Asked by Anonymous
The altitude (i.e., height) of a triangle is increasing at a rate of 2 cm/minute while the area of the triangle is increasing at a rate of 4 square cm/minute. At what rate is the base of the triangle changing when the altitude is 10.5 centimeters and the area is 97 square centimeters?
Answers
Answered by
Jai
Area of triangle:
A = (1/2)bh
When A = 97 and h = 10.5,
A = (1/2)bh
97 = (1/2)(b)(10.5)
b = 18.48
Differentiate area with respect to time:
A = (1/2)bh
dA/dt = (1/2)( h*db/dt + b*dh/dt )
Substitute the values:
4 = (1/2)( 10.5*db/dt + 18.48*2)
Solve for db/dt.
hope this helps~ `u`
A = (1/2)bh
When A = 97 and h = 10.5,
A = (1/2)bh
97 = (1/2)(b)(10.5)
b = 18.48
Differentiate area with respect to time:
A = (1/2)bh
dA/dt = (1/2)( h*db/dt + b*dh/dt )
Substitute the values:
4 = (1/2)( 10.5*db/dt + 18.48*2)
Solve for db/dt.
hope this helps~ `u`
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