Question

what did i do wrong here
The altitude of a triangle is increasing at a rate of 1.5 centimeters/minute while the area of the triangle is increasing at a rate of 4.5 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 7 centimeters and the area is 90 square centimeters?

Area = A, altiitude = h and base = b
=> A = (1/2)bh
=> dA/dt = (1/2) (b*dh/dt + h*db/dt) ... ( 1 )

A = 90 sq. cm and h = 7 cm
=> 90 = (1/2) b*7
=> b = 180/7

Plugging, dA/dt = 4.5, dh/dt = 1.5, h = 7 and b = 180/7 in eqn. ( 1 ),
4.5 = (1/2) [(180/7) 1.5 + 7 db/dt)
=> 9

Answer 1

=> 9 = (270/7) + 7 db/dt
=> db/dt
= (1/7) (9 - 270/7)
= - 207/49 cm/minute
=> base is decreasing at the rate of 207/49 cm/minute.

Answer 2

Area = A, altiitude = h and base = b
=> A = (1/2)bh
=> dA/dt = (1/2) (b*dh/dt + h*db/dt) ... ( 1 )

A = 90 sq. cm and h = 7 cm
=> 90 = (1/2) b*7
=> b = 180/7

Plugging, dA/dt = 4.5, dh/dt = 1.5, h = 7 and b = 180/7 in eqn. ( 1 ),
4.5 = (1/2) [(180/7) 1.5 + 7 db/dt)
=> 9 = (270/7) + 7 db/dt
=> db/dt
= (1/7) (9 - 270/7)
= - 207/49 cm/minute
=> base is decreasing at the rate of 207/49 cm/minute.

Answer 3

did u just copy pate? wth?

Answer 4

*paste

Answer 5

what?

Answer 6

What are you saying yo

Answer 7

the final ans in that solution is wrong...whats the cause?

Answer 8

altitude-rate = 1.5 cm/min
area-rate = 4.5 cm^2/min

altitude = 7 cm
area = 90 cm^2

Let, base = b cm, SO,
(1/2)(b)(7) = 90,
b = 90 * 2 / 7 = 180/7 cm
Now,
altitude-time, t1 = 7/1.5 = 14/3 min
area-time, ta = 90/4.5 = 20 min
Let,
base-time = tb, SO,
(1/2)(tb)(t1) = ta, SO,
tb = 20*2 / (14/3) = 60/7 min
Hence,
base-rate = base/base-time = (180/7) / (60/7),

base-rate = 3 cm/min >==================< ANSWER

Answer 9

okay... uhmm lets see...

Answer 10

=> 9 = (270/7) + 7 db/dt wrong..

Answer 11

you had a half already... remember? but u're multiplying the inside out again.. just drop the half and that's it

Answer 12

wait nevermind.. i was dumb

Answer 13

A = (1/2) * h * b

Chain rule

dA/dt = (1/2) * dh/dt * b + (1/2) * h * db/dt

We know the following

dA/dt = 4.5 cm^2/min
dh/dt = 1.5 cm/min
h = 7 cm
b = 2A/h = 2 * 90 cm^2 / 7 cm = 25 5/7 cm

dA/dt = (1/2) * dh/dt * b + (1/2) * h * db/dt
4.5 cm^2/min = (1/2) * 1.5 cm/min * 25 5/7 cm + (1/2) * 7 cm * db/dt
9 cm^2/min = 38 4/7 cm^2/min + 7 cm * db/dt
9 cm^2/min - 38 4/7 cm^2/min = 7 cm * db/dt
-29 4/7 cm^2/min = 7 cm * db/dt
-29 4/7 cm^2/min / 7 cm = db/dt
-5 25/49 cm/min = db/dt

Answer 14

= (1/7) (9 - 270/7)
!= - 207/49 cm/minute
it is = -261/49 cm/min

Answer 15

hmm, webwork says thats not right either

Answer 16

Bahrom what are you doing

Answer 17

kumar what do you mean

Answer 18

ok u guys finally forced me to write this out...

Answer 19

haha sorry bahrom

Answer 20

Sorry for what?

Answer 21

sorry for forcing him to write it out

Answer 22

Wow

Answer 23

whats the wow for?

Answer 24

Wow for u saying sorry what else can a wow be fore.

Answer 25

hes just voluntarily helping me out. hes not getting paid so...

Answer 26

Oh okay

Answer 27

jinnie, your original answer was correct. all modifications were wrong. webworks sucks, tell the professor to go and commit a suicide for assigning these problems

Answer 28

Bahrom it's fine don't worry about it

Answer 29

man this sucks. this thing is due tomorrow. I cant meet the professor before then. f*******