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Please help. Calculus

Please help. Calculus-example-1
User Donald
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1 Answer

15 votes
15 votes

Answer:

0.339 m/min

Explanation:

You want to know the rate of change of the third side in a triangle with sides 10 m and 14 m and an angle between them of 60°. The angle is increasing at 2°/min.

Law of Cosines

The law of cosines tells you the third side (c) satisfies the relation ...

c² = a² +b² -2ab·cos(C)

Filling in the given values, we have ...

c² = 10² +14² -2(10)(14)cos(C) = 296 -280cos(C)

Rate of change

Taking the square root and differentiating with respect to time, we have ...

c = √(296 -280cos(C))

c' = 280sin(C)·C'/(2√(296 -280cos(C)))

We want the value of this when C=60°, and C' = 2°/min = π/90 rad/min.

c' = 280(sin(60°))·(π/90)/(2√(296 -280cos(60°))) = (14π√3/9)/(2√156)

c' ≈ 0.339 . . . . m/min

The third side is increasing at a rate of about 0.339 meters per minute.

Please help. Calculus-example-1
User Jayeshkv
by
2.8k points
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