Math Problem Statement

найти производную второго порядка

найти производную второго порядка

Solution

To solve this, we are given parametric equations:

x=1sint,y=tantx = \frac{1}{\sin t}, \quad y = \tan t

We need to find the second derivative, d2ydx2\frac{d^2y}{dx^2}.

Step-by-Step Solution:

  1. Find dxdt\frac{dx}{dt}: x=1sint    dxdt=costsin2tx = \frac{1}{\sin t} \implies \frac{dx}{dt} = -\frac{\cos t}{\sin^2 t}

  2. Find dydt\frac{dy}{dt}: y=tant    dydt=sec2ty = \tan t \implies \frac{dy}{dt} = \sec^2 t

  3. Find dydx\frac{dy}{dx}: Using the chain rule: dydx=dydtdxdt=sec2tcostsin2t=sec2tsin2tcost\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\sec^2 t}{-\frac{\cos t}{\sin^2 t}} = -\frac{\sec^2 t \sin^2 t}{\cos t} Simplify: dydx=sin2tcos3t\frac{dy}{dx} = -\frac{\sin^2 t}{\cos^3 t}

  4. Find the second derivative d2ydx2\frac{d^2y}{dx^2}: Differentiate dydx\frac{dy}{dx} with respect to tt and divide by dxdt\frac{dx}{dt}: d2ydx2=ddt(sin2tcos3t)dxdt\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(-\frac{\sin^2 t}{\cos^3 t}\right)}{\frac{dx}{dt}}

    First, compute the derivative of sin2tcos3t-\frac{\sin^2 t}{\cos^3 t}: ddt(sin2tcos3t)=ddt(sin2t)1cos3tsin2tddt(1cos3t)\frac{d}{dt}\left(-\frac{\sin^2 t}{\cos^3 t}\right) = -\frac{d}{dt}(\sin^2 t) \cdot \frac{1}{\cos^3 t} - \sin^2 t \cdot \frac{d}{dt}\left(\frac{1}{\cos^3 t}\right) Use the product rule: =(2sintcost1cos3t)sin2t(3sintcos4t)= -\left(2\sin t \cos t \cdot \frac{1}{\cos^3 t}\right) - \sin^2 t \cdot \left(-3\frac{\sin t}{\cos^4 t}\right) Simplify: =2sintcos2t+3sin3tcos5t= -\frac{2\sin t}{\cos^2 t} + \frac{3\sin^3 t}{\cos^5 t}

    Combine with dxdt\frac{dx}{dt}: d2ydx2=2sintcos2t+3sin3tcos5tcostsin2t\frac{d^2y}{dx^2} = \frac{-\frac{2\sin t}{\cos^2 t} + \frac{3\sin^3 t}{\cos^5 t}}{-\frac{\cos t}{\sin^2 t}}

    Simplify: d2ydx2=2sin3tcos3t3sin5tcos6t\frac{d^2y}{dx^2} = \frac{2\sin^3 t}{\cos^3 t} - \frac{3\sin^5 t}{\cos^6 t}


Let me know if you'd like a further breakdown or have specific questions! Here are some related questions for exploration:

  1. How is dydx\frac{dy}{dx} related to the parametric representation of a curve?
  2. What are the conditions for d2ydx2\frac{d^2y}{dx^2} to be undefined?
  3. Can this method be generalized to other parametric equations?
  4. How does the second derivative relate to the concavity of the curve?
  5. What are the geometrical implications of higher-order derivatives for parametric equations?

Tip: Always simplify trigonometric terms wherever possible to make derivatives easier to interpret.

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Math Problem Analysis

Mathematical Concepts

Parametric Equations
Calculus
Second Derivatives
Trigonometry

Formulas

x = 1 / sin(t), y = tan(t)
dx/dt = -cos(t) / sin^2(t)
dy/dt = sec^2(t)
dy/dx = (dy/dt) / (dx/dt)
d^2y/dx^2 = (d(dy/dx)/dt) / (dx/dt)

Theorems

Chain Rule
Product Rule

Suitable Grade Level

Undergraduate Calculus or Advanced High School (Grades 11-12)