Calculus of Single & Multiple Variables -Assignment

# Calculus of Single & Multiple Variables -Assignment | Mathematical Methods - Physics PDF Download

Q.1. If f (x) be a twice differentiable function such that f"(x) = -f (x) and f'(x) = g (x ). If h'(x) = [f(x)]2+[g(x)]2 ’h (1) = 8 and h (0) = 2 then find the value of h (2) .
Ans. h'( X ) = [f ( x )]2+[ g ( x )]2
h'' ( x ) = 2 f (x ) f' ( x ) + 2 g ( x ) g ' (x )
h'' ( x ) = 2 f(x) g (x ) + 2g ( x ) f'' (x )
h'' ( x ) = 2 f(x) g (x ) + 2g ( x ) (-f (x )) = 0
Thus, h'(x) = c, a constant for all x.
⇒ h (x) = cx + c1
It is given that h (l) = 8 and h (0) = 2, therefore c1 = 2 and c2 = 6
∴ h (x ) = 6 x + 2 ⇒ h (2) = 6 x 2 + 2 = 14

Q.2. If y = ax then find the value of dy/dx as a function of x and y ( a is a constant).
Ans. We can write y = axy
Taking Logrithm of both sides, we obtain ln y = xy ln a.
⇒ ln (ln y) = y ln x + ln (ln a)
Differentiating with respect to x gives

Q.3. Find the co-ordinates of the point P on the curve y2 = 2x3 such that the tangent at P is perpendicular to the line 4 x - 3 y + 2 = 0.
Ans.
Let, P(x0,y0) be the point at which the tangent is perpendicular to the line 4x - 3y + 2 = 0.
Thus slope at P is equal to m =
Thus (i)
From the equation of curve  (ii)
From equation (i) and (ii)

Thus there are two points  and (0,0) at which the tangent line is perpendicular
to the line 4x - 3y + 2 = 0.

Q.4. If the tangent to the curve xy + ax + by = 0 at (1,1) is inclined at an angle tan-1 2 with x axis then find the values of a and b .
Ans. The point (1,1) lies on the curve xy + ax + by = 0.
Hence, a+b = -1    (i)
Differentiating xy + ax + by = 0 with respect to x we get

Since the tangent at (1,1) makes an angle of tan-1 2 with the x - axis hence

Solving equations (i) and (ii) we obtain
a = 1 and b = -2

Q.5. If f (x) = a log |x| + bx2 + x has its extreme values (local maximum or minimum value) at x = -1 and x = 2 , then find the values of a and b .
Ans. The logarithmic function is defined for all x > 0. Hence the domain of f (x) is (0, ∞) . Using this fact we can write
f (x) = a log x + bx2 + x

We see that the derivative is defined at all points of the domain of the function. Hence the function can attain extreme values only at points where f'(x) = 0. Since f (x) attains its extreme values at x = -1,2.
∴ f'(-1) = 0 and f'(2) = 0

Q.6. If y = sin-1  where α is a constant then find the value of y'(0) .

Ans.

Q.7. Find the equation of the tangent to the curve x = t cos t and y = t sin t at the origin of xy plane.
Ans.

At the origin x = 0, y = 0
⇒ t cos t = 0 and t sin t = 0
t = 0 and cos t = 0 and t = 0 and sin t = 0
Since there is no value of t for which cos t and sin t are simultaneously 0 , hence t = 0.Thus the slope of the tangent at (0,0) is Thus the equation of the tangent at the origin is y = 0.

Q.8. A curve with equation of the form y = ax4 + bx3 + c + cx + d has slope 0 at the point (0,1) and also touches the x -axis at the point (-1,0) then find the values of x for which the curve has negative slope.
Ans:

Hence, 4a • 0+3b • 0+c = 0 ⇒ c = 0
And, 4a (-1) + 3b = 0    (i)

Also, the curve passes through (0,1) and (-1,0), hence
d = 1 and 0 = a - b - c + d
⇒ a - b - c + 1 = 0    (ii)
From (i) and (ii), we get
a = 3,b = 4,c = 0 d = 1

Q.9. Find the point at which the local maximum or local minimum value of the function

occurs. Also find the local minimum and for maximum values of the function.
Ans. We have y = f (x) = sinx + cos4 x

= -4 cos x sin x (cos2 x - sin2 x)
= -2 sin 2x cos 2x = - sin 4x
Over the given interval  is defined for all x , hence for local maximum or minimum value Where we have used the fact that.

Thus  is the only point where local maximum or minimum value of the function occurs. Here we can use either first derivative test or the second derivative test to check for local maxima or minimum. Using the second derivative test, we obtain

Thus at a local minimum occurs. The local minimum value is
Note: Students should check using the first derivative test that a local minimum indeed occurs at x =

Q.10. If A > 0,B > 0  then find the maximum value of tan A + tan B.
Ans. We have A + B =

Let, z = tan A+tan B

F or maximum value of  because which implies that x = tan A > 0. It can be earily checked that  Hence, z is maximum for x =  For this value of x, z =

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## Mathematical Methods

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## FAQs on Calculus of Single & Multiple Variables -Assignment - Mathematical Methods - Physics

 1. What is calculus of single and multiple variables?
Ans. Calculus of single and multiple variables is a branch of mathematics that deals with the study of change and motion. It involves the concepts of differentiation and integration, which are used to analyze functions and their properties in one or more variables.
 2. What is the difference between calculus of single and multiple variables?
Ans. The main difference between calculus of single and multiple variables lies in the number of variables involved. Single-variable calculus focuses on functions of a single variable and deals with concepts like limits, derivatives, and integrals in one dimension. On the other hand, multiple-variable calculus deals with functions of multiple variables and involves concepts like partial derivatives, multiple integrals, and vector calculus.
 3. What are the applications of calculus of single and multiple variables?
Ans. The applications of calculus of single and multiple variables are extensive and diverse. In single-variable calculus, it is used to model and analyze physical phenomena such as motion, growth, and decay. It is also fundamental in the fields of physics, engineering, economics, and computer science. In multiple-variable calculus, it is applied to study functions of multiple variables, optimization problems, and vector fields, which find applications in fields like fluid mechanics, electromagnetism, and optimization algorithms.
 4. What are the prerequisites for studying calculus of single and multiple variables?
Ans. To study calculus of single and multiple variables, a strong foundation in algebra, trigonometry, and basic calculus concepts is essential. It is important to have a solid understanding of functions, limits, derivatives, and integrals in one variable before diving into the multi-variable calculus. Additionally, knowledge of vectors and matrices can also be beneficial in understanding certain concepts in multiple-variable calculus.
 5. How can I prepare for the IIT JAM exam in calculus of single and multiple variables?
Ans. To prepare for the IIT JAM exam in calculus of single and multiple variables, it is recommended to follow a structured study plan. Start by thoroughly understanding the fundamental concepts of single-variable calculus, including limits, derivatives, and integrals. Practice solving a variety of problems from textbooks and previous year question papers. Once comfortable with single-variable calculus, move on to studying concepts of multiple-variable calculus, such as partial derivatives, multiple integrals, and vector calculus. Regularly solve mock tests and sample papers to assess your progress and identify areas where you need more practice.

## Mathematical Methods

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