# Thread: Check my calculations ^_^

1. ## Check my calculations ^_^

I keep trying this question over and over again, and keep getting the wrong answer. I was wondering if somebody would be nice enough to check my calculations, and see where I went wrong (or if I’m going about it the wrong way!)

Find the minimum distance from the parabola y=x^2 to the point (5,0).

D = √[(x2-x1)^2 + (y2-y1)^2]

I substituted in points (5,0) and (x, x^2)

D = √[(x-5)^2 + x^2
D = √2x^2 - 10x + 25

d/dx = (4x-10)/ [blahblah not important]
0 = 4x-10
4x=10
x=10/4

Substituting this into the D formula doesn't give the right answer ... the answr is apparently 4.06

Thanks!

2. Originally Posted by freswood
I keep trying this question over and over again, and keep getting the wrong answer. I was wondering if somebody would be nice enough to check my calculations, and see where I went wrong (or if I’m going about it the wrong way!)

Find the minimum distance from the parabola y=x^2 to the point (5,0).

D = √[(x2-x1)^2 + (y2-y1)^2]

I substituted in points (5,0) and (x, x^2)

D = √[(x-5)^2 + x^2
The above should be:

$
D =\sqrt{ (x-5)^2 + x^4}
$

Also it is easier to minimise $D^2$

RonL

3. Oops silly me Even with that in mind, I still can't seem to solve it.

d/dx = (2x^3 + x - 5) / [blah]

so how do you solve:

0 = 2x^3 + x - 5

4. Originally Posted by freswood
Oops silly me Even with that in mind, I still can't seem to solve it.

d/dx = (2x^3 + x - 5) / [blah]

so how do you solve:

0 = 2x^3 + x - 5
Unless your a dab hand with the cubic formula you solve it numericaly,
the (real) solution is $x \approx 1.235$, which you now
need to substitute back into the distance formula to find the distance
correspomding to this x.

RonL

5. Originally Posted by freswood
Oops silly me Even with that in mind, I still can't seem to solve it....
so how do you solve:
0 = 2x^3 + x - 5
Hi, freswood,

I can only think of 2 ways:

1. Use an iteration (for instance Newtons Method)

2. Use Cardanos Formula to solve a reduced cubic equation.

Never mind which method you use, there is only one real solution:

$x=\sqrt[3]{\frac{5}{4}+\sqrt{\frac{677}{432}}}+\sqrt[3]{\frac{5}{4}-\sqrt{\frac{677}{432}}}$ $\approx 1.23477$

Best wishes

EB

6. Thanks for that guys! It's pretty bizarre though - we haven't learnt that method of solving... after the holidays I'll go and ask my teacher how we're supposed to do it.