Question

DIRECT VARIATION QUESTION!
If y varies directly as \[x^2\], find k when x=2 and y=8.
The answer I got was k=4... Is this correct?

Answer 1

\[y = kx^2\]\[8=k(2)^2\]\[8=k(4)\]\[k=\]

Answer 2

You're up to your old trick of forgetting to square x, I think...

Answer 3

Wait, wouldn't the answer be k=2? I think I remembered to square x...

Answer 4

Yes, \(k = 2\) is correct.

\[y = 2x^2\]\[y = 2*(2)^2\]\[y = 2*4\]\[y = 8\]so that checks out

Answer 5

you can test your answer by plugging in the original point and verifying that it gives you the original y from the original x. However, if you consistently make a mistake in your calculations (like forgetting to square x) and make the same mistake when checking your work, the error will probably slip past you unnoticed...

Answer 6

This is an example of where the units would be helpful in catching mistakes. As I mentioned in our previous problem, position varies directly with the square of time, with acceleration as the constant of variation (actually 1/2 of acceleration).

Let's say we had \(y\) as position (in units of meters), and \(x\) as time in units of seconds

\[y = kx^2\]\[8\text{ m} = k (2\text{ s})^2\]\[8\text{ m} = k*(4 \text{ s}^2)\]\[k = \frac{8\text{ m}}{4\text{ s}^2} = 2 \frac{\text{m}}{\text{s}^2}\]

And m/s^2 is the appropriate unit for acceleration.

However, if you'd forgotten to square x while working the problem, you'd get this:

\[y = kx^2\]\[8\text{ m} = k(2 \text{ s})\]\[k = \frac{8\text{ m}}{2\text{ s}} = 4\frac{\text{m}}{\text{s}}\]But that's not the right unit for acceleration! And if you used that in to calculate a new value of \(y\), you would end up with the wrong units as well:

\[y = kx^2\]\[y = (4\frac{\text{m}}{\text{s}})*(1 \text{ s})^2\]\[y = (4\frac{\text{m}}{\text{s}})*(1 \text{ s}^2) = 4 \text{ m*s}\]
Uh, meters*seconds is not a unit we expect to see for position! Something must be wrong with our calculation.

So, rather than being an annoyance, units are a help that keep us from making mistakes!