Answer:
[tex]\frac{dy}{dt}=\sin t(-b+2t)+t^2\cos t[/tex]
Step-by-step explanation:
Given : Equation [tex]y=b\cos t+t^2\sin t[/tex]
To find : Differentiate with respect to t ?
Solution :
Equation [tex]y=b\cos t+t^2\sin t[/tex]
Differentiate with respect to t,
[tex]\frac{dy}{dt}=\frac{d(b\cos t+t^2\sin t)}{dt}[/tex]
[tex]\frac{dy}{dt}=\frac{d(b\cos t)}{dt}+\frac{d(t^2\sin t)}{dt}[/tex]
We know, [tex]\frac{d(\cos x)}{dx}=-\sin x[/tex]
Applying product rule, [tex]\frac{d}{dy}(u\cdot v) =u'v+v'u[/tex]
[tex]\frac{dy}{dt}=-b\sin t+\frac{d(t^2)}{dt}\times \sin t+\frac{d(\sin t)}{dt}\times t^2[/tex]
[tex]\frac{dy}{dt}=-b\sin t+2t\times \sin t+\cos t\times t^2[/tex]
[tex]\frac{dy}{dt}=-b\sin t+2t\sin t+t^2\cos t[/tex]
[tex]\frac{dy}{dt}=\sin t(-b+2t)+t^2\cos t[/tex]
Therefore, [tex]\frac{dy}{dt}=\sin t(-b+2t)+t^2\cos t[/tex]
