Question

Find the point closest to the coordinate (7,0) on the graph y = sqrt(x)

Answer 1

So, what's your best guess, or what is your approach when you see this? I'll help give you the intuition here so you can solve more problems on your own.

Answer 2

We'll also solve it haha, as long as you're trying and guessing, don't worry about being wrong.

Answer 3

I know I need to use the distance formula: \[d = \sqrt{(x-x_{1})-(y-y_{1})}\]
put in 7 and 0, and then take the derivative. I am just not sure what to do after wards to get a point? Just set the derivative to 0?

Answer 4

Oops I wrote the distance formula wrong

Answer 5

That's fine, I know what'cha mean.

Answer 6

So why might you set the derivative =0? It is the right thing to do here.

Answer 7

Also, before you take any derivatives, can you show me what your specific distance formula will look like?

Answer 8

I know when the derivative is 0 it is a critical point. Which means it is either a max or a min, in this case meaning the min distance between the point?

Answer 9

Exactly. If you were to graph your distance formula, with respect to one variable (hint hint), you'll see that as your line changes, it will get shorter and shorter as you move closer to the point (7,0). But once you get as close as you possibly can, then that line will start increasing and the distance will increase as well.

Answer 10

\[d = \sqrt{{(x-7)^2+(y-0)^2}}\] SInce we know \[y = \sqrt{x}\] we can replace y and make it \[d = \sqrt{(x-7)^2+x^2}\]

Answer 11

Your last equation is almost correct.

Answer 12

x not x^2

Answer 13

Ok, so as a visual aid, you should be sort of visualizing this for your own understanding: This is a graph of the distance from (7,0) depending on where you are on the x-axis.

http://www.wolframalpha.com/input/?i=d%3Dsqrt%28%28x-7%29%5E2%2Bx%29

Now you can see that the point where the distance is smallest is also where the slope =0.

Answer 14

If anything about derivatives being used here does not make sense to you, I'll gladly explain it. Now that you're actually getting to using them with some real examples is when most people actually begin to understand what a derivative really _is_.

Answer 15

Alright that helps a lot. So for my derivative I got: \[d' = (2x - 13)\div(2(\sqrt{(x-7)^2+x})\]

Answer 16

\[\frac{ 2x-13 }{ 2\sqrt{(x-7)^2+x} }\]

Answer 17

The bottom part is correct, but the top part is wrong, although almost right. It might be easier to multiply out the square and combine it into three terms under the square root.

Answer 18

It's wrong? I get the right x answer. 13/2

Answer 19

Wait, I'm wrong you're right haha.

Answer 20

I wrote -15 and realized I subtracted +x whoops lol.

Answer 21

Haha no worries! Thanks for talking this out with me. I really appreciate it and helped me understand what I didn't in class.
Much appreicaited

Answer 22

appreciated

Answer 23

Yeah, no problem, calculus is fun once you realize you can do some interesting things that are semi-useful. Like when you get into integrals, you can derive some cool stuff you probably have been wondering about for a long time.

For instance, ever notice the derivative of a circle's area formula is the circumference? =D

A=pi*r^2
C=2pi*r

Answer 24

Haha yup! This is my second year of calculus 1 and I am learning stuff via word problems I never knew I could solve before. Next semester is Calc 2, looking kinda forward to that :P. Thanks again!

Answer 25

(Second year meaning High School and then again in College, haha)

Answer 26

What I really looking forward to is how all of this math ties itself into Computer Science (my major) will make math a lot more fun and seem more useful.

Answer 27

Cool, good luck! It can be tough seeing how some of calculus is relevant, and even as a computer science major it might be weird at times. I have a friend who's a comp sci major and he couldn't seem to see how power series were useful to him since it basically amounts to turning functions like sine or cosine into an infinite number of polynomial terms like x, x^2, x^3, etc... but once you realize that you don't need an infinite number of terms to come to an approximately good answer, there might be some use in just turning cos(x) into 1-x^2/2+x^4/24.