Question

Find the rate of change of the distance between the origin and a moving
point on the graph of y=x^2+1 if dx/dt=7 centimeters per second

Answer 1

use the distance formula, derive it, and plug in the values

Answer 2

ok

Answer 3

@amistre64 I would love to see you solve it.

Answer 4

d^2 = x^2 + y^2

dd' = xx' + yy', x' = 7

Answer 5

and x and y are equal to zero?

Answer 6

so,\[d' = \frac{7x+yy'}{\sqrt{x^2+y^2}}\]
y' can be determined from the given setup for y

Answer 7

What values do we plug in for x and y?

Answer 8

whatever values satisfy the definition for y of course.

Answer 9

the rate of change is a function of time, they dont give you a specific point to asses so you have to let it be assessed for any point that satisifies the setup of course

Answer 10

but if we plug in (0,0) the answer would be all zero.

Answer 11

y=x^2+1

0,0 is not a point of concern

Answer 12

now, if we are creating a vector field, then yeah we would run into that situation, but in this case the domain is not RxR, but a subset of it that does not include 0,0

Answer 13

okay can you solve it please then? what values do you plug in for x and y? You derive them from the equation?

Answer 14

x is independant, y is already given

x' is stated, and y' can be solved

d' is defined by the derivative of the distance formula ... which part is not working out?

Answer 15

So after you derived the formula, what values do you plug in for x and y? This is my question.

Answer 16

The subset of RxR (the ordered pairs) that satisfy y = x^2 + 1 as is stated in the problem

Answer 17

\[D=\{(x,y)\in RxR:y=x^2+1\}\]

Answer 18

i spose \cross would be more appealing, or the lazier R^2

Answer 19

okay so for y we plug in the formula and for x you plug in the x?

Answer 20

yep

Answer 21

and y' = 2xx' = 14x

Answer 22

@amistre64 Thank you. I had a lot of trouble with these types of questions too. I really appreciate your help.

Answer 23

its all about reading the information, and not being blinded by the whatifs :)

Answer 24

dd' = xx' + yy', x' = 7, y' = 2xx' = 14x

\[d' = \frac{7x+14x}{\sqrt{x^2+(x^2+1)^2}}\]

Answer 25

opps forgot the y up top

Answer 26

\[d' = \frac{7x+14x(x^2+1)}{\sqrt{x^2+(x^2+1)^2}}\]
thats better

Answer 27

@lamp12345 btw its a really nice question I really like it.

Answer 28

its a thinker, doable but coy

Answer 29

@amistre64 I was understanding it too but thats not an answer choice. Here they are

Answer 30

simplify is needed of course. i figure that by the time your doing this stuff that you already are familiar and sufficient with your algebra

Answer 31

also, if you could put all the pics in one file it would do better than trying to hunt and peck thru 5 seperate files

Answer 32

yeah sorry I typed that out before you gave the other answer

Answer 33

when i do implicits i like to keep my x' in there since its a place holder for dx/dt. IF this had been a dx/dx setup then x'=1, but dx/dt = 7 is their definition

Answer 34

for the double chk

http://www.wolframalpha.com/input/?i=+simplify+%5Cfrac%7B7x%2B14x%28x%5E2%2B1%29%7D%7B%5Csqrt%7Bx%5E2%2B%28x%5E2%2B1%29%5E2%7D%7D

14x^3 + 21x
------------
sqrt(.....)

seems to fit