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
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
2 answers
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
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
0 answers