Which equation does not represent a direct variation?

1 answer:

4 0

Okay I think there has been a transcription issue here because it appears to me there are two answers. However I can spot where some brackets might be missing, bear with me on that.

A direct variation, a phrase I haven't heard before, sounds a lot like a direct proportion, something I am familiar with. A direct proportion satisfies two criteria:

The gradient of the function is constant s the independent variable (x) varies

The graph passes through the origin. That is to say when x = 0, y = 0.

Looking at these graphs, two can immediately be ruled out. Clearly A and D pass through the origin, and the gradient is constant because they are linear functions, so they are direct variations.

This leaves B and C. The graph of 1/x does not have a constant gradient, so any stretch of this graph (to y = k/x for some constant k) will similarly not be direct variation. Indeed there is a special name for this function, inverse proportion/variation. It appears both B and C are inverse proportion, however if I interpret B as y = (2/5)x instead, it is actually linear.

This leaves C as the odd one out.

I hope this helps you :)

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Step-by-step explanation:

The interior of the square is the region D = { (x,y) : 0 ≤ x,y ≤1 }. We call L(x,y) = 7y²x, M(x,y) = 8x²y. Since C is positively oriented, Green Theorem states that

$\int\limits_C {L(x,y)} \, dx + {M(x,y)} \, dy = \int\limits^1_0\int\limits^1_0 {(Mx - Ly)} \, dxdy$

Lets calculate the partial derivates of M and L, Mx and Ly. They can be computed by taking the derivate of the respective value, treating the other variable as a constant.

Mx(x,y) = d/dx 8x²y = 16xy
Ly(x,y) = d/dy 7y²x = 14xy

Thus, Mx(x,y) - Ly(x,y) = 2xy, and therefore, the line ntegral is equal to the double integral

$\int\limits^1_0\int\limits^1_0 {2xy} \, dxdy$

We can compute the double integral by applying the Barrow's Rule, a primitive of 2xy under the variable x is x²y, thus the double integral can be computed as follows

$\int\limits^1_0\int\limits^1_0 {2xy} \, dxdy = \int\limits^1_0 {x^2y} |^1_0 \,dy = \int\limits^1_0 {y} \, dy = \frac{y^2}{2} \, |^1_0 = 1/2$