Multivariable Calculus: Steps, Applications & Examples

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Multivariable calculus or multivariate calculus is the branch of mathematics that is useful to find the relationship between the input and output variables. In mathematics, multivariable calculus is the extension of calculus that deals with multiple variables. Multivariable calculus includes two or more variables instead of a single variable.

Moreover, multivariable calculus consists of limits and derivatives, partial differentiation, multiple integrations, fundamental theorems with multivariable dimensions, vector fields, etc. Apart from this, multivariable calculus plays an essential role in many fields such as engineering, natural and social science, weather forecasting, astronomy, etc. Also, the two- dimensional(2D) and three-dimensional(3D) models are studied with the help of multivariable calculus.

Read Also: Methods of Integration

Table of Content

What is Multivariable Calculus?
Steps to Solve Multivariable Calculus
Important Topics of Multivariable Calculus
Applications of Multivariable Calculus
Things to Remember
Sample Questions

Key Takeaways: Calculus, Multivariable Calculus, Partial Differentiation, Integration, Vector Space, Limits and Derivatives

Multivariable Calculus

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Multivariable calculus is the function of two or more variables while single calculus deals with the function of a single variable. The relationship between the input and output variable can be found using multivariable calculus. Also, the integration and differentiation are solved in the same way as in single calculus. Multivariable functions are the functions of two or more variables. Mathematically,

f(z) = f(x,y)

Where, f(z) is an output function, x and y are the input variables. Also, z is dependent on x and y.

Diagrams describe multivariable calculus

Diagrams describe multivariable calculus

The important operations are involved in the multivariable calculus are mentioned below:

Multiple integrations
Partial differentiation
Limits and continuity
Curves and surfaces etc.

Check Important Notes for Conditional Probability

Steps to Solve Multivariable Calculus

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There are some basic steps to be followed to solve the multivariable calculus. Such as:

When the function is dependent on two or more variables then a partial derivative is used to determine multivariable calculus. To find the derivative of the function concerning one of those variables, keep all the variables constant.

Mathematically,

If z = f(x,y) then the derivative is given as,

dz = f_xdx + f_ydy

Or, If there is a function ‘w’ which is dependent on three variables like x,y,z then,

w = f(x,y,z)

dw = f_xdx + f_ydy + f_xdz

When we change all the variables and find the derivative then it is referred to as the total derivative. When there are two functions like f(x) and g(x) then the derivative of both the functions is the sum of these two functions. Mathematically,

(f + g)’ = f ’ + g’

Also Check: Three Dimensional Geometry

Topics of Multivariable Calculus

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There are many topics in multivariable calculus such as:

S.No.	Multivariable Topics	Advanced Calculus Topics
1.	Differential Calculus	Differentiation Partial derivatives Chain rule Directional derivatives and gradient Applications of differential calculus
2.	Integral Calculus	Integration Double integrals Triple integrals Changing variables
3.	Vector Field	Basics of vector field Vector operators
4.	Curves and Surfaces	Parameterized curves Parameterized surfaces Length of curve Surface area of Parameterized surfaces
5.	Fundamental Theorem	Divergence theorem Stokes theorem Green’s theorem Gradient theorem for line integral
6.	Integration over Curves and Surfaces	Line integrals Surface integrals Integration synopsis

Read More: Calculus Formula

Applications of Multivariable Calculus

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Multivariable calculus is widely accepted in many fields such as science, engineering, finance, research etc. some of the major applications of multivariable are mentioned below:

In a dynamic system, multivariable calculus is the core tool used which is used for optimal control.
Multivariable calculus is helpful in evaluating empirical data in regression analysis.
In the field of finance, quantitative analysts use multivariable calculus which is helpful in predicting the future trends in the stock market.
Multivariable calculus is also used to determine the 2D and 3D motion of fluid particles and streams in mechanical engineering.
Multivariable calculus is used to study the high dimensional model system which has deterministic nature in engineering as well as social science.

Check More: Increasing and Decreasing Functions in Calculus

Things to Remember

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Multivariable calculus deals with the multivariable functions and variables.
Multivariable calculus functions are also known as multivariate functions.
In mathematics, Multivariable calculus is the extension of calculus.
Multivariable calculus is also useful in the study and modelling of two-dimensional and three-dimensional systems in engineering.
Multivariable calculus is also useful in determining the motion of rotating bodies in astronomies.

