IDZ - 6.2. It is necessary to find y' and y" for the following equations:
№ 1.9. tgy = 3x + 5y.
In order to find the derivatives of y' and y", you need to differentiate the equation with respect to the variable x. We get:
t(dy/dx) + y(dt/dx) = 3 + 5(dy/dx)
Let's express y' in terms of t and x:
(dy/dx) = (3 - ty) / (5 - t)
Now let's find y". To do this, we differentiate the resulting expression for y' with respect to x:
(d²y/dx²) = (d/dx) [(3 - ty) / (5 - t)]
(d²y/dx²) = [(d/dt)(3 - ty)(dt/dx) - (d/dt)(5 - t)(dy/dx)] / (5 - t)²
(d²y/dx²) = [-t(dy/dx) - (5 - t)(d²y/dx²)] / (5 - t)²
Now let's substitute the expression for y' and solve the equation for y":
(d²y/dx²)[1 + (5 - t) / (5 - t)²] = (-t(3 - you) - (5 - t)(3 - you) / (5 - t)) / ( 5 - t)²
(d²y/dx²) = (-2t - 2y + t²y) / (5 - t)³
Thus, y' = (3 - ty) / (5 - t), and y" = (-2t - 2y + t²y) / (5 - t)³.
№ 2.9. { x = 4t + 2t²; y = 5t³ - 3t² }.
To find y' and y" we differentiate the equations with respect to the variable t:
x' = 4 + 4t, y' = 15t² - 6t
x" = 4, y" = 30t - 6
Thus, y' = 15t² - 6t and y" = 30t - 6.
No. 3.9. For a given function y and argument x0, it is necessary to calculate y‴(x0), where y = Ln(x + 1), x0 = 2.
In order to find y‴(x0), you need to differentiate the function y three times and substitute the value of x0. We get:
y' = 1 / (x + 1), y'' = -1 / (x + 1)², y''' = 2 / (x + 1)³
I substitute x0 = 2 and get:
y‴(2) = 2 / (2 + 1)³ = 2 / 27
Thus, y‴(2) = 2 / 27.
No. 4.9. Let us write the formula for the nth order derivative for the function y = √x.
In order to find the nth order derivative of the function y = √x, you can use the Leibniz formula:
y^(n) = (1/2^n) * (1/√x) * (d/dx - √x)^n
Thus, the formula for the nth order derivative of the function y = √x will look like:
y^(n) = (1/2^n) * (1/√x) * (d/dx - √x)^n.
No. 5.9. Let us write the equation of the tangent to the curve y = x² – 6x + 2 at the point with the abscissa x = 2.
In order to find the equation of the tangent to a curve at a given point, you must first find the value of the derivative of the function at this point:
y' = 2x - 6
Note that at x = 2, the value of the derivative y' is -2. Now let's find the slope of the tangent:
k = -2
Since the tangent passes through the point (2, 2), its equation will look like:
y - 2 = k(x - 2)
y - 2 = -2(x - 2)
y = -2x + 6
Thus, the equation of the tangent to the curve y = x² – 6x + 2 at the point with abscissa x = 2 will be y = -2x + 6.
No. 6.9. It is necessary to find the speed of the material point S = 4sin(t/3 + π/6) at the moment of time t = π/2 s.
In order to find the speed of a material point, you need to differentiate the equation S with respect to time t:
S' = (dS/dt) = (4/3)cos(t/3 + π/6)
Let's substitute the value t = π/2:
S'(π/2) = (4/3)cos(π/6) = (4/3)√3/2 = (2√3)/3
Thus, the speed of the material point at the moment of time t = π/2 s is equal to (2√3)/3.
This product is an electronic version of tasks on mathematical analysis completed by Ryabushko A.P. in option 9. This digital product is available for download in a convenient format, which allows you to use it on any device such as a computer, tablet or smartphone.
The product includes tasks No. 1.9, No. 2.9, No. 3.9, No. 4.9, No. 5.9 and No. 6.9, which will develop skills in solving mathematical problems and deepen knowledge in the field of mathematical analysis.
This product is designed in a beautiful and convenient html format, which makes it easy and quick to familiarize yourself with the information and start solving tasks. Also, thanks to the electronic format, you can easily print out assignments and work with them in any convenient place and at any time.
By purchasing this product, you get a unique opportunity to improve your knowledge and skills in mathematical analysis, as well as save your time searching for materials and analyzing tasks. If you have any questions or difficulties with the product, you can always contact us at the email address provided in the seller information.
This product is an electronic version of tasks on mathematical analysis completed by Ryabushko A.P. in option 9. This digital product is available for download in a convenient format, which allows you to use it on any device such as a computer, tablet or smartphone.
The product includes tasks No. 1.9, No. 2.9, No. 3.9, No. 4.9, No. 5.9 and No. 6.9, which will develop skills in solving mathematical problems and deepen knowledge in the field of mathematical analysis.
This product is designed in a beautiful and convenient html format, which makes it easy and quick to familiarize yourself with the information and start solving tasks. Also, thanks to the electronic format, you can easily print out assignments and work with them in any convenient place and at any time.
By purchasing this product, you get a unique opportunity to improve your knowledge and skills in mathematical analysis, as well as save your time searching for materials and analyzing tasks. If you have any questions or difficulties with the product, you can always contact us at the email address provided in the seller information.
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Ryabushko A.P. IDZ 6.2 version 9 is a textbook in mathematics containing tasks and solutions on the following topics:
This manual will be useful for students and schoolchildren studying mathematics at a deeper level, as well as for teachers who can use its tasks to test students' knowledge. If you have any questions, you can contact the seller at the specified mail.
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