Related Rates-The altitude of a triangle

[wind]

The altitude of a triangle is incresing at a rate of 1cm/min while the area of the triangle is incresing at a rate of 2cm^2/min. At what rate is the base of the triangle changing when the altitude is 10cm and the area is 100 cm^2.

let A rep the area
let a rep the altitude

dA/dt = 2
db/dt= ? when a=10 and A= 100

altitude is the same as height right?

A= (b)(a) / 2
100=b(10)/3
300=10b
10=b


A= (b)(a) / 2

f(x)= ba
f'(x)= a + b

g(x)=2
g'(x)=0

d/dt A(x)= d/dt [( a +b)(2) - (0)(ba)] / 2^2

d/dt A(x)= d/dt a + d/dt b / 2
(2)(100)= (1)(10) + d/dt (10) / 2
(2)(100)= 5+ d/dt 5
200-5= d/dt5
195/5=d/dt
39=d/dt

can someone check this over, thanks

 

[soroban]

Hello, wind!

You don't seem to have an understanding of Related Rates.
Let me walk you through this one . . .

The altitude of a triangle is increasing at a rate of 1 cm/min
while the area of the triangle is increasing at a rate of 2 cm²/min.
At what rate is the base of the triangle changing
when the altitude is 10 cm and the area is 100 cm².

The area formula is: $\displaystyle \,A\;=\;\frac{1}{2}bh$ . . . where $\displaystyle b$ = base, $\displaystyle h$ = height (altitude).

Differentiate with respect to time: \(\displaystyle \L\:\frac{dA}{dt}\;=\;\frac{1}{2}\left[b\left(\frac{dh}{dt}\right) + h\left(\frac{db}{dt}\right)\right]\;\;\) [1]

We are given: $\displaystyle \:\frac{dh}{dt}\,=\,1,\;\frac{dA}{dt} \,=\,2,\;h\,=\,10,\;A\,=\,100$

From $\displaystyle A\,=\,\frac{1}{2}bh$, we have: $\displaystyle \:100\,=\,\frac{1}{2}b(10)\;\;\Rightarrow\;\;b\,=\,20$

Substitute into [1]: $\displaystyle \:2\;=\;\frac{1}{2}\left[(20)(1)\,+\,(10)\left(\frac{db}{dt}\right)\right]$

Then: $\displaystyle \:20\,+\,10\left(\frac{db}{dt}\right)\;=\;4$

. . . . . . . . $\displaystyle 10\left(\frac{db}{dt}\right) \;=\;-16$

. . . . . . . . . . .$\displaystyle \frac{db}{dt} \;= \;-1.6$


Therefore, the base is decreasing at 1.6 cm/min.

 

[wind]

Thanks soroban