Find $\frac{d z}{d \theta}$ if
(i) $z=\sin ^n \theta$
(ii) $z=\cos ^m \theta$
(iii) $z=\sin ^n \theta \cos ^m \theta$.
Answer (1)
For z = sin n θ , apply the chain rule to get d θ d z = n ( sin θ ) n − 1 cos θ .
For z = cos m θ , apply the chain rule to get d θ d z = − m ( cos θ ) m − 1 sin θ .
For z = sin n θ cos m θ , use the product rule and chain rule to find d θ d z = n ( sin θ ) n − 1 ( cos θ ) m + 1 − m ( sin θ ) n + 1 ( cos θ ) m − 1 .
The derivatives are d θ d z = n ( sin θ ) n − 1 cos θ , − m ( cos θ ) m − 1 sin θ , n ( sin θ ) n − 1 ( cos θ ) m + 1 − m ( sin θ ) n + 1 ( cos θ ) m − 1 .
Explanation
Problem Analysis We are given three expressions for z in terms of θ , and we need to find d θ d z for each case. We will use the chain rule and product rule as necessary.
Finding derivative of z = sin^n(theta) (i) z = sin n θ Let u = sin θ , so z = u n . Then d u d z = n u n − 1 and d θ d u = cos θ . Thus, by the chain rule, d θ d z = d u d z d θ d u = n ( sin θ ) n − 1 cos θ .
Finding derivative of z = cos^m(theta) (ii) z = cos m θ Let v = cos θ , so z = v m . Then d v d z = m v m − 1 and d θ d v = − sin θ . Thus, by the chain rule, d θ d z = d v d z d θ d v = m ( cos θ ) m − 1 ( − sin θ ) = − m ( cos θ ) m − 1 sin θ .
Finding derivative of z = sin^n(theta)cos^m(theta) (iii) z = sin n θ cos m θ Let f ( θ ) = sin n θ and g ( θ ) = cos m θ . Then z = f ( θ ) g ( θ ) , so by the product rule, d θ d z = f ′ ( θ ) g ( θ ) + f ( θ ) g ′ ( θ ) .
From parts (i) and (ii), f ′ ( θ ) = n ( sin θ ) n − 1 cos θ and g ′ ( θ ) = − m ( cos θ ) m − 1 sin θ . Thus, d θ d z = n ( sin θ ) n − 1 cos θ ( cos m θ ) + ( sin n θ ) ( − m ( cos θ ) m − 1 sin θ ) d θ d z = n ( sin θ ) n − 1 ( cos θ ) m + 1 − m ( sin θ ) n + 1 ( cos θ ) m − 1 .
Final Answer Therefore, the derivatives are: (i) d θ d z = n ( sin θ ) n − 1 cos θ (ii) d θ d z = − m ( cos θ ) m − 1 sin θ (iii) d θ d z = n ( sin θ ) n − 1 ( cos θ ) m + 1 − m ( sin θ ) n + 1 ( cos θ ) m − 1
Examples
Understanding derivatives of trigonometric functions is crucial in physics, especially when dealing with oscillatory motion. For example, when analyzing the motion of a pendulum, the position can be described using sine and cosine functions. Finding the derivative helps determine the velocity and acceleration of the pendulum at any given time, which is essential for understanding its dynamics and energy transfer.
