To find the derivatives dy/dx and d²y/dx² for the equation x² sin(t) + y³ cos(t) = 0, we will need to use implicit differentiation and the chain rule. Let’s break this down step by step:
Step 1: Implicit Differentiation
We start with the equation:
x² sin(t) + y³ cos(t) = 0
Taking the derivative with respect to t, we differentiate each term:
- For x² sin(t), we use the product rule:
Derivative: 2x sin(t) (dx/dt) + x² cos(t) - For y³ cos(t), we again apply the product rule:
Derivative: 3y² (dy/dt) cos(t) – y³ sin(t)
Setting the total derivative to zero:
2x sin(t) (dx/dt) + x² cos(t) + (3y² (dy/dt) cos(t) – y³ sin(t)) = 0
Step 2: Solve for dy/dx (Chain Rule Application)
Using the chain rule, we have:
dy/dx = (dy/dt) / (dx/dt)
Rearranging the equation to isolate dy/dt:
(3y² (dy/dt) cos(t)) = – (2x sin(t) (dx/dt) + x² cos(t) + y³ sin(t))
Now substituting (dx/dt) with (1) for simplification:
So, we have:
dy/dt = (-2x sin(t) – x² cos(t) – y³ sin(t)) / (3y² cos(t))
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Step 3: Compute dy/dx
The derivatives can simplify and we can evaluate at specific t values. For example, at t=0:
– sin(0) = 0 and cos(0) = 1, substituting into our derivatives.
Repeat this with t=2 to compute dy/dx.
Step 4: Second Derivative d²y/dx²
For the second derivative, we start from:
dy/dx = f(t)
and differentiate once more with respect to t:
d²y/dx² = (d/dt(dy/dx))/(dx/dt)
Here we apply similar rules as before and substitute back into our expression to find the second derivative.
In conclusion, this process allows for evaluating both the first and second derivatives of the given implicit relation within the specified limits. Ensure to find numeric values at key points to derive practical conclusions.