Part 1:. Solution 1a): Given that the voltage waveform.

Use calculus to determine engine parts quality Missing values A=1.5 B=1.4 C=1.1 D=100

Part 1:
Solution 1a): Given that the voltage waveform . We know that the rate of
change of function is .
So, rate of change of is
(By using product rule of derivative)
At seconds, the rate of change is
Hence, rate of change is volts per second decrease.
Solution 1b): Given . We know that
So,
Differentiating with respect to x we get
At , the gradient is
Solution 2: The displacement function is .
The velocity of the end of the spring is
At
… (1)
After seconds, the velocity of the end of the spring is
The acceleration of the end of the spring is
Differentiating equation (1) with respect to t we get
After seconds, acceleration of the end of the spring is
Solution 3: the noise of an engine in decibels is given by , where t is
the time in seconds.
Now, the rate of change of noise is .
So,
At time , the rate of change of noise is
Part 2:

Solution 1: Given, the force , where x is the displacement. The work
done is
Solution 2: Given that . Aim is to find E.
Using integration by part method
Where C is constant of integration.
Solution 3: The alternating voltage v is given as
.
The mean value of the voltage over one cycle is
And RMS of the voltage over one cycle is
Part 3: To find the solution of the differential equation with initial
condition .
Now,
Simplify further we get
Since this implies that . Then the required solution is

a): After ,
b): Suppose that the current take t second to decay 50% of its initial value, then
Solution 2: Given that also given that
So,
Since, this implies that . So
After