Sample Questions

Ques. Evaluate the following limit: [2 marks]

\(\lim\limits_{x \to 1} |x^3 - x^2 +1|\)

Ans. Given,

\(\lim\limits_{x \to 1} |x^3 - x^2 +1|\)

Now, \(\lim\limits_{x \to 1} |x^3 - x^2 +1| = (1)3 - (1)2 + 1\)

\(= 1 - 1 + 1 = 1\)

Ques. Find the derivative of the function f(x) = 3x at x= 2. [2 marks]

Ans. Given, \(f(x) = 3x\)

Derivative of f(x) is given as,

f’(x) = 3

When x = 2 then,

f(2) = 3.

Therefore, f(x) at x = 2 is 3.

Ques. When the function z = f(x,y) = x⁴ + y³ + sin xy. Then find out the first partial derivative. [2 marks]

Ans. GIven, z = f(x,y) = x⁴ + y³ + sin xy.

Here, function z is dependent on two variables x and y. So, a partial derivative is found with respect to x as well as y separately.

Now, first partial derivative w.r.t to x is given as,

∂ z/ ∂ x = ∂ f/ ∂ x = 4x³ +0 + cos(xy) . y

Similarly, first partial derivative w.r.t to y is given as,

∂ z/ ∂ y = ∂ f/ ∂ y = 0+3y² + cos(xy) . x

Ques. If two vectors and b→ are such that |a→ | = 2 and |b→ | = 3 and a→ . b→ = 4. then Determine | a→ - b→| . [3 marks]

Ans.

a-b

Ques. When w = xy² + yz - 3 , x = r²s - t , y = r + 2st , z = rs - 2t. Find ∂ w/ ∂ t by using chain rule. [5 marks]

Ans. Given , w = xy² + yz - 3, x = r²s - t, y = r + 2st, z = rs - 2t.

Here, f(x,y,z) and x= g(r,s,t) , y=h(r,s,t), z=j(r,s,t) and also f,g,h,j are differentiable. Then find the partial derivative using chain rule such as:

∂ w/ ∂ t = (∂ w/ ∂ x)(∂ x/∂ t) + (∂ w/ ∂ y)(∂ y/∂ t) + (∂ w/ ∂ z)(∂ z/∂ t)

Now,

∂ w/ ∂ x = y²

∂ w/ ∂ y = 2xy+z

∂ w/ ∂ z = y

∂ x/∂ t = -1

∂ y/∂ t) = 2s

∂ z/∂ t) = -2

According to Chain rule,

∂ w/ ∂ t = (∂ w/ ∂ x)(∂ x/∂ t) + (∂ w/ ∂ y)(∂ y/∂ t) + (∂ w/ ∂ z)(∂ z/∂ t)

∂ w/ ∂ t = y² (-1) + (2xy+z)(2s) + y(-2)

∂ w/ ∂ t = -y² + 4 sxy + 2sz – 2y

∂ w/ ∂ t = -( r+2st )²+ 4s(r²s - t)(r+2st) + 2s(rs - 2t) - 2(r+2st)

∂ w/ ∂ t = -r² - 4s²t² - 4rst + (4r²s² - 4st)(r + 2st) + 2s²r - 4st - 2r - 4st

∂ w/ ∂ t = -r² - 4s²t² - 4rst + 4r³s² + 8 r².s³t - 4srt - 8s²t² + 2s².r - 8st - 2r

∂ w/ ∂ t = -r² -12s²t² - 8rst + 4r³s² + 8 r².s³t + 2s².r - 8st - 2r

Hence, ∂ w/ ∂ t = -r² -12s²t² - 8rst + 4r³s² + 8 r².s³t + 2s².r - 8st - 2r

Ques. Solve: ∫( x³ - 1 /x²) dx. [2 marks]

Ans. Given, ∫( x³ - 1 /x²) dx

∫( x³ - 1 /x²) dx = ∫( x³ / x² - 1/x²) dx

= ∫(x - x^-2) dx

= ( x¹⁺¹/1+1 - x^-2+1/-2+1) + c

= (x²/2 -x^-1/-1) +c

= x²/2 -1/x + c

Hence, ∫( x³ - 1 /x²) dx = x²/2 -1/x + c.

Ques. Determine the total derivative if the function z = 5x²y² + 4x sin y. [2 marks]

Ans. GIven,

z = 5x²y² + 4x sin y

The total differentiation of the given function is,

dz = (∂ z/ ∂ x) dx + (∂ z/ ∂ y) dy

dz = (10xy² + 4 siny)dx + (10x²y+4x cosy) dy

Therefore, dz =(10xy² + 4 siny)dx + (10x²y+4x cosy) dy

Ques. Determine the area of a parallelogram whose adjacent sides are given by the vectors such as: [2 marks]

\(\overrightarrow{a} = 3\hat{i} + \hat{j} + 4\hat{k} \text{ and } \overrightarrow{b} = \hat{i} - \hat{j} + \hat{k}\)

Ans. Given, \(\overrightarrow{a} \times \overrightarrow{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\[0.3em] 3 & 1 & 4 \\[0.3em] 1 & -1 & 1 \end{vmatrix} = 5\hat{i} + \hat{j} - 4\hat{k}\)
Now, \(|\overrightarrow{a} \times \overrightarrow{b}| = \sqrt{25+1+16} = \sqrt{42}\)
Hence the required area is √42.

