Answer:
[tex]\frac{dT}{dt} = -10cos(t)sin(t)e^{sin(t)} + 5cos^3(t)e^{sin(t)} + 6sin^4(t) - 18cos^2(t)sin^2(t)[/tex]
Step-by-step explanation:
(a) By chain rule:
[tex]\frac{dx}{dt} = \frac{dcos(t)}{dt} = -sin(t)[/tex]
[tex]\frac{dy}{dt} = \frac{dsin(t)}{dt} = cos(t)[/tex]
[tex]\frac{dT}{dt} = d\frac{5x^2e^y}{dt} - d\frac{6xy^3}{dt}[/tex]
We can first use product rule to split up
[tex]\frac{dT}{dt} = 5(\frac{dx^2}{dt}e^y + x^2\frac{de^y}{dt}) - 6(\frac{dx}{dt}y + x\frac{dy^3}{dt}[/tex]
From here apply chain rule for dx/dt and dy/dt we get
[tex]\frac{dT}{dt} = 5(2x(-sin(t))e^y + x^2e^ycos(t)) - 6(-sin(t)y^3 + x3y^2cos(t))[/tex]
[tex]\frac{dT}{dt} = -10cos(t)sin(t)e^{sin(t)} + 5cos^3(t)e^{sin(t)} + 6sin^4(t) - 18cos^2(t)sin^2(t)[/tex]
(b) by expressing T in term of t, we can substitute t for x and y in T
[tex]T = 5x^2e^y - 6xy^3 = 5cos^2(t)e^{sin(t)} - 6cos(t)sin^3(t)[/tex]
[tex]\frac{dT}{dt} = 5(2cos(t)(-sin(t))e^{sin(t)} + cos^3(t)e^{sin(t)}) - 6(-sin^4(t) + cos^2(t)3sin^2(t))[/tex]
[tex]\frac{dT}{dt} = -10cos(t)sin(t)e^{sin(t)} + 5cos^3(t)e^{sin(t)} + 6sin^4(t) - 18cos^2(t)sin^2(t)[/tex]
