Question

Please help will fan and medal!!! The altitude of a triangle is increasing at a rate of 1 cm/min while the area is increasing at a rate of 2 cm^2/min. At what rate is the base of the triangle changing when the altitude is 10 cm and the area is 100cm^2 okay for this one I made a sketch I had a triangle and the 10 cm on the y side 1 cm/min there too and than the 100 cm^2 in the middle with the 2cm^2/min and than I used the area formula and differentiated A=1/2bh

Answer 1

yeah sketch it, and list all the given things , rates, and values

Answer 2

Okay

Answer 3

da/dt = 1 cm/min
dA/dt = 2 cm^2/min

db/dt = ??
when a=10 cm, and A=100cm^2
--------------------------------

Given the rates of change for the altitude and the area.
Find the rate of change of the base, at the instant the altitude is 10 and the area is 100

Answer 4

I wish I could send a picture of my sketch

Answer 5

i can picture it, the altitude is getting larger, and in tern the area also increases

Answer 6

Right

Answer 7

So to find side b the base can you do...? Area formula? 1/2bh=100cm^2 when h=10?

Answer 8

And would you do that formula for the first part? Differentiation for the area equation?

Answer 9

I'm not very good at these, its a pretty concept for us

Answer 10

the general way to solve these related rate probs, is to understand what you have and what to find

Here the Area function relates all three things , area , altitude

Answer 11

you need a relationship between the different variabes,
in this prob, we need the derivative w.r.t. time
d/dt
of the total area function,

none of the terms are in terms of a time, base, height, area

have to use the chain rule to get to the time derivative

Answer 12

Okay

Answer 13

I used A=1/2bh to get da/dt=1/2(b(dh/dt)+h(db/bt))

Answer 14

Idk if that is right

Answer 15

d/dt [A] = d/dt [ b * h]

yah looks good

Answer 16

Yay :)

Answer 17

Than I multiplied that 1/2 through

Answer 18

So when I started plugging in numbers I got 2=1/2b(dh/dt)+q/2(10)(1)

Answer 19

1/2 after the plus sign

Answer 20

And now would I use A=1/2bh when h=10 and A=100 to find b?

Answer 21

yeah you have

\[\frac{ dA }{ dt }= \frac{ 1 }{ 2 }*[b*\frac{ dh }{ dt } + h*\frac{ db }{ dt }]\]

you know the height and the area at a moment in time,

A=1/2 b * h
100 = 1/2 * b * 10
so the b=20 then

Answer 22

the rates of change in area and height are also given, one thing left to solve for

Answer 23

Db/dt

Answer 24

any prob like these related rates ones,

use some relationship for the variables you need, some may be in terms of others.
derivative with respect to what is needed

solve

Answer 25

I got db/dt is -1.6 m/s?

Answer 26

does that make sense, if the area changes by 2 cm^2/min, and the height increases at 1 cm/min

the base has to become smaller?

Answer 27

Yes

Answer 28

yeah , it does

you can increase time by one min, and see increase area by 2 and h by 1

A = 1/2 b * h

102 = 1/2 * b * 11

b is less than 20

Answer 29

that it, goodluck

Answer 30

Thats much easier than trying to think of it from a picture standpoint ill for sure have to write that! So ill remember for the quiz tomorrow

Answer 31

Thank you! It's Slowly coming day by day! Thank you for the help! :)