# Thread: Help with a Derivative question that involves Sin

1. ## Help with a Derivative question that involves Sin

Hey i have this question given to me by teacher that i just cant wrap my head around i know the derivative of sin(x) is cos(x) but the way its done confuses me here's the question:

The self induced e.m.f (measured in volts, V) in an inductor can be found by using: e= L𝒅𝒊 𝒅𝒕 V, where L is the inductance in henrys. Calculate the induced e.m.f. when an inductor of 270mH passes a current: i = Bsin(314t+𝜋 6 ) ampere at the instant when time t = 10ms.(See table for your value of B)

2. ## Re: Help with a Derivative question that involves Sin

Yes, the derivative of sin(x) is cos(x) so the derivative of $\displaystyle Bsin(314t+ 6\pi)$ is $\displaystyle cos(314t+ 6\pi)$ times the derivative of $\displaystyle 314t+ 6\pi$ which is just 314. The derivative of $\displaystyle Bsin(314t+ 6\pi)$ is $\displaystyle 314B cos(314t+ 6\pi)$.

(That "$\displaystyle \pi 6$" looks strange. Did you mean "$\displaystyle \pi/6$" and the "/" got lost? If so the derivative of $\displaystyle Bsin(314t+ \pi/6)$ is $\displaystyle 314B cos(314t+ \pi/6)$.)

3. ## Re: Help with a Derivative question that involves Sin

Oh yeah sorry i didnt notice it yeah its mean to be π/6

so would it be 314Bcos(314t+π/6) * 314t + 6π or just leave it at the cos part and what would happen if you replace the t with 10/10^3 for the ms and B with 12.2

4. ## Re: Help with a Derivative question that involves Sin

No, you multiply the cosine by the derivative of $\displaystyle 314t+ \pi/6$. You do NOT multiply it by $\displaystyle (314t+ \pi/6)$ again!

You now want to evaluate that at t= 10/10^3? That is the same as 1/100= 0.01. Setting t= 0.01 and B= 12.2 we would have $\displaystyle (314)(12.2) cos(3.14+ \pi/6)$. 3.14 is approximately $\displaystyle \pi$ itself so we can approximate $\displaystyle 3.14+ \pi/6$ as $\displaystyle 7\pi/6$ and [math]7 cos(7\pi/6)= -0.8660 ($\displaystyle -\sqrt{3}/2$). (3.14)(12.2)(-0.8660)= -33.17.