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

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 8 Differentials and Partial Derivatives Ex 8.6 Textbook Questions and Answers, Notes.

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 \(\frac { du }{ dt }\) and evaluate if at t = 0
Solution:
u (x, y) = x²y + 3xy⁴, x = e^t, y = sin t

∴ \(\frac { du }{ dt }\) = (2xy + 3y⁴) e^t + (x² + 12xy³) cos t
= (2e^t sin t + 3 sin⁴t) e^t + (e^2t + 12 e^t sin³t) cos t
\(\frac { du }{ dt }\) = e^t(2 e^t sin t + 3 sin⁴t + e^t cos t + 12 sin³t cos t)
at t = 0
\(\frac { du }{ dt }\) = e⁰(2 e⁰ sin 0 + 3 sin⁴ 0 + e⁰ cos 0 + 12 sin³0 cos 0)
= 1(0 + 0 + 1 + 0) (cos (0) = 1, sin(0) = 0, e⁰ = 1)
\(\frac { du }{ dt }\) = 1

Question 2.
Let u (x, y, z) = xy²z³ x = sin t, y = cos t, z = 1 + e^2t, Find \(\frac { du }{ dt }\).
Solution:
u (x, y, z) = xy²z³ x = sin t, y = cos t, z = 1 + e^2t

= y²z³ cos t – 2xyz³ sin t + 6 xy²z² e^2t
= cos²t (1 + e^2t)³ cost – 2 (sin t) cost (1 + e^2t)³ sin t + 6 (sin t) cos²t) (1 + e^2t)² e^2t
= (1 + e^2t)² [cos³t (1 + e^2t) – sin t sin2t (1 + e^2t) + 6 e^2t sin t cos²t]

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

\(\frac { dw }{ dt }\) = 2x e^t + 2y (e^t sin t + e^t cos t) + 2z (e^t cos t – e^t sin t)
= 2 e^2t + 2 (e^t sin t) (e^t sin t + e^t cos t) + 2 (e^t cos t) (e^t cos t – e^t sin t)
= 2 e^2t [1 + sin²t + sin t cos t + cos²t – sin t cos t]
= 2 e^2t (1 + sin²t + cos²t)
[∵ sin²t + cos²t = 1]
= 2 e²t (1 + 1) = 4 e^2t
\(\frac { dw }{ dt }\) = 4 e^2t

Question 4
Let U(x, y, z) = xyz, x = e^-t, y = e^-t cos t, z – sin t, t ∈ R, find \(\frac { dU }{ dt }\)
Solution:
U(x, y, z) = xyz, x = e^-t, y = e^-t cos t

\(\frac { dU }{ dt }\) = -(e^-t cos t sin t) e^-t + e^-t sin t [ e^-t (cos t – sin t )] + e^-2t cos t (cos t)
= -e^-2t cos t sin t – e^-2t sin t cos t – e^-2t sin²t + e^-2t cos²t
= -e^-2t (2 sin t cos t + sin²t – cos²t)
= -e^-2t [sin 2t – (cos²t – sin²t)]
= -e^-2t (sin 2t + cos 2t)

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

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

Question 7.
Let U (x, y) = e^x sin y where x = st², y = s²t, s, t ∈ R. Find \(\frac{\partial U}{\partial s}\), \(\frac{\partial U}{\partial t}\) and evaluate then s = t = 1. dt
Solution:
U (x, y) = e^x sin y, x = st², y = s²t

= e^st2 sin (s²t) t² + e^st2 cos(s²t) 2st
= e^st2 [t² sin (s²t) + 2st cos (s²t)]
= t e^x [t sin(s²t) + 2s cos (s²t)]
\(\frac{\partial U}{\partial t}\) = e^x sin y 2st + e^x cos y (s²)
= e^st2 sin(s²t) 2st + e^st2 cos(s²t) s²
= s e^st2 [2t sin (s²t) + s cos(s²t)]
At s = t = 1
\(\frac{\partial U}{\partial s}\) = e[sin(1) + 2 cos (1)]
= e[sin(1) + 2 cos (1)]
\(\frac{\partial U}{\partial t}\) = e[2 sin(1) + cos (1)]

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:
z (x, y) = x³ – 3x²y³

(3x² – 6xy³) e^t – 9 (xy)² e^-t
[3 (s e^t)² – 6 (s e^t) (se^-t)³]e^t
-9 (s e^t s e^-t)² × e^-t
= (3 s² e^2t – 6 s⁴ e^-2t) e^t – 9 s⁴ e^-t
= 3s² [(e^2t – 2s² e^-2t) e^t -3 e^-t s²]
\(\frac{\partial z}{\partial s}\) = 3s² e^t (e^2t -2s² e^-2t – 3 e^-2t s²)
\(\frac{\partial z}{\partial t}\) = \(\frac{\partial z}{\partial x}\) \(\frac{\partial x}{\partial t}\) + \(\frac{\partial z}{\partial y}\) \(\frac{\partial y}{\partial t}\)
= (3x² – 6xy³) (s e^t)+ (-9x²y²) (- s e^-t)
= [3 (s e^t)² – 6 (s e^t) (s e^-t)³] s e^t + 9(se^t s e^-t)² s e^-t
= (3 s² e^2t – 6 s⁴ e^-2t) s e^t + 9 s⁵ e^-t
= 3 s³ e^3t – 6 s⁵ e^-t + 9 s⁵ e^-t
= 3 s³ e^3t + 3 s⁵ e^-t
\(\frac{\partial y}{\partial t}\) = 3 s³ (e^3t + s² e^-t)

Question 9.
W(x, y, z) = xy + yz + zx, x = u – v, y = uv, z = u + v, u, v ∈ R. Find \(\frac{\partial W}{\partial u}\), \(\frac{\partial W}{\partial v}\) evaluate then at (\(\frac{1}{2}\), 1)
Solution:

= (y + z) × 1 + (x + z) × v + (y + x) × 1
= uv + u + v + (u – v + u + v) v+ (uv + u – v)
= uv + u + v + uv + uv + uv + u – v
= 4 uv + 2u
\(\frac{\partial w}{\partial u}\) (\(\frac{1}{2}\), 1) = 4 × \(\frac{1}{2}\) × 1 + 2 × \(\frac{1}{2}\) = 2 + 1 = 3
\(\frac{\partial w}{\partial v}\) = \(\frac{\partial W}{\partial x}\) \(\frac{\partial x}{\partial v}\) + \(\frac{\partial W}{\partial y}\) \(\frac{\partial y}{\partial v}\) + \(\frac{\partial W}{\partial z}\) \(\frac{\partial z}{\partial v}\)
= (y + z) (-1) + (x + z) u + (y + x) × 1
= -y – z + xu + zu + y + x
= -u – v + (u – v) u + (u + v) u + u – v
= -u – v + u² – vu + u² + vu + u – v
= 2u² – 2v
\(\frac{\partial W}{\partial v}\) (\(\frac{1}{2}\), 1) = 2 × \(\frac{1}{4}\) – 2 × 1
= \(\frac{1}{2}\) – 2 = –\(\frac{3}{2}\)