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

∴ $$\frac { du }{ dt }$$ = (2xy + 3y4) et + (x² + 12xy³) cos t
= (2et sin t + 3 sin4t) et + (e2t + 12 et sin3t) cos t
$$\frac { du }{ dt }$$ = et(2 et sin t + 3 sin4t + et 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 + e2t, Find $$\frac { du }{ dt }$$.
Solution:
u (x, y, z) = xy²z³ x = sin t, y = cos t, z = 1 + e2t

= y²z³ cos t – 2xyz³ sin t + 6 xy²z² e2t
= cos²t (1 + e2t)³ cost – 2 (sin t) cost (1 + e2t)³ sin t + 6 (sin t) cos2t) (1 + e2t)² e2t
= (1 + e2t)² [cos³t (1 + e2t) – sin t sin2t (1 + e2t) + 6 e2t sin t cos²t]

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

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

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 = es, 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 et, 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) = ex 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) = ex sin y, x = st², y = s²t

= est2 sin (s²t) t² + est2 cos(s²t) 2st
= est2 [t² sin (s²t) + 2st cos (s²t)]
= t ex [t sin(s²t) + 2s cos (s²t)]
$$\frac{\partial U}{\partial t}$$ = ex sin y 2st + ex cos y (s²)
= est2 sin(s²t) 2st + est2 cos(s²t) s²
= s est2 [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 = set, 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) et – 9 (xy)² e-t
[3 (s et)² – 6 (s et) (se-t)³]et
-9 (s et s e-t)² × e-t
= (3 s² e2t – 6 s4 e-2t) et – 9 s4 e-t
= 3s² [(e2t – 2s² e-2t) et -3 e-t s²]
$$\frac{\partial z}{\partial s}$$ = 3s² et (e2t -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 et)+ (-9x²y²) (- s e-t)
= [3 (s et)² – 6 (s et) (s e-t)³] s et + 9(set s e-t)² s e-t
= (3 s² e2t – 6 s4 e-2t) s et + 9 s5 e-t
= 3 s³ e3t – 6 s5 e-t + 9 s5 e-t
= 3 s³ e3t + 3 s5 e-t
$$\frac{\partial y}{\partial t}$$ = 3 s³ (e3t + 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 = –$$\frac{3}{2}$$