1/tan(x) is right

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Posted on 2005-04-15

I have two quick questions. The first is, if a ball is thrown from a bridge 25 meter above the ground at 49m/s, is the acceleration negative? I thought since it was thrown up, it would be positive, but the book has a similar problem, a ball is thrown upward at a speed of 48ft/s from a cliff 432ft above the ground, and it says the acceleration is -32.

Second, what is the antiderivative of (x^2/sqrt(x)) + (1/sin^2(x))?

I can got:

((2/5)x^(5/2)) + (1/tan(x)) + c

But I am pretty sure (1/tan(x)) isn't right.

Second, what is the antiderivative of (x^2/sqrt(x)) + (1/sin^2(x))?

I can got:

((2/5)x^(5/2)) + (1/tan(x)) + c

But I am pretty sure (1/tan(x)) isn't right.

7 Comments

1/tan(x) is right

integral( 1/sin^2x )dx =integral( csc^2x )dx = - cot x + C

cosecant is 1/sin.

The integral of 1/sin^2(x) is -cot(x) + c or -1/tan(x) + c

The integral of 1/cos^2(x) is tan(x) + c.

The acceleration can be thought of as either positive or negative depending on whether the initial velocity is thought of as positive or negative.

If the initial velocity is negative then the acceleration is positive. If the initial velocity is positive then the acceleration is negative.

The choice of whether the initial velocity of the ball is positive or negative is entirely up to the author though it is standard to say that acceleration due to gravity is (positive) 9.8 m/s² so if the ball was thrown upwards the initial velocity is negative.

Just in case you're interested gravity at the North Pole is 9.832 m/s², Java (near the equator) it's 9.782 m/s².

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