Math / CalculusQuestionSuppose that w=x2⋅exp(2y)⋅cos(6z)w=x^{2} \cdot \exp (2 y) \cdot \cos (6 z)w=x2⋅exp(2y)⋅cos(6z) with x=sin(t+π2)y=ln(t+7)z=t\begin{array}{c} x=\sin \left(t+\frac{\pi}{2}\right) \\ y=\ln (t+7) \\ z=t \end{array}x=sin(t+2π)y=ln(t+7)z=t a. Find dw dt\frac{\mathrm{d} w}{\mathrm{~d} t} dtdw in terms of ttt. dw dt=\frac{\mathrm{d} w}{\mathrm{~d} t}= dtdw= aba^{b}ab sin(a)∂∂xf\sin (a) \quad \frac{\partial}{\partial x} fsin(a)∂x∂f : ∞\infty∞ α\alphaα Ω\OmegaΩStuddy Solutiondwdt=−2(t+7)2sin(t)cos(t)cos(6t)+2(t+7)cos2(t)cos(6t)−6(t+7)2cos2(t)sin(6t) \frac{dw}{dt} = -2(t+7)^2\sin(t)\cos(t)\cos(6t) + 2(t+7)\cos^2(t)\cos(6t) - 6(t+7)^2\cos^2(t)\sin(6t) dtdw=−2(t+7)2sin(t)cos(t)cos(6t)+2(t+7)cos2(t)cos(6t)−6(t+7)2cos2(t)sin(6t)View Full Solution - FreeWas this helpful?
