Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.6

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.6

Question 1.
If u(x, y) = x²y + 3xy⁴, x = e^t and y = sin t, find \(\frac{d u}{d t}\) and evaluate it at t = 0.
Solution:

= (2xy + 3y⁴) (e^t) + (x² + 12xy³) (cos t)
= (2e^t sin t + 3 sin⁴ t) e^t + [e^2t + 12e^t sin³ t] cos t
= e^t [2e^t sin t + 3 sin⁴ t + e^t (cos t) + 12 sin³t cos t]
at t = 0

Question 2.

Solution:

Question 3.
If w(x, y, z) = x² + y² + z², x = e^t, y = e^t sin t, and z = e^tcos t, find \(\frac{d w}{d t}\)
Solution:
w(x, y, z) = x² + y² + z² ; x = e^t ; y = e^t sin t, z = e^t cos t

Question 4.

Solution:

(Replace x, y, z value)

Question 5.
If w(x, y) = 6x³ – 3xy + 2y², x = e^s, y = cos s, s ∈ R, find \(\frac{d w}{d s}\), and evaluate at s = 0.
Solution:

Question 6.
If z(x, y) = x tan^-1 (x y), x = t², y = s e^t, s, t ∈ R, Find \(\frac{\partial z}{\partial \mathbf{t}}\) and \(\frac{\partial z}{\partial \mathbf{t}}\) at s = t = 1
Solution:
z (x, y) = x tan^-1 (xy) ; x = t² ; y = se^t

Question 7.
Let U (x, y) = e^x sin y, where x = st², y = s²t, s, t ∈ R. Find them at s = t = 1.
Solution:

(Replace x, y values)

Question 8.
Let z(x, y) = x³ – 3x²y³, where x = se^t, y = se^-t, s, t ∈ R. Find \(\frac{\partial z}{\partial s}\) and \(\frac{\partial z}{\partial t}\)
Solution:

Question 9.
W(x,y, z) = xy + yz + zx, x = u -v, y = uv, z = u + v, u, v e R. Find \(\frac{\partial \boldsymbol{w}}{\partial \boldsymbol{u}}\), \(\frac{\partial \boldsymbol{w}}{\partial \boldsymbol{v}}\) them at \(\left(\frac{1}{2}, 1\right)\)
Solution:

Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.6 Additional Problems

Question 1.
Suppose that z = \(y e^{x^{2}}\) where x = 2t and y = 1 – t then find \(\frac{d z}{d t}\).
Solution:


(Since x = 2t and y = 1 – t)

Question 2.
If w = x + 2y + z² and x = cos t ; y = sin t ; z = t. Find \(\frac{d w}{d t}\).
Solution: