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Magazine Chapter 11: Problem 1
Calculate \(\frac{\mathrm{d} y}{\mathrm{~d} t}\) and \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} t^{2}}\) given (a) \(y=t^{2}+t\) (b) \(y=2 t^{3}-t^{2}+1\) (c) \(y=\sin 2 t\) (d) \(y=\sin k t \quad k\) constant (e) \(y=2 \mathrm{e}^{3 t}-t^{2}+1\) (f) \(y=\frac{t}{t+1}\) (g) \(y=4 \cos \frac{t}{2}\) (h) \(y=\mathrm{e}^{t} t\) (i) \(y=\sinh 4 t\) (j) \(y=\sin ^{2} t\)
Short Answer
Expert verified
Each function has been differentiated twice, yielding its first and second derivatives accordingly.
Step by step solution
01
Differentiate with Respect to t (Basic Derivative)
To find the first derivative \( \frac{\mathrm{d} y}{\mathrm{d} t} \), apply the standard differentiation rules for polynomials, trigonometric, exponential, and hyperbolic functions. Use the chain rule and product rule as necessary.
02
Differentiate Again for Second Derivative
To find the second derivative \( \frac{\mathrm{d}^2 y}{\mathrm{d} t^2} \), differentiate the expression obtained from Step 1 once more. Apply differentiation rules as before.
03
Sub-step for (a): Compute Derivatives of a Polynomial
For \( y = t^2 + t \), the first derivative is \( \frac{\mathrm{d} y}{\mathrm{d} t} = 2t + 1 \). The second derivative is \( \frac{\mathrm{d}^2 y}{\mathrm{d} t^2} = 2 \).
04
Sub-step for (b): Compute Derivatives of a Cubic Polynomial
For \( y = 2t^3 - t^2 + 1 \), the first derivative is \( \frac{\mathrm{d} y}{\mathrm{d} t} = 6t^2 - 2t \). The second derivative is \( \frac{\mathrm{d}^2 y}{\mathrm{d} t^2} = 12t - 2 \).
05
Sub-step for (c): Compute Derivatives of a Sine Function
For \( y = \sin 2t \), the first derivative is \( \frac{\mathrm{d} y}{\mathrm{d} t} = 2\cos 2t \). The second derivative is \( \frac{\mathrm{d}^2 y}{\mathrm{d} t^2} = -4\sin 2t \) by applying the chain rule.
06
Sub-step for (d): Differentiate Sine with a Constant Multiplier
For \( y = \sin kt \), \( k \) constant, the first derivative is \( \frac{\mathrm{d} y}{\mathrm{d} t} = k\cos kt \). The second derivative is \( \frac{\mathrm{d}^2 y}{\mathrm{d} t^2} = -k^2\sin kt \).
07
Sub-step for (e): Differentiate an Exponential and Polynomial
For \( y = 2e^{3t} - t^2 + 1 \), the first derivative is \( \frac{\mathrm{d} y}{\mathrm{d} t} = 6e^{3t} - 2t \). The second derivative is \( \frac{\mathrm{d}^2 y}{\mathrm{d} t^2} = 18e^{3t} - 2 \) by using the chain rule.
08
Sub-step for (f): Differentiate a Rational Function using Quotient Rule
For \( y = \frac{t}{t+1} \), the first derivative is \( \frac{\mathrm{d} y}{\mathrm{d} t} = \frac{1}{(t+1)^2} \). The second derivative is \( \frac{\mathrm{d}^2 y}{\mathrm{d} t^2} = -\frac{2}{(t+1)^3} \) by applying the quotient rule.
09
Sub-step for (g): Compute Derivatives of a Cosine Function with a Constant Multiplier
For \( y = 4 \cos \frac{t}{2} \), the first derivative is \( \frac{\mathrm{d} y}{\mathrm{d} t} = -2\sin \frac{t}{2} \). The second derivative is \( \frac{\mathrm{d}^2 y}{\mathrm{d} t^2} = -\cos \frac{t}{2} \).
10
Sub-step for (h): Differentiate Product of Exponential and Linear Function
For \( y = e^t t \), use the product rule: the first derivative is \( \frac{\mathrm{d} y}{\mathrm{d} t} = e^t + te^t \). The second derivative is \( \frac{\mathrm{d}^2 y}{\mathrm{d} t^2} = 2e^t + te^t \).
11
Sub-step for (i): Differentiate Hyperbolic Sine
For \( y = \sinh 4t \), the first derivative is \( \frac{\mathrm{d} y}{\mathrm{d} t} = 4\cosh 4t \). The second derivative is \( \frac{\mathrm{d}^2 y}{\mathrm{d} t^2} = 16\sinh 4t \).
12
Sub-step for (j): Use Chain Rule for Sine Squared
For \( y = \sin^2 t \), the first derivative is \( \frac{\mathrm{d} y}{\mathrm{d} t} = 2\sin t \cos t = \sin 2t \). The second derivative is \( \frac{\mathrm{d}^2 y}{\mathrm{d} t^2} = 2\cos 2t \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polynomial Differentiation
Polynomial differentiation is one of the fundamental parts of calculus. It's the process of finding how the polynomial function's output changes in response to changes in its input. This is achieved by applying simple rules of differentiation directly.
- For a basic polynomial like \( y = t^n \), the derivative with respect to \( t \) is \( y' = nt^{n-1} \). This is known as the power rule.
- Extending this concept, for a polynomial like \( y = t^2 + t \), we differentiate each term separately: the derivative becomes \( y' = 2t + 1 \).
- Higher degree terms follow similar rules: differentiate each term using the power rule, like in \( y = 2t^3 - t^2 + 1 \), where the derivative becomes \( y' = 6t^2 - 2t \).
Trigonometric Differentiation
Trigonometric functions, such as sine and cosine, often appear in calculus. Differentiating these functions involves understanding their specific derivative rules.
- The derivative of \( y = \sin x \) is \( y' = \cos x \). Consequently, for \( y = \sin 2t \), the chain rule gives us \( y' = 2\cos 2t \).
- Similarly, for \( y = \sin kt \) where \( k \) is a constant, the derivative is \( y' = k\cos kt \).
- The differentiation of cosine follows a similar pattern. For example, for \( y = 4\cos \frac{t}{2} \), the derivative is \( y' = -2\sin \frac{t}{2} \).
Exponential Functions Differentiation
Exponential functions involve constant terms raised to the power of a variable. Differentiating them follows specific rules that highlight exponential growth or decay.
- For the function \( y = e^x \), its derivative is \( y' = e^x \); the function differentiates to itself. This unique property simplifies many calculus problems.
- If the exponent is a function of \( x \), such as \( y = e^{3t} \), the chain rule is applied giving \( y' = 3e^{3t} \).
- In cases where exponential functions are combined with other functions like polynomials, such as \( y = 2e^{3t} - t^2 + 1 \), each part is differentiated separately, leveraging the specific rules for each function type.
Hyperbolic Functions Differentiation
Hyperbolic functions resemble trigonometric functions but relate to the hyperbola, and they have their unique differentiation rules.
- The function \( y = \sinh x \) has a derivative \( y' = \cosh x \), following its innate relationship with the hyperbolic cosine.
- For \( y = \sinh 4t \), the chain rule yields \( y' = 4\cosh 4t \), illustrating an enhanced growth in response to input changes.
- The derivatives for hyperbolic functions remain consistent with their base identities, making them predictable for analyses involving hyperbolic forms or in solving differential equations.
