing3 / (thing4)^2

Answer 2

y=160?

Answer 3

that's what I get. \[80 = k*(5)(2)\]\[k=8\]\[y=8*4*5= 160\]

Answer 4

Sweet. Thank you. Best response goes to you.

Answer 5

you're welcome. really, the only thing to remember here is the form of each kind of variation, and that there's always that constant to find (it may turn out to be 1, but usually isn't).

Answer 6

once you understand the concept, you'll start seeing that it appears everywhere. slope of a line — that's just the y value varying directly with the x value. physics is of course full of such equations.

Answer 7

Ugh, I'm taking Physics next year. Perhaps you could help me on one more question?

Answer 8

bring it on :-)

Answer 9

What is the equation of the line passing through (8, –6) and perpendicular to the line y = x – 2 in slope-intercept form?
a. y = –x + 2
b. y = x + 2
c. y = –x – 2
d. y = x – 2

Answer 10

brb - nature calls

Answer 11

lol

Answer 12

okay. here's the general drill: to find the slope of the first line, rearrange it into slope-intercept form, which is:
\[y=mx+b\]where \(m\) is slope, and \(b\) is y-intercept
Perpendicular lines have the property that the product of their slopes = -1 (unless one of them is parallel to the y-axis, in which case it has no slope, and then the other line would be parallel to the x-axis, giving you equations of the form y=constant, x=constant)

So, armed with the slope of the first line, we'll call it \(m_1\), we can find the slope of the second line \(m_2\) by doing \(m_2=-1/m_1\)

Now, we use the newly found slope (\(m_2)\) in the point-slope equation for a line with slope \(m\) passing through a known point \((x_0,y_0)\):
\[y-y_0=m(x-x_0)\]

(being sure to use the value of \(m_2\) as the slope) and rearranging into whatever form the problem requests (slope-intercept, standard, etc.)

Answer 13

give it a try, and I'll check your work.

Answer 14

You lost me... lol

Answer 15

Okay. you've got a line: \[y=x-2\]That's already in slope-intercept form:\[y = mx + b\]You don't have to do any rearranging, so just read off the values of \(m\) and \(b\). \[m=\]\[b=\]

Answer 16

well m is 1 and b is -2
is that correct?

Answer 17

Yes! So what is the slope of this line?

Answer 18

no deep thinking required here...m is the slope

Answer 19

Okay, if the slope of this line is 1, and the product of the slopes of perpendicular lines = -1, what is the slope of our new line going to be?

Answer 20

y=x-2?

Answer 21

no, the slope is going to be a number.

Answer 22

a number such that newslope * oldslope = -1

Answer 23

oldslope = 1
newslope * 1 = -1
newslope =