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2 answers:
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I'll assume that's an exponent at the end there:
The first derivative gives the velocity. The second derivative gives the acceleration; increasing velocity is the same as positive acceleration. So we want to find when the second derivative is positive.
Let's see if we can use to avoid multiplying this out.
That worked; let's do it again:
That's a nice parabola. It's zero or negative for so positive everywhere else:
Answer: Increasing velocity when t < 1 or t > 3
Answer:
The solutions are 0° and 3π
Step-by-step explanation:
On solving the equation given;
<u>Since sin is negative in the 3rd and 4th quadrant, </u>
In the 3rd quadrant;
x =180°+2π
x = π + 2π
x = 3π
In the 4th quadrant;
x = 360°-2π
x = 2π-2π
x = 0°