If y varies directly as x^2 and y=12 when x=2, find y when x=5. Help please!
okay, if \(u\) varies directly with \(v\), then we can write \[u = kv\] Here you have \(y\) varying directly with \(x^2\). What will the relationship look like, given what I just stated?
\(k\) is the constant of variation, btw.
You wrote u=kv...
@whpalmer4
Yes, I'm aware of that :-) If you have "thing 1" varying directly with "thing 2", "thing 1" = k* "thing 2"
Yes, but it's the x^2 that's tripping me up. Would direct variation work the same way?
"thing 1" is y "thing 2" is x^2
So I'd get y=k(x^2)?
it can be a considerably more complicated expression than \(x^2\), but it still works the same way: \(u = k * v\) if \(u\) varies directly with \(v\) yes! \[y = kx^2\]Now, you know a pair of \((x,y)\) that go together. Use them to find the value of \(k\)...
I just don't know how... I don't know the step that comes after y=kx^2.
@whpalmer4
" y=12 when x=2" plug those numbers into your new formula and solve for \(k\)
And what I get will be my correct answer?
no, but it will be a necessary step to finding the final answer.
\[12 = k(2)^2\]\[12 = k*4\]\[k =\]
When I come up with it I'll put it on here.
I got the answer of y=30. Is this correct?
Uh, no. Can you show me what you got for \(k\)?
I got 6=k :/
\[12 = k(2)^2\]\[12 = k*4\]\[k=\]
it looks like you're forgetting to square \(x\)...
both in finding \(k\) and then in finding the value of \(y\) that goes with the new value of \(x\)
at least you're consistent — we just have to fix one problem and then you'll be doing it right :-)
Oh, k=3?
yes! so the new, improved value of \(y\) when \(x=5\) is?
So the second answer would be y=15?
I was afraid you would answer that. the equation is \[y = kx^2 = k*x*x\]
\[y = 3*5*5 = \]
75!
yes
Thank you so much for your help and patience, I appreciate it so much!
what would the value of \(y\) be if \(x = -1\)?
3?
yes, very good! I made a graph of the curve that you have here, and put on crossing lines at the various points we visited.
You are unbelievable helpful. Thank you!!!
You're welcome!
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