Question

If y=x^3+2x and dx/dt=6. Find dy/dt when x=5

Answer 1

dy/dt = 3x^2(dx/dt)+2(dx/dt)
dy/dt = 3*5^2*6 + 2*6 = 3*25*6+12 = 75*6 + 12 = 450+12 = 462

Answer 2

If z^2=x^2+y^2,dx/dt=9, and dy/dt=6, find dz/dt when x=2 and y=4

Answer 3

do the same thing Jinnie

Answer 4

haha yeah

Answer 5

2z(dz/dt)...........

Answer 6

im a little confuse since there are two different d's

Answer 7

z^2=x^2+y^2
2z*(dz/dt) = 2x(dx/dt) + 2y(dy/dt)
plug in all values.

Answer 8

ok understood

Answer 9

can you take a look at this one, its a different concept

Answer 10

sure

Answer 11

A street light is mounted at the top of a 11 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 7 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 35 ft from the base of the pole?

Answer 12

Typical related rates problem. And you have to use similar triangles.

Answer 13

BE/AB = DE/CD => plug in all values that you have, and take derivative

Answer 14

ok thanks
are you like a math teacher?

Answer 15

i did work as a tutor lol, but no i'm just a teenager haha

Answer 16

what?!!! how old are you?

Answer 17

what?!!! how old are you?

Answer 18

18

Answer 19

actually turning 19 in 5 months

Answer 20

whoa how did you learn all this at age 18

Answer 21

haha good question. anyway back to your homework.. That's a typical shadow woman walking away from a lamp post question lol

Answer 22

im really urious im a sophmore in college and idont know all this lol

Answer 23

im a junior in college, and i've taken calc1-3, diff equations, topology, lin alg, so i know all this lol.

Answer 24

you're 18 and a junior in college?

Answer 25

Wow, bahrom just 18 with all Cal stuff and topology! Look like you aim to network security major. I wish my cousin just 1/100th as mature as you :)

Answer 26

y' = 3x^2 x' + 2 x'
= 3 * 5^2 * 6 + 2 * 6
= 6( 75 + 2) = 6 * 77 = 462

Answer 27

lol im a math major

Answer 28

chlorophyll thanks for the help. i was focused on this question
A street light is mounted at the top of a 11 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 7 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 35 ft from the base of the pole?

you can scroll up and see an attachment i posted along with it

Answer 29

Jinnie, it's much better for you to learn from bahrom!

Answer 30

Anyway, I'm going to be back after feeding my precious kitten :)

Answer 31

alright thanks for all the help

Answer 32

Suppose X is the total distance ( the leg in red) in which:
S is shadow of the woman,
x is distance from woman to the base of street light

-> Tip of the shadow: X = s + x
=> Rate of tip shadow moving: X' = s' + x'
Given x' = +7ft/sec

By ratio: s/6 = ( s + x)/ 11 (similar triangle)
-> s = 6x/5
=> s' = (6/5) x' = (6/5) * (7) = 42/5
Thus X' = 42/5 + 7 = (42 + 35) / 5 = 15.4 ft/sec

Answer 33

I keep considering the distance 35 ft, and don't know how to apply in this situation ???