The height 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 square cm/min. At what rate is the base of the triangle changing when the height is 7 centimeters and the area is 91 square centimeters?
Well, it sounds like this triangle is really growing tall! It's like a tree on stilts!
To find the rate at which the base is changing, we can use some calculus. We know that the area of a triangle is given by the formula A = 1/2 * base * height.
Now, we have the height increasing at a rate of 1 cm/min and the area increasing at a rate of 2 square cm/min.
So, let's differentiate the area formula with respect to time (t):
dA/dt = 1/2 * (dbase/dt) * height + 1/2 * base * (dheight/dt)
Since we're looking for the rate at which the base is changing (dbase/dt), we can rearrange the formula:
dbase/dt = (2 * dA/dt - base * dheight/dt) / height
Now, at the given time, the height is 7 cm and the area is 91 square cm.
So, let's fill in those values:
dbase/dt = (2 * 2 - base * 1) / 7
Simplifying:
dbase/dt = (4 - base) / 7
And there you have it! The rate at which the base of the triangle is changing when the height is 7 cm and the area is 91 square cm is (4 - base) / 7. But remember, this rate will vary depending on the value of the base. So keep an eye out for those changing bases!
To find the rate at which the base of the triangle is changing, we can use the formula for the area of a triangle and apply the chain rule in calculus.
Let's denote the height of the triangle as h (in cm) and the base as b (in cm). The formula for the area of a triangle is:
A = (1/2) * b * h
Given that the height is increasing at a rate of 1 cm/min, we have dh/dt = 1 cm/min. Also, the area is increasing at a rate of 2 square cm/min, so dA/dt = 2 square cm/min.
We are asked to find db/dt (the rate at which the base is changing) when h = 7 cm and A = 91 square cm.
To find db/dt, we need to differentiate the formula for the area of a triangle with respect to time (t):
dA/dt = (1/2) * (b * dh/dt + h * db/dt)
Substituting the known values into the equation:
2 = (1/2) * (b * 1 + 7 * db/dt)
Simplifying:
2 = (1/2) * b + (7/2) * db/dt
Rearranging the equation to solve for db/dt:
(7/2) * db/dt = 2 - (1/2) * b
db/dt = (4 - b/2) * (2/7)
To find db/dt when h = 7 cm and A = 91 square cm, substitute these values into the equation:
db/dt = (4 - 7/2) * (2/7)
= (8/2 - 7/2) * (2/7)
= 1/2 * (2/7)
= 1/7
Therefore, when the height of the triangle is 7 cm and the area is 91 square cm, the base of the triangle is changing at a rate of 1/7 cm/min.
To find the rate at which the base of the triangle is changing, we need to use the formula for the area of a triangle.
The formula for the area of a triangle is: Area = (1/2) * base * height.
Given that the height is increasing at a rate of 1 cm/min (dh/dt = 1 cm/min) and the area is increasing at a rate of 2 square cm/min (dA/dt = 2 square cm/min), we want to find the rate at which the base is changing (db/dt) when the height is 7 centimeters and the area is 91 square centimeters.
We can start by differentiating the area formula with respect to time (t):
dA/dt = (1/2) * (base * dh/dt + height * db/dt).
Now, we substitute the given values into the equation:
2 = (1/2) * (base * 1 + 7 * db/dt).
Simplifying the equation, we get:
4 = base + 7 * db/dt.
Rearranging the equation to solve for db/dt:
db/dt = (4 - base) / 7.
To find db/dt when the height is 7 centimeters and the area is 91 square centimeters, we substitute these values into the equation:
db/dt = (4 - base) / 7.
We are given that the height is 7 centimeters (height = 7) and the area is 91 square centimeters (Area = 91).
Substituting these values into the equation, we get:
db/dt = (4 - base) / 7 = (4 - base) / 7.
Now, we solve for db/dt:
db/dt = (4 - base) / 7 = (4 - 7) / 7 = -3/7.
Therefore, the base of the triangle is changing at a rate of -3/7 cm/min when the height is 7 centimeters and the area is 91 square centimeters.
A = (1/2) b h
dA/dt = (1/2)(b dh/dt + h db/dt)
given: dA/dt = 2 , dh/dt = 1 , A = 91 , h = 7 ,
we need the base b,
from A = 1/2)bh
91 = (1/2)(7b)
182 = 7b
b = 26
2 = (1/2)(26(1) + 7db/dt)
4 = 23 + 7db/dt
db/dt = -19/7 cm/min
At that moment, the base is decreasing at 19/7 cm/min