Ryabushko A.P. IDZ 6.2 version 10

IDZ - 6.2. Problem solving.

No. 1.10. The equation xy = cot x is given. It is necessary to find the value of the derivative y' and the second derivative y".

Answer:

Let's find the derivative of the function y with respect to x using the product derivative rule:

y' = (xy)' = x'y + y'x = y + xy'

Replace y in the equation xy = cot x:

xy = ctg x

y = ctg x / x

Then:

y' = (ctg x / x) + x(-ctg^2 x / (sin^2 x)) = (ctg x / x) - ctg^2 x / sin^2 x

To find the second derivative y" we differentiate the resulting expression for y':

y" = (-ctg^2 x / sin^2 x)' = -2ctg x / (sin^2 x * cos x)

Ответ: y' = (ctg x / x) - ctg^2 x / sin^2 x, y" = -2ctg x / (sin^2 x * cos x).

No. 2.10. The equation of the parametric curve is given: x = L(t) / t, y = t ln t. It is necessary to find the value of the derivative y' and the second derivative y".

Answer:

Let's differentiate the equation x = L(t) / t with respect to t:

x' = (L(t) / t)' = (L'(t) * t - L(t)) / t^2

Replace L(t) with tx in the equation for x:

x = L(t) / t = tx / t = x

Then:

x' = (L'(t) * t - L(t)) / t^2 = (L'(t) - x) / t

Let's differentiate the equation y = t ln t with respect to t:

y' = ln t + 1

Now let's find the value of the derivative y':

y' = (dy/dt) / (dx/dt) = (ln t + 1) / ((L'(t) - x) / t)

To find the second derivative y" we differentiate the resulting expression for y':

y" = [(d/dt)((ln t + 1) / ((L'(t) - x) / t))] / ((dx/dt) / t)

y" = (t(ln t + 2) - (L'(t) - x)(ln t + 1)) / ((L'(t) - x)^2)

Ответ: y' = (ln t + 1) / ((L'(t) - x) / t), y" = (t(ln t + 2) - (L'(t) - x)(ln t + 1)) / ((L'(t) - x)^2).

No. 3.10. Given the equation of the function y = x^2 e^x and the argument x0 = 0. It is necessary to calculate the value of the third derivative y‴(x0).

Answer:

Let's find the first, second and third derivatives of the function y with respect to x:

y' = 2x e^x + x^2 e^x

y" = 2e^x + 4x e^x + x^2 e^x

y‴ = 6e^x + 12x e^x + 2x^2 e^x

Let's substitute the value x0 = 0 into the resulting expressions:

y'(0) = 0

y"(0) = 2

y‴(0) = 6

Answer: y‴(0) = 6.

No. 4.10. The equation for the function y = x e^(3x) is given. It is necessary to write down the formula for the nth order derivative.

Answer:

Let's differentiate the function y by x n times:

y^(n) = (x e^(3x))^(n)

Using the product derivative rule, we get:

y^(n) = (x^(n) e^(3x)) + n(x^(n-1) e^(3x) * 3) + n(n-1)(x^(n-2) e^(3x) * 3^2) + ... + 3^n(x e^(3x))

Thus, the formula for the nth order derivative of the function y = x e^(3x) has the form:

y^(n) = e^(3x) * P_n(x),

where P_n(x) is a polynomial of the nth degree, expressed in terms of derivatives of x^n.

No. 5.10. Given the equation of the curve y = x^2/4 - 4x + 5 and a point with the abscissa x = 4. It is necessary to write down the equation of the tangent to this curve at a given point.

Answer:

Let's find the value of the derivative of the function y with respect to x:

y' = x/2 - 4

Derivative value at point x = 4:

y'(4) = 2 - 4 = -2

Thus, the equation of the tangent to the curve y = x^2/4 - 4x + 5 at the point with the abscissa x = 4 is:

y - (16/4 - 16 + 5) = -2(x - 4),

or

y = -2x + 13.

No. 6.10. The law of motion of a material point is given: S = -3 cos(t/4+π/12). It is necessary to find the speed of this point at time t = 2π/3c.

Answer:

Let us find the derivative of the law of motion S with respect to time t:

v = dS/dt = (d/dt)(-3cos(t/4 + π/12)) = 3/4 sin(t/4 + π/12)

Let's substitute the value t = 2π/3c:

v = 3/4 sin(π/6 + π/12) = 3/4 sin(π/4) = 3/8√2 м/с.

Answer: the speed of a material point at time t = 2π/3c is equal to 3/8√2 m/s.

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This digital product is a solution to problems from IPD 6.2 version 10, compiled by the author Ryabushko A.P. The product includes solutions to problems with detailed step-by-step explanations and formulas necessary to solve them.

A beautifully designed html product design ensures ease of reading and allows you to conveniently navigate through the content, which will make the process of studying the material even more convenient and effective.

By purchasing this product, you get access to reliable and high-quality solutions to math problems that will help you prepare for exams or develop your math problem solving skills.

The product is a solution to problems from IDZ 6.2 version 10, compiled by the author Ryabushko A.P. Solutions to problems are included with detailed step-by-step explanations and necessary formulas. Beautiful html product design ensures ease of reading and navigation through the content.

In the first problem (No. 1.10) it is necessary to find the value of the derivative y' and the second derivative y" for the equation xy = cot x.

In the second problem (No. 2.10) it is necessary to find the value of the derivative y' and the second derivative y" for the parametric curve x = L(t) / t, y = t ln t.

In the third problem (No. 3.10) it is necessary to calculate the value of the third derivative y‴(x0) for the function y = x^2 e^x and the argument x0 = 0.

In the fourth problem (No. 4.10), it is necessary to write down the formula for the nth-order derivative of the function y = x e^(3x).

In the fifth problem (No. 5.10) it is necessary to write down the equation of the tangent to the curve y = x^2/4 - 4x + 5 at the point with the abscissa x = 4.

In the sixth problem (No. 6.10) it is necessary to find the speed of a material point moving according to the law S = -3 cos(t/4+π/12) at time t = 2π/3s.

By purchasing this product, you get a reliable and high-quality solution to mathematical problems that will help you prepare for exams or develop problem-solving skills. If you have any questions, you can contact the seller by email.

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Ryabushko A.P. IDZ 6.2 option 10 is an educational task or test as part of a course related to mathematical analysis and differential equations. The assignment presents six problems, each of which requires solving a specific mathematical problem.

The first problem is to find the first and second order derivatives for the function given by the equation xy = cot x.

In the second problem, you need to find the equation of a curve defined parametrically: x = L(t)/t and y = t Ln t.

The third task involves calculating the third derivative for the function y = x²eˣ at the point x0 = 0.

The fourth problem requires writing a formula for the nth order derivative of the function y = x e³ˣ.

In the fifth problem, you need to write down the equation of the tangent to the curve y = x²/4 – 4x + 5 at the point with the abscissa x = 4.

The sixth task is related to calculating the speed of a material point according to the given law of motion S = -3 cos(t/4+π/12) at the time t= 2π/3 s.

This product is intended for students and anyone interested in mathematical analysis and differential equations.

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