Question

Assume that x and y are both differentiable functions of t and find the required values of dy/dt and dx/dt.
x^2 + y^2 = 100
(a) Find dy/dt, when x = 6, y = 8 given that dx/dt
= 4.
(b) Find dx/dt, when x = 8, y = 6, given that
dy/dt = −2.

Answer 1

What do I do?

Answer 2

differentiate with respect to t

Answer 3

\[\frac{ d }{ dt }\left( x^2 \right)=2 x \frac{ dx }{ dt }\]

Answer 4

So I just have to plug in the 2(6)+8^2?

Answer 5

\[2 x \frac{ dx }{ dt }+2 y \frac{ dy }{ dt }=0,x \frac{ dx }{ dt }+y \frac{ dy }{ dt }=0\]

Answer 6

but they are giving me these values of x=6 and y=8... and then values of x=8 and y=6..

Answer 7

SOMEBODY please help!!???

Answer 8

surji wrote out most of the solution for you.

you start with
x^2 + y^2 = 100

you differentiate both sides (and all terms) with respect to t
as posted above you get
2 x dx/dt + 2y dy/dt =0
or (dividing by 2)
x dx/dt + y dy/dt =0

now replace the variables with the given values and evaluate.

Answer 9

So... is it 2(6)+2(8)=0??? and 2(8)+2(6)??

Answer 10

I'm taking this course online and I have no idea what I'm doing

Answer 11

(a) Find dy/dt, when x = 6, y = 8 given that dx/dt
= 4.

they give you 3 out of the 4 unknowns in:
x dx/dt + y dy/dt =0

dy/dt is the 4th unknown that you have to solve for.

Answer 12

huh? sooooooo..... what does it mean when x=6 and y=8 and dx/dt=4? and what do I do here... I am lost to this stuff since I have no person to ask and walk me through the steps

Answer 13

I get part of it =0 but I don't understand where to plug in what and when..

Answer 14

It means replace x with 6, replace y with 8 and dx/dt with 4 in the equation
x dx/dt + y dy/dt =0

what do you get ?

Answer 15

6^2+8^2=100 it says it =4 on the next part, how do they get 4?

Answer 16

2(6)=12 not 4

Answer 17

if the function changes to 2x

Answer 18

Let's start at the beginning.

they give you an equation
x^2 + y^2 = 100
which is the equation of a circle with center at (0,0) and radius = 10

they tell you both x and y are "functions of t"
in other words, where you are on the circle is determined by time.
Think of a satellite in orbit, or a spot on a wheel that is turning.

Answer 19

ok

Answer 20

i picture an x y axis here right?

Answer 21

as you move along the circle, the velocity in the x direction (denoted as dx/dt )
changes. for example, at the top of the circle you are moving mostly left/right
at the side of the circle, you are moving mostly up/down.

the idea is the velocity in the x direction (dx/dt) and the velocity in the y direction (dy/dt)
change depending on where you are

Answer 22

using implicit differentiation we can find an equation that tells us some info about the velocity dx/dt and dy/dt

Hopefully you know how to find
\[ x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \]

if we are at the point (6,8), and are moving in the x direction at speed 4, how fast are we moving in the y direction?