The altitude of a triangle is increasing at a rate of
2
centimeters/minute while the area of the triangle is increasing at a rate of
1.5
square centimeters/minute. At what rate is the base of the triangle changing when the altitude is
11.5
centimeters and the area is
91
square centimeters?
im a little confused on the formula so the answer is always wrong
Area of triangle is (1/2) base * height
given: dh/dt = 2 , dA/dt = 1.5,
find: db/dt when A = 91 and h = 11.5 = 23/2
(skipping the units, since that will all work out)
at A = 91 and h = 11.5
91 = (1/2)(b)(11.5)
b = 364/23 = appr 15.826
A = (1/2)bh
dA/dt = (1/2)b dh/dt + (1/2)h db/dt
1.5 = (1/2)(364/23)(2) + (1/2)(23/2) db/dt
3/2 = 364/23 + 23/4 db/dt
23/4 db/dt = -659/46
db/dt = - 1318/526 = appr - 2.49 cm/sec
the base is decreasing at appr 2.5 cm/s
dA = 1.5 cm² / min
dh = 2 cm / min
h = 11.5 cm
A = 91 cm²
Area of a triangle:
A = b ∙ h / 2
91 = b ∙ 11.5 / 2
Multiply both sides by 2
182 = 11.5 b
b = 182 / 11.5
Differentiate equation for area of a triangle to find rate of change of the area of a triangle ( dA ):
A = ( 1 / 2) b ∙ h
dA / dt = ( 1 / 2 ) ( b ∙ dh / dt + h ∙ db / dt )
The altitude of the triangle is increasing at a rate of 2 cm / min while the area of a triangle is increasing at a rate of 1.5 cm² / min means:
dh / dt = 2
When the altitude is 11.5 cm and the area is 88 cm², the base is b = 182 / 11.5
This gives:
1.5 = ( 1 / 2 ) [ ( 182 / 11.5 ) ∙ 2 + 11.5 ∙ db / dt )
1.5 = ( 1 / 2 ) ( 364 / 11.5 + 11.5 db / dt )
Multiply both sides by 2
3 = 364 / 11.5 + 11.5 db / dt
Subtract 364 / 11.5 to both sides
3 - 364 / 11.5 = 11.5 db / dt
34.5 / 11.5 - 364 / 11.5 = 11.5 db / dt
Multiply both sides by 11.5
34.5 - 364 = 132.25 db / dt
- 329.5 = 132.25 db / dt
- 329.5 / 132.25 = db / dt
db / dt = - 329.5 / 132.25 = - 32950 / 13225 = - 25 ∙ 1318 / 25 ∙ 529 = - 1318 / 529 cm / min
The length of the base of the triangle is decreasing at the rate 2.4915 cm / min ≈ 2.5 cm / min
To solve this problem, we can use the formula for the area of a triangle:
Area = (1/2) * base * altitude
First, differentiate both sides of the equation with respect to time (t) using the chain rule:
d(Area)/dt = (1/2) * (d(base)/dt * altitude + base * d(altitude)/dt)
The problem provides:
- d(altitude)/dt = 2 cm/min (rate at which the altitude is increasing)
- d(Area)/dt = 1.5 cm²/min (rate at which the area is increasing)
- altitude = 11.5 cm
- Area = 91 cm²
Let's plug in the given values into the equation:
1.5 = (1/2) * (d(base)/dt * 11.5 + base * 2)
Simplify the equation:
3 = d(base)/dt * 11.5 + 2base
Now we have an equation with two variables, d(base)/dt (rate of change of the base) and base. We need to find the value of d(base)/dt when base = ?.
To proceed further, we need more information about the triangle, specifically the value of the base.
To solve this problem, we need to use the formula for the area of a triangle, which is given by:
Area = (1/2) * base * altitude
We are given that the altitude is increasing at a rate of 2 centimeters/minute and the area is increasing at a rate of 1.5 square centimeters/minute.
Let's set up the known rates:
d(altitude)/dt = 2 cm/min
d(area)/dt = 1.5 cm^2/min
We are asked to find the rate at which the base of the triangle is changing, which is d(base)/dt. To find this, we can use the chain rule of differentiation.
First, differentiate the area equation with respect to time (t):
d(area)/dt = (1/2) * d(base)/dt * altitude + (1/2) * base * d(altitude)/dt
Now, we can substitute the given values:
1.5 = (1/2) * d(base)/dt * 11.5 + (1/2) * base * 2
Simplify the equation:
1.5 = 5.75 * d(base)/dt + base
Rearrange the equation to solve for d(base)/dt:
5.75 * d(base)/dt = 1.5 - base
Now, substitute the given area value to find base:
91 = (1/2) * base * 11.5
base = 182/11.5 = 15.8261
Now, substitute the base value back into the equation:
5.75 * d(base)/dt = 1.5 - 15.8261
Simplify:
d(base)/dt = (1.5 - 15.8261) / 5.75
Calculate:
d(base)/dt ≈ -2.82 cm/min
Therefore, the rate at which the base of the triangle is changing when the altitude is 11.5 cm and the area is 91 cm^2 is approximately -2.82 cm/min.