Ques. Using the section formula, prove that the three points (-4,6,10),(2,4,6) and (14,0,-2) are collinear. [5 marks]

Ans. Let us assume that A (-4,6,10),B(2,4,6) and C(14,0,-2) are the given points.Also, consider that C divides the point |AB| in the ratio of k:1.

By using section formula(internally), the coordinates of C are given by,

C(14,0,-2) = (2k-4/k+1 ,4k+6/k+1 ,6k+10/k+1)....(i)

Now, equate the x coordinates, we get,

2k-4/k+1 = 14

⇒ 2k-4 = 14(k+1)

⇒ 2k-4 = 14k+14

⇒2k -14k = 14+4

⇒ -12k = 18

⇒k = - 18/12

⇒ k= -3/2 .

Put the value of k in equation (i) , we get,

⇒ 2(-3/2) -4/(-3/2)+1, 4(-3/2)+6/(-3/2)+1 ,6(-3/2)+10/(-3/2)+1 )

⇒-3-4/-½ , -6+6/-½ , -9+10/-½

⇒-7*2/-1 , 0*2/-1, 1*2/-1

⇒14, 0 , -2 which is equal to point C.

Therefore, C is the point which divides the points |AB| in the ratio of 3:2.

Hence, A, B and C are collinear.

⇒ x= 15 , y= -10 , z= 16

Therefore, C(15,-10,16) is the third vertex.

Mathematics Related Links:
Tangents and Normals	Integral Calculus	Approximations
Types of Functions	Binary Operations	Coplanar Vector
Signum Function	Properties of Determinants	Maxima and Minima
Methods of Integration	Integration by Partial Fractions	Integration by Parts
Definite Integral	Fundamental Theorem of Calculus	Application of Integrals
Differential Equations	Polynomial	Bayes Theorem

Multivariable Calculus: Steps, Applications & Examples

Multivariable Calculus

Steps to Solve Multivariable Calculus

Topics of Multivariable Calculus

Applications of Multivariable Calculus

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.
At a birthday party, children are being served orange juice in conical cups, as shown in the figure.

Each cup is 15 cm deep and has a radius 5 cm. The juice is being poured into this cup at a rate of 0·1 cm3/s.
On the basis of the above information, answer the following questions :

2.
If \( xy = e^{x - y} \), then find \( \frac{dy}{dx} \).

3.
Find:

The principal value of \[ \sec^{-1}(\sqrt{2})+2\csc^{-1}(-2) \] is:

4.
Find:

If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

5.
Differentiate \( \tan^{-1}\left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \) with respect to \( \cos^{-1}(x^2) \).

6.
A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length.

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Multivariable Calculus: Steps, Applications & Examples

Multivariable Calculus

Steps to Solve Multivariable Calculus

Topics of Multivariable Calculus

Applications of Multivariable Calculus

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.At a birthday party, children are being served orange juice in conical cups, as shown in the figure. Each cup is 15 cm deep and has a radius 5 cm. The juice is being poured into this cup at a rate of 0·1 cm3/s.On the basis of the above information, answer the following questions :

2.If \( xy = e^{x - y} \), then find \( \frac{dy}{dx} \).

3. Find: The principal value of \[ \sec^{-1}(\sqrt{2})+2\csc^{-1}(-2) \] is:

4. Find: If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

5.Differentiate \( \tan^{-1}\left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \) with respect to \( \cos^{-1}(x^2) \).

6.A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length.

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
At a birthday party, children are being served orange juice in conical cups, as shown in the figure.

Each cup is 15 cm deep and has a radius 5 cm. The juice is being poured into this cup at a rate of 0·1 cm3/s.
On the basis of the above information, answer the following questions :

2.
If \( xy = e^{x - y} \), then find \( \frac{dy}{dx} \).

3.
Find:

The principal value of \[ \sec^{-1}(\sqrt{2})+2\csc^{-1}(-2) \] is:

4.
Find:

If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

5.
Differentiate \( \tan^{-1}\left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) \) with respect to \( \cos^{-1}(x^2) \).

6.
A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length.