Question

find the point in the first quadrant on the graph of y=x^2 that is closest to the point (0,4)

Answer 1

If you graph it, you can see it more easily, (or jot down a few points for the equation.) The shortest distance would be straight across from (0,4)......Can you see it? :)

Answer 2

but i have to show method how i got for exam not allowed to just graph it and see needs to show work

Answer 3

Maybe you could say, since (2,4) is a point on y=x², and (0,4) has the same y value, that must be the closest point?

Answer 4

ok

Answer 5

NO, you can't "just say" that.

lovingnilam: Do you know how to calculate the distance between 2 points?

Answer 6

The distance between two points \[A(a_1,a_2) and B(b_1,b_2)\]is given by the formula\[d(A,B)=\sqrt{(a_1-b_1)^2 + (a_2-b_2)^2}\]This is just Pythagoras.
Now here we could say that A = (0,4) and B = (x, y²).
When we put these in the distance formula we get:\[d(A,B)=\sqrt{(0-x)^2+(4-x^2 )^2}=\sqrt{x^2+16-8x^2+x^4}=\sqrt{x^4-7x^2+16}\]This distance of A and B is a function of x.
To calculate the minimum of this function, we need to differentiate it.

However, the function looks quite complicated, because of the root of a 4th degree polynomial. We can reduce the work, by realising that we don't need the root at all!
The whole thing is minimal when the number under the root sign is minimal, so we just need to find the minimum of \[f(x)=x^4-7x^2+16\]Therefore, differentiate it:\[f'(x)=4x^3-14x\]Solve \[f'(x)=0\]\[4x^3-14x=0\]\[2x(2x^2-7)=0\]\[2x=0, or, 2x^2-7=0\]This gives solutions\[x=0\]\[x=\pm \sqrt{\frac{ 7 }{ 2 }}=\pm \frac{ 1 }{ 2 } \sqrt{14}\]We only need \[x=\frac{ 1 }{ 2 }\sqrt{14}\]because we are looking in the first quadrant.
(At x = 0, the distance function has a (local) maximum, which is 4).

Put this x in y=x² to find y = 7/2.
The point we are looking for is:\[\left( \frac{ 1 }{ 2 }\sqrt{14},\frac{ 7 }{ 2 } \right)\]
Hope this helps!

ZeHanz

Answer 7

Here is a drawing of the situation:

Answer 8

There is also a different way to do this, looking at the drawing (see above).
One could argue that the shortest path from A(0, 4) to a point P(p, p²) on the graph of y=x² is perpendicular to the tangent line in P.
There is this well-known little formula: \[m_1 * m_2 = -1\]that says: two lines are perpendicular if the product of their respective slopes equals -1.
We just have to calculate these slopes, set their product to -1 and solve the equation.

The slope of the line AP is \[\frac{ \Delta y}{ \Delta x }=\frac{ y_P-y_A }{ x_P-x_A }=\frac{ p^2-4 }{ p-0 }=\frac{ p^2-4 }{ p }\]
To calculate the slope of the tangent line in P, we need to differentiate y = x²:
\[\frac{ dy }{ dx }=2x\]
This has value 2p in x = p.
Now set the product to -1:\[\frac{ p^2-4 }{ p }*2p=-1\]
This simplifies to:\[p^2-4=-\frac{ 1 }{ 2 },\]provided that p > 0.
Now we see that \[p^2=3\frac{ 1 }{ 2 }=\frac{ 7 }{ 2 }\]
Just like the answer above we see that the point we are looking for is\[P(\frac{ 1 }{ 2 }\sqrt{14}, \frac{ 7 }{ 2 })\]

You can judge for yourself which method you prefer...

ZeHanz