Respuesta :
Answer:
The value of [tex]\frac{\partial w}{\partial s}[/tex] is [tex]e^{3t}-18e^{2s+t}[/tex]
The value of [tex]\frac{\partial w}{\partial t}[/tex] is [tex]3e^{3t}-9e^{2s+t}[/tex]
The partial derivative at s=-5 and t=10 is [tex]\frac{\partial w}{\partial s}[/tex] is [tex]e^{30}-18[/tex]
The partial derivative at s=-5 and t=10 is [tex]\frac{\partial w}{\partial t}=3(e^{30}-3)[/tex]
Step-by-step explanation:
Given that the Function point are [tex]w=y^3-9x^2y[/tex]
[tex]x=e^s[/tex], [tex]y=e^t[/tex] and s = -5, t = 10
To find [tex]\frac{\partial w}{\partial s}[/tex] and [tex]\frac{\partial w}{\partial t}[/tex]using the appropriate Chain Rule :
[tex]w=y^3-9x^2y[/tex]
Substitute the values of x and y in the above equation we get
[tex]w=(e^t)^3-9(e^s)^2(e^t)[/tex]
[tex]w=e^{3t}-9e^{2s}.e^t[/tex]
Now partially differentiating w with respect to s by using chain rule we have
[tex]\frac{\partial w}{\partial ∂s}=e^{3t}-9(e^{2s}).2(e^t)[/tex]
[tex]=e^{3t}-18e^{2s}.(e^t)[/tex]
[tex]=e^{3t}-18e^{2s+t}[/tex]
Therefore the value of [tex]\frac{\partial w}{\partial s}[/tex] is [tex]e^{3t}-18e^{2s+t}[/tex]
[tex]w=e^{3t}-9e^{2s}.e^t[/tex]
Now partially differentiating w with respect to t by using chain rule we have
[tex]\frac{\partial w}{\partial t}=e^{3t}.(3)-9e^{2s}(e^t).(1)[/tex]
[tex]=3e^{3t}-9e^{2s+t}[/tex]
Therefore the value of [tex]\frac{\partial w}{\partial t}[/tex] is [tex]3e^{3t}-9e^{2s+t}[/tex]
Now put s-5 and t=10 to evaluate each partial derivative at the given values of s and t :
[tex]\frac{\partial w}{\partial s}=e^{3t}-18e^{2s+t}[/tex]
[tex]=e^{3(10}-18e^{2(-5)+10}[/tex]
[tex]=e^{30}-18e^{-10+10}[/tex]
[tex]=e^{30}-18e^0[/tex]
[tex]=e^{30}-18[/tex]
Therefore the partial derivative at s=-5 and t=10 is [tex]\frac{\partial w}{\partial s}[/tex] is [tex]e^{30}-18[/tex]
[tex]\frac{\partial w}{\partial t}=3e^{3t}-9e^{2s+t}[/tex]
[tex]=3e^{3(10)}-9e^{2(-5)+10}[/tex]
[tex]=3e^{30}-9e{-10+10}[/tex]
[tex]=3e^{30}-9e{0}[/tex]
[tex]=3e^{30}-9[/tex]
[tex]\frac{\partial w}{\partial t}=3(e^{30}-3)[/tex]
Therefore the partial derivative at s=-5 and t=10 is [tex]\frac{\partial w}{\partial t}=3(e^{30}-3)[/tex]
Using the Chain Rule, it is found that:
[tex]\frac{\partial W}{\partial s}(-5,10) = -18[/tex]
[tex]\frac{\partial W}{\partial s}(-5,10) = 3e^{30} - 9[/tex]
w is a function of x and y, which are functions of s and t, thus, by the Chain Rule:
[tex]\frac{\partial W}{\partial s} = \frac{\partial W}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial W}{\partial y}\frac{\partial y}{\partial s}[/tex]
[tex]\frac{\partial W}{\partial t} = \frac{\partial W}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial W}{\partial y}\frac{\partial y}{\partial t}[/tex]
Then, the derivatives are:
[tex]\frac{\partial W}{\partial x} = -18xy[/tex]
[tex]\frac{\partial x}{\partial s} = e^s[/tex]
[tex]\frac{\partial W}{\partial y} = 3y^2 - 9x^2[/tex]
[tex]\frac{\partial y}{\partial s} = 0[/tex]
Then
[tex]\frac{\partial W}{\partial s} = \frac{\partial W}{\partial x}\frac{\partial x}{\partial s}[/tex]
[tex]\frac{\partial W}{\partial s} = -18xy(e^s)[/tex]
[tex]\frac{\partial W}{\partial s} = -18e^se^t(e^s)[/tex]
[tex]\frac{\partial W}{\partial s} = -18e^{2s}e^t[/tex]
[tex]\frac{\partial W}{\partial s} = -18e^{2s + t}[/tex]
At s = -5 and t = 10:
[tex]\frac{\partial W}{\partial s}(-5,10) = -18e^{2(-5) + 10} = -18[/tex]
Then, relative to t:
[tex]\frac{\partial x}{\partial t} = 0[/tex]
[tex]\frac{\partial y}{\partial t} = e^t[/tex]
[tex]\frac{\partial W}{\partial t} = \frac{\partial W}{\partial y}\frac{\partial y}{\partial t}[/tex]
[tex]\frac{\partial W}{\partial t} = (3y^2 - 9x^2)e^t[/tex]
[tex]\frac{\partial W}{\partial t} = (3e^{2t} - 9e^{2s})e^t[/tex]
At s = -5 and t = 10:
[tex]\frac{\partial W}{\partial s}(-5,10) = (3e^{20} - 9e^{-10})e^{10} = 3e^{30} - 9[/tex]
A similar problem is given at https://brainly.com/question/10309252
