Vector calculus, ODE, PDE engineering mathematics 15 solution

Below we have discussed few examples on this topic that can clear your basic concept of vector calculus, ODE, PDE engineering mathematics problems. Also will show you how to deal with such problems as listed.

Vector Calculus

Q 1. If r = ƒ(t) has a constant magnitude. Then prove that r.(dr/dt )= 0.

Ans.
Given r = ƒ(t) has a constant magnitude.
Then r.r = |r|² = constant.
⇒ (d/dt).(r.r) = 0 [∴ a.b = b.a]
⇒ (dr/dt).r + r.(dr/dt) = 0
⇒ 2r.(dr/dt) = 0
⇒ r.(dr/dt) = 0 (hence proved)

Q 2. Find the unit normal vector to the surface x²yz³ = 4 at point (-1, -1, 2).

Ans.
ƒ(x, y, z) = x²yz³ – 4 = 0
The Gradient of ƒ(x, y, z) at point x,y,z is a vector normal to the surface at this point.
The gradient is obtain as follows,

∇ƒ(x, y, z) = (ƒx, ƒy, ƒz)
= (xyz³, x²z³, x²yz²) at point (-1, -1, 2) has the value (8, 8, -4) .

the unit normal vector will be (8, 8, -4)/√(8² + 8² + (-4)²)
= (8i + 8j – 4k)(1/√144)
= (2/3)i + (2/3)j – (1/3)k.

Q 3. If F and G are irrrotational, then show that F×G is solenoid.

Ans
According to question ∇×F = 0 and ∇×G = 0
Also
G(∇×F) = 0 — (i)
F(∇×G) = 0 — (ii)
By (i) – (ii)
G(∇×F) – F(∇×G) = 0
Now,
∇(F×G) = G(∇×F) – F(∇×G)
Therefore ∇(F×G) will also be zero.
So F×G will be solenoid.

Q 4. Prove Curl(Curl(V)) = Grad(Div(V) – (Grad)²(V).

Ans.
V = X²î + Y²j + XYZk

V = X²î + Y²j + XYZk

Div V = 2X + 2Y + XY
Grad(Div V) = (2+y)i + (2+x)j
(Grad)²V = 2i + 2j +0k

Grad(Div V) – (Grad)²V = (2+y)i + (2+x)j – 2i – 2j
= Yi + Xj = Curl(CurlV) (Hence Proved)

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Q 5. Define Gradient, Curl, and Divergence and give their physical significance.

Ans.

Because the gradient, divergence, and curl are primarily mathematical operators that tell you something about either a scalar function (the gradient) or a vector field .

1.Gradient.

Assume that height is a function of x and y. The gradient would then show you the direction of greatest height gain (at a given place). It would, in other words, indicate in the direction of the sharpest gradient.
The gradient is the slope of a line tangent to the graph with magnitude equal to the slope.

2.Divergence

This indicates how ‘spiky’ your vector field is at a given position.
The divergence indicates the volume density of a vector field’s outward flux from an infinitesimal volume around a particular point,”.

3. Curl

This informs you how’swirly’ your vector field is at a given position.
The curl indicates the rotation of the vector field (around a single point you’re interested in).

Ordinary differential equation(ODE)

Q1. Evaluate (xdy/dx) – y = √(x² + y²).

Ans.
dy/dx = [√(x² + y²) + y]/x
It is a homogeneous equation. Putting y = vx and dy/dx = v = xdv/dx.
we get
v + xdv/dx = [√(x² + v²x²) + vx]/x
= √(1+v²) +v
⇒ v + xdv/dx = √(1+v²) +v
⇒ dv/√(1+v²) = dx/x

On integration
∫dv/√(1+v²) = ∫dx/x = log(v+√(1+v²)) = logx + logc
v + √(1+v²) = cx (Now putting v = y/x)
y/x + √((x² + y²)/x²) = cx
now by further evaluating this equation by taxing x as a common denominator and moving it to RHS we will get.
y + √(x² + y²) = cx² (is the required solution).

