An isosceles triangle has a base that is 1.5 times the height. If the area is increasing at 9cm2/s, how fast is the height increasing when it is 12 cm high?
Known:
base=1.5H
dA/dt=9.0cm2/s
h=12cm
dh/dt= ?
Area for a triangle: LxW/2
After that, I am confused.
ok, let's go with your definitions:
height: --- h
base --- 1.5h
area ---- a
given: d(a)/dt = 9 cm/s
find dh/dt, when h = 12
(I see variables a for area and h for height.
So I will need an equation that contains a and h)
a = (1/2)base x height
a = (1/2)(h)(1.5h) = .75h^2
d(a)/dt = 1.5h dh/dt
plug in our values
9 = 1.5(12)dh/dt = 18dh/dt
dh/dt = 9/18 = 1/2 cm/s
A common error is to use the given case of h=12 too soon.
You have to wait until you have found the derivative before subbing it in.
To find how fast the height is increasing (dh/dt) when it is 12 cm high, we first need to find the values of the base and the area.
Let's start by deducing the base of the triangle. It is given that the base is 1.5 times the height, so we have:
base = 1.5 * height
base = 1.5 * 12 cm
base = 18 cm
Next, we need to find the area of the triangle. The formula for the area of a triangle is (base * height) / 2. Substituting the values of the base and height, we have:
area = (18 cm * 12 cm) / 2
area = 216 cm^2
Now that we have the area and its rate of change, we can differentiate the area with respect to time (t) to find how fast it's increasing. Differentiating the area equation gives us:
dA/dt = (1/2) * (db/dt) * h + (1/2) * b * (dh/dt)
Where dA/dt is the rate of change of area, db/dt is the rate of change of the base, and dh/dt is the rate of change of the height.
Since the base is not changing in this scenario (it remains constant at 18 cm), db/dt is 0. Therefore, the equation simplifies to:
dA/dt = (1/2) * b * (dh/dt)
We are given that dA/dt is 9 cm^2/s and we want to find dh/dt when h is 12 cm. Substituting the known values into the equation and solving for dh/dt, we have:
9 cm^2/s = (1/2) * 18 cm * (dh/dt)
9 cm^2/s = 9 cm * (dh/dt)
dh/dt = 1 cm/s
Therefore, the height is increasing at a rate of 1 cm/s when it is 12 cm high.