Question

$\large\rm { \bullet}$ Given that,

$\large\rm { \nabla \phi = \bigg ( 5x - \frac{9}{2} \bigg ) \bar{i} +(-z) \bar{j} + (-y) \bar{k} }$ and $\large\rm { \nabla \Psi = 8xy \bar{i} + 4x^{2} \bar{j} +3z^{2} \bar{k} }$

To find:-

The angle between the surfaces $\large\rm { \phi = \frac{5}{2} x^{2} -yz- \frac{9}{2} x = 0}$ and $\large\rm { \Psi = 4x^{2} y + z^{3} - 4 = 0}$ at (1,1,1)

All the best.​

Answer 1

SOLUTION

GIVEN

$\sf{ \nabla \phi = \bigg ( 5x - \frac{9}{2} \bigg ) \hat{i} +(-z) \hat{j} + (-y) \hat{k} }$

and

$\sf{ \nabla \Psi = 8xy \hat{i} + 4x^{2} \hat{j} +3z^{2} \hat{k} }$

To find

The angle between the surfaces

$\sf { \phi = \frac{5}{2} x^{2} -yz- \frac{9}{2} x = 0}$

and

$\sf { \Psi = 4x^{2} y + z^{3} - 4 = 0}$

at (1,1,1)

CONCEPT TO BE IMPLEMENTED

The angle between the surfaces at the point is the angle between the normals to the surfaces at the point

EVALUATION

Here it is given that the two surfaces are

$\sf { \phi = \frac{5}{2} x^{2} -yz- \frac{9}{2} x = 0}$

and

$\sf { \Psi = 4x^{2} y + z^{3} - 4 = 0}$

Therefore

$\sf{ \nabla \phi = \bigg ( 5x - \frac{9}{2} \bigg ) \hat{i} +(-z) \hat{j} + (-y) \hat{k} }$

and

$\sf{ \nabla \Psi = 8xy \hat{i} + 4x^{2} \hat{j} +3z^{2} \hat{k} }$

Now

$\displaystyle \sf{ \nabla \phi \bigg|_{(1, 1, 1)} = \frac{1}{2} \hat{i} - \hat{j} - \hat{k} }$

$\displaystyle \sf{ \nabla \Psi \bigg|_{(1, 1, 1)} = 8 \hat{i} + 4 \hat{j} + 3 \hat{k} }$

$\sf{Let \: \theta \: be \: the \: angle }$

Then

$\displaystyle \sf{ \cos \theta = \frac{ \nabla \phi . \nabla \Psi}{ |\nabla \phi| |\nabla \Psi| } }$

$\implies \displaystyle \sf{ \cos \theta = \frac{8 - 4 - 3}{ \sqrt{ \frac{1}{4} + 1 + 1 } \times \sqrt{64 + 16 + 9} } }$

$\implies \displaystyle \sf{ \cos \theta = \frac{1}{ \sqrt{ \frac{9}{4} } \times \sqrt{89} } }$

$\implies \displaystyle \sf{ \cos \theta = \frac{2}{ 3 \sqrt{89} } }$

$\implies \displaystyle \sf{ \theta = { \cos}^{ - 1} \bigg( \frac{2}{ 3 \sqrt{89} } \bigg) }$

FINAL ANSWER

The required angle is

$\displaystyle \sf{ \theta = { \cos}^{ - 1} \bigg( \frac{2}{ 3 \sqrt{89} } \bigg) }$

━━━━━━━━━━━━━━━━

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Divergence of r / r^3 is

(a) zero at the origin

(b) zero everywhere

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Answer 2

CONCEPT TO BE IMPLEMENTED

The angle between the surfaces at the point is the angle between the normals to the surfaces at the point

EVALUATION

Here it is given that the two surfaces are

\sf { \phi = \frac{5}{2} x^{2} -yz- \frac{9}{2} x = 0}ϕ=

2

5

x

2

−yz−

2

9

x=0

and

\sf { \Psi = 4x^{2} y + z^{3} - 4 = 0}Ψ=4x

2

y+z

3

−4=0

Therefore

\sf{ \nabla \phi = \bigg ( 5x - \frac{9}{2} \bigg ) \hat{i} +(-z) \hat{j} + (-y) \hat{k} }∇ϕ=(5x−

2

9

)

i

^

+(−z)

j

^

+(−y)

k

^

and

\sf{ \nabla \Psi = 8xy \hat{i} + 4x^{2} \hat{j} +3z^{2} \hat{k} }∇Ψ=8xy

i

^

+4x

2

j

^

+3z

2

k

^

Now

\displaystyle \sf{ \nabla \phi \bigg|_{(1, 1, 1)} = \frac{1}{2} \hat{i} - \hat{j} - \hat{k} }∇ϕ

∣

∣

∣

∣

∣

(1,1,1)

=

2

1

i

^

−

j

^

−

k

^

\displaystyle \sf{ \nabla \Psi \bigg|_{(1, 1, 1)} = 8 \hat{i} + 4 \hat{j} + 3 \hat{k} }∇Ψ

∣

∣

∣

∣

∣

(1,1,1)

=8

i

^

+4

j

^

+3

k

^

\sf{Let \: \theta \: be \: the \: angle }Letθbetheangle

Then

\displaystyle \sf{ \cos \theta = \frac{ \nabla \phi . \nabla \Psi}{ |\nabla \phi| |\nabla \Psi| } }cosθ=

∣∇ϕ∣∣∇Ψ∣

∇ϕ.∇Ψ

\implies \displaystyle \sf{ \cos \theta = \frac{8 - 4 - 3}{ \sqrt{ \frac{1}{4} + 1 + 1 } \times \sqrt{64 + 16 + 9} } }⟹cosθ=

4

1

+1+1

×

64+16+9

8−4−3

\implies \displaystyle \sf{ \cos \theta = \frac{1}{ \sqrt{ \frac{9}{4} } \times \sqrt{89} } }⟹cosθ=

4

9

×

89

1

\implies \displaystyle \sf{ \cos \theta = \frac{2}{ 3 \sqrt{89} } }⟹cosθ=

3

89

2

\implies \displaystyle \sf{ \theta = { \cos}^{ - 1} \bigg( \frac{2}{ 3 \sqrt{89} } \bigg) }⟹θ=cos

−1

(

3

89

2

)

FINAL ANSWER

The required angle is

\displaystyle \sf{ \theta = { \cos}^{ - 1} \bigg( \frac{2}{ 3 \sqrt{89} } \bigg) }θ=cos

−1

(

3

89

2

)