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Q 2. Evaluate y(xy + 2x²y²)dx + x(xy – x²y²)dy = 0

Ans.
xy²dx + 2x²y³dx + x²ydy – x³y²dy = 0 (by solving the given equation)

Now dividing it by x³y³ we get
(dx/x²y) + (2dx/x) + (dy/xy²) – dy/y = 0
Now on Integrating above equation we will get
∫(dx/x²y) + ∫(dy/xy²) + ∫2dx/x – ∫dy/y = 0
-1/xy + 2ln|x| – ln|y| = c.

Q 3. Evaluate d²x/dt² – 3dx/dt + 2x = 0.

Ans
It is in a form of D² x-3Dx+2x=0
=(D²-3D+2)x=0
x = eF + FI

The auxiliary equation is,
m² – 3m + 2 = 0
⇒ m² – 2m -m + 2 = 0
⇒ m = 1,2
Common factor (CF) = C1(eˆt) + C2(eˆ(2t))
PI = 0
∴ x = C1(eˆt) + C2(eˆ(2t)).

Q 4. Evaluate (D² − 4D + 1)y = eˆ(2x)sinx.

Q 5. Evavulate (D² – 2D + 1)y = xeˆ(2x)sinx

partial differential equation problems

Q 1. Evaluate (p²+q²)x = pq

Ans.
ƒ=(p²+q²)x-pq=0
we will solve this problem using charpit’s equation,

dƒ/dx = p²+q²,
dƒ/dp=2qx-p,
dƒ/dy=0,
dƒ/dz=0,
dƒ/dq=2qx-p
Now according to charpit’s equation we get
dp/(p²+q²)=dq/0–(i)
We get q=a where a is constant because of dq/0
By integrating both the side of eq(i) and taking q=a we get

∫0dp = ∫(p²+q²)dq
or c = (p²+a²)q–(ii)

According to question we have (p²+q²)x = pq putting the constant a instead of q we will get,
(p²+a²)x = pa
Now from eq(ii)
(c/q)x=pa
or p = cx/qa
or p = cx/a²

Now taking dz=pdx+qdy and integrating both the sides,
∫dz=∫(cx/a²)dx+∫ady

z=(cx²/2a²)+ay where c,a being an arbitrary constant.

Q 2. Evaluate (dˆ2)z/dxdy = (xˆ2)y subject to the condition z(x, 0) = xˆ2 and z(1, y) = cosy.

Q 3. Solve the wave equation,

Q 4. State Heat and Laplace equation and give its applications.

Ans.
The Heat equation is a prototypical parabolic partial differential equation, that illustrate the variation of temperature in a given region over a period of time.
${\frac {\partial u}{\partial t}}={\frac {\partial ^{2}u}{\partial x_{1}^{2}}}+\cdots +{\frac {\partial ^{2}u}{\partial x_{n}^{2}}},$
Application –
1. Heat equation in image processing
2. Heat equation in cancer model and spatial
ecological model

Laplace’s equation states that the sum of the second-order partial derivatives of R, the unknown function, with respect to the Cartesian coordinates, equals zero:

The sum on the left often is represented by the expression ∇²R, in which the symbol ∇² is called the Laplacian, or the Laplace operator.

Applications –
It is widely used in problems of electrical, magnetic, hydrodynamics, gravitational potential and of steady-state temperature.
Also it is use to analysis of linear time-invariant system such as harmonic oscillators, signal processing, optical device, mechanical system etc.

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Q5. State wave equation, and why wave equation is hyperbolic.

Ans.
The wave equation is a linear second – order partial differential equation which illustrate the propagation of oscillation at a fixed speed in some quantity is,
${\frac {\partial ^{2}u}{\partial t^{2}}}\;=\;c^{2}\left({\frac {\partial ^{2}u}{\partial x_{1}^{2}}}+{\frac {\partial ^{2}u}{\partial x_{2}^{2}}}+\cdots +{\frac {\partial ^{2}u}{\partial x_{n}^{2}}}\right)$
The partial differential equation is known as one-dimensional wave equation, because the motion is only in vertical direction not in horizontal direction.
consider this equation,

where A, B, C can be constants or functions of x and y.
As B²-4AC > 0 satisfied the way equation which makes it hyperbolic.