Final answer:
To calculate dz/dt, the chain rule and product rule for differentiation were applied to the given functions of z, x, and y. This resulted in dz/dt = 6e^(1-t^2) - 30t^2e^(1-t^2) when expressed in terms of t.
Step-by-step explanation:
To find dz/dt in terms of t when z=(x y)e^y, x=6t, and y=1-t^2, we must use the chain rule and the product rule., we differentiate z with respect to t:dz/dt = d/dt [(6t)(1-t^2)e^(1-t^2)Using the product rule, dz/dt = (d/dt [(6t)(1-t^2)] ) e^(1-t^2) + (6t)(1-t^2) d/dt [e^(1-t^2)].
Next, we continue by differentiating each part:d/dt [(6t)(1-t^2)] = 6(1-t^2) - 12t^2d/dt [e^(1-t^2)] = -2te^(1-t^2) (because of the chain ruleCombining and simplifyingdz/dt = [(6(1-t^2) - 12t^2) e^(1-t^2)] + [-2t(6t)(1-t^2)e^(1-t^2)]dz/dt = [6e^(1-t^2) - 18t^2e^(1-t^2) - 12t^2e^(1-t^2)dz/dt = 6e^(1-t^2) - 30t^2e^(1-t^Therefore, dz/dt in terms of t is 6e^(1-t^2) - 30t^2e^(1-t^2).To find dz/dt, we first need to express z in terms of t. Given:z = (x y)e^yx = 6ty = 1-t^2Substituting the values of x and y into the expression for zz = (6t)(1-t^2)e^(1-t^2)Next, we can differentiate z with respect to t using the product rule and chain rule:dz/dt = d/dt[(6t)(1-t^2)e^(1-t^2)]Using the product rule and chain rule, we obtain:dz/dt = (6(1-t^2)e^(1-t^2) + (6t)(-2t)e^(1-t^2)) + (6t)(1-t^2)(-2t)e^(1-t^2)Simplifying the expression gives:dz/dt = 6e^(1-t^2) - 12t(1-t^2)e^(1-t^2) - 12t^2(1-t^2)e^(1-t^2)