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 :)

4 0

3 years ago

Find the exact length of the curve. x=et+e−t, y=5−2t, 0≤t≤2 For a curve given by parametric equations x=f(t) and y=g(t), arc len

Rama09 [41]

The length of a curve C parameterized by a vector function r(t) = x(t) i + y(t) j over an interval a ≤ t ≤ b is

$\displaystyle\int_C\mathrm ds = \int_a^b \sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2} \,\mathrm dt$