Melody

Are you both doing it by trial and error?  The question does say ALL

How did you work that out Chris?

A lesson in understanding definite integrations

Think about what you are finding when you get the area under a curve.

In this instance from t=0 to t=1 you are finding the area of a triangle which is 4.  But 4 what?

Look at the units.  They are     t  time (seconds)   in the horizonal directions  and  velocity (feet/sec) in the vertical direction.

When you find the area you can find the units the same way.  The unit of the area will be      \(seconds\times \frac{feet}{sec}=feet\)

So that is why the area under the velocity curve gives displacement.

And youcan find the area with integration.

NOW

Lets think about integration for a moment.

The integration sign.    \(\int\)    is a stylized S and it stands for   SUM

What you are doing with any definite integral is finding the sum of all the rectangles (with infinitely small width) that lie between the curve and the axis that you are integrating with respect to (usually the horizonal axis).

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A lesson in understanding differentiation

With differentiation the  d   stands for difference.  So    \(\frac{dx}{dt}=\frac{\text{difference in x}}{\text{difference in t }} \)

Since this is the answer as t tends to 0, dx/dt results in the instantaneous velocity.

The gradient of the tangent to a curve IS the differential to the curve at that point.

If you have questions then please ask them    

What about it?

Obviously I cannot see how your graph interacts with you.

But here is the graph of the first region. Perhaps from mine you can work out how to move the points (and what you are doing it for)

(x^2+y^2)^2+2x^2y^2-y^4=x^2(x^2+4y^2)

\((x^2+y^2)^2+2x^2y^2-y^4=x^2(x^2+4y^2)\\ \)

\(LHS\\=x^4+y^4+2x^2y^2+2x^2y^2-y^4\\ =x^4+4x^2y^2\\ =x^2(x^2+4y^2)\\ =RHS\\~\\ QED\)

I have no idea how to do this but I would be interested if anyone else has an approach.   

Hi Dana :)

2^(x+y) * 3^(x-y) * 6^(2x+2y) = 72

\(2^{x+y} * 3^{x-y} * 6^{2x+2y}= 72\\ 2^{x+y} * 3^{x-y} * 2^{2(x+y)}*3^{2(x+y)}= 3^2*2^3\\ 3^{x-y+2x+2y} * 2^{3(x+y)}= 3^2*2^3\\ 3^{3x+y} * 2^{3x+3y}= 3^2*2^3\\ so\\ 3x+y=2\quad(1a)\\ 3x=2-y \quad (1b)\\ 3x+3y=3\quad(2a)\\ x+y=1\quad(2b)\\ y=1-x\quad(2c)\\ sub (2c) into (1b)\\ 3x=2-(1-x)\\ 3x=1+x\\ 2x=1\\ {\bf{x=0.5,\quad y=0.5}} \\ check\\ 2^1*3^0*6^{2}=72\qquad \text{Das ist ausgezeichnet!} \)

See my answer here