Asked by Jermaine
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.
Known:
base=1.5H
dA/dt=9.0cm2/s
h=12cm
dh/dt= ?
Area for a triangle: LxW/2
After that, I am confused.
Answers
Answered by
Reiny
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.
<b>So I will need an equation that contains a and h)</b>
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.
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.
<b>So I will need an equation that contains a and h)</b>
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.
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