Suppose z = x^2 sin y, x = 4s^2 + 5t^2, y = -6st. A. Use the chain rule to find

Question

Suppose z = x^2 sin y, x = 4s^2 + 5t^2, y = -6st. A. Use the chain rule to find partial differential z/partial differential s and partial differential z/partial differential t as functions of x, y, s and t. partial differential z/partial differential s = 4x8siny(-6t)+ x^2cosy(-8s) partial differential z/partial differential t = -12xtsiny-8x^2scos7 B. Find the numerical values of partial differential z/partial differential z and partial differential z/partial differential z when (s, t) = -3, -2). partial differential z/partial differential s (-3, -2) = 2197.5 partial differential z/partial differential y = (-3, -2) = 2403.9

Explanation / Answer

z = x^2*sin y

x = 4s^2 + 5t^2

y = -6*s*t

using derivatives

dz/dx = 2*x*sin y

dz/dy = x^2*cos y

dx/ds = 2*4*s + 0 = 8*s

dx/dt = 0 + 2*5*t = 10*t

dy/ds = -6*1*t = -6*t

dy/dt = -6*s*1 = -6*s

Now

dz/ds = (dz/dx)*(dx/ds) + (dz/dy)*(dy/ds) = (2*x*sin y)*(8*s) + (x^2*cos y)*(-6*t)

dz/ds = 16*x*s*sin y - 6*x^2*t*cos y

And

dz/dt = (dz/dx)*(dx/dt) + (dz/dy)*(dy/dt) = (2*x*sin y)*(10*t) + (x^2*cos y)*(-6*s)

dz/dt = 20*x*t*sin y - 6*x^2*s*cos y

B.

when (s, t) = (-3, -2)

at (-3, -2), x = 4*s^2 + 5*t^2

x = 4*(-3)^2 + 5*(-2)^2 = 56

y = -6*(-3)*(-2) = -36

So,

dz/ds(-3, -2) = 16*x*s*sin y - 6*x^2*t*cos y

= 16*56*(-3)*sin (-36) - 6*(56)^2*(-2)*cos (-36)

= -7481.43

And

dz/dt(-3, -2) = 20*x*t*sin y - 6*x^2*s*cos y

= 20*56*(-2)*sin (-36) - 6*(56)^2*(-3)*cos (-36) = -9444.88

Let me know if you have any doubt.

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.