IDNLearn.com is your trusted platform for finding reliable answers. Get the information you need from our community of experts, who provide detailed and trustworthy answers.
Sagot :
To find the line integral of [tex]\( f(x, y) = y e^{x^2} \)[/tex] along the curve [tex]\( r(t) = -4t \mathbf{i} + 3t \mathbf{j} \)[/tex] for [tex]\( -2 \leq t \leq -1 \)[/tex], follow these steps:
### Step 1: Parameterize the Curve
Given the parameterization [tex]\( r(t) \)[/tex]:
- [tex]\( x(t) = -4t \)[/tex]
- [tex]\( y(t) = 3t \)[/tex]
### Step 2: Compute [tex]\( dx \)[/tex] and [tex]\( dy \)[/tex]
Differentiate [tex]\( x(t) \)[/tex] and [tex]\( y(t) \)[/tex] with respect to [tex]\( t \)[/tex]:
- [tex]\( \frac{dx}{dt} = \frac{d}{dt}(-4t) = -4 \)[/tex]
- [tex]\( \frac{dy}{dt} = \frac{d}{dt}(3t) = 3 \)[/tex]
### Step 3: Express [tex]\( f(x, y) \)[/tex] in Terms of [tex]\( t \)[/tex]
Substitute [tex]\( x(t) \)[/tex] and [tex]\( y(t) \)[/tex] into [tex]\( f(x, y) \)[/tex]:
[tex]\[ f(x(t), y(t)) = y(t) e^{(x(t))^2} = 3t e^{(-4t)^2} = 3t e^{16t^2} \][/tex]
### Step 4: Formulate the Line Integral
The line integral of [tex]\( f(x, y) \)[/tex] along the curve is given by:
[tex]\[ \int_{a}^{b} f(x(t), y(t)) \left( \frac{dx}{dt} dt \right) \][/tex]
Here, [tex]\( a = -2 \)[/tex] and [tex]\( b = -1 \)[/tex].
Compute:
[tex]\[ \frac{dx}{dt} = -4 \][/tex]
Substitute into the integral:
[tex]\[ \int_{-2}^{-1} 3t e^{16t^2} (-4) dt = -12 \int_{-2}^{-1} t e^{16t^2} dt \][/tex]
### Step 5: Evaluate the Integral
To evaluate the integral [tex]\( -12 \int_{-2}^{-1} t e^{16t^2} dt \)[/tex]:
Use substitution [tex]\( u = 16t^2 \)[/tex], then [tex]\( du = 32t dt \)[/tex] or [tex]\( \frac{du}{32} = t dt \)[/tex].
Change the limits of integration with respect to [tex]\( u \)[/tex]:
- When [tex]\( t = -2 \)[/tex], [tex]\( u = 16(-2)^2 = 64 \)[/tex]
- When [tex]\( t = -1 \)[/tex], [tex]\( u = 16(-1)^2 = 16 \)[/tex]
Thus, the integral becomes:
[tex]\[ -12 \int_{64}^{16} e^u \frac{du}{32} = -\frac{12}{32} \int_{64}^{16} e^u du = -\frac{3}{8} \int_{64}^{16} e^u du \][/tex]
### Step 6: Solve the Integral
[tex]\[ -\frac{3}{8} \int_{64}^{16} e^u du = -\frac{3}{8} \left[ e^u \right]_{64}^{16} \][/tex]
Evaluate the definite integral:
[tex]\[ -\frac{3}{8} \left( e^{16} - e^{64} \right) = -\frac{3}{8} e^{16} + \frac{3}{8} e^{64} \][/tex]
### Step 7: Write the Final Answer
The value of the line integral is:
[tex]\[ = \frac{3}{8} e^{64} - \frac{3}{8} e^{16} \][/tex]
Simplifying, we get:
[tex]\[ = -3 \frac{e^{16}}{8} + 3 \frac{e^{64}}{8} \][/tex]
Thus, the final result of the line integral is:
[tex]\[ -\frac{3}{8} e^{16} + \frac{3}{8} e^{64} \][/tex]
Therefore, the line integral of [tex]\( f(x, y) \)[/tex] along the curve [tex]\( r(t) \)[/tex] from [tex]\( t = -2 \)[/tex] to [tex]\( t = -1 \)[/tex] is:
[tex]\[ -3 \frac{e^{16}}{8} + 3 \frac{e^{64}}{8} \][/tex]
### Step 1: Parameterize the Curve
Given the parameterization [tex]\( r(t) \)[/tex]:
- [tex]\( x(t) = -4t \)[/tex]
- [tex]\( y(t) = 3t \)[/tex]
### Step 2: Compute [tex]\( dx \)[/tex] and [tex]\( dy \)[/tex]
Differentiate [tex]\( x(t) \)[/tex] and [tex]\( y(t) \)[/tex] with respect to [tex]\( t \)[/tex]:
- [tex]\( \frac{dx}{dt} = \frac{d}{dt}(-4t) = -4 \)[/tex]
- [tex]\( \frac{dy}{dt} = \frac{d}{dt}(3t) = 3 \)[/tex]
### Step 3: Express [tex]\( f(x, y) \)[/tex] in Terms of [tex]\( t \)[/tex]
Substitute [tex]\( x(t) \)[/tex] and [tex]\( y(t) \)[/tex] into [tex]\( f(x, y) \)[/tex]:
[tex]\[ f(x(t), y(t)) = y(t) e^{(x(t))^2} = 3t e^{(-4t)^2} = 3t e^{16t^2} \][/tex]
### Step 4: Formulate the Line Integral
The line integral of [tex]\( f(x, y) \)[/tex] along the curve is given by:
[tex]\[ \int_{a}^{b} f(x(t), y(t)) \left( \frac{dx}{dt} dt \right) \][/tex]
Here, [tex]\( a = -2 \)[/tex] and [tex]\( b = -1 \)[/tex].
Compute:
[tex]\[ \frac{dx}{dt} = -4 \][/tex]
Substitute into the integral:
[tex]\[ \int_{-2}^{-1} 3t e^{16t^2} (-4) dt = -12 \int_{-2}^{-1} t e^{16t^2} dt \][/tex]
### Step 5: Evaluate the Integral
To evaluate the integral [tex]\( -12 \int_{-2}^{-1} t e^{16t^2} dt \)[/tex]:
Use substitution [tex]\( u = 16t^2 \)[/tex], then [tex]\( du = 32t dt \)[/tex] or [tex]\( \frac{du}{32} = t dt \)[/tex].
Change the limits of integration with respect to [tex]\( u \)[/tex]:
- When [tex]\( t = -2 \)[/tex], [tex]\( u = 16(-2)^2 = 64 \)[/tex]
- When [tex]\( t = -1 \)[/tex], [tex]\( u = 16(-1)^2 = 16 \)[/tex]
Thus, the integral becomes:
[tex]\[ -12 \int_{64}^{16} e^u \frac{du}{32} = -\frac{12}{32} \int_{64}^{16} e^u du = -\frac{3}{8} \int_{64}^{16} e^u du \][/tex]
### Step 6: Solve the Integral
[tex]\[ -\frac{3}{8} \int_{64}^{16} e^u du = -\frac{3}{8} \left[ e^u \right]_{64}^{16} \][/tex]
Evaluate the definite integral:
[tex]\[ -\frac{3}{8} \left( e^{16} - e^{64} \right) = -\frac{3}{8} e^{16} + \frac{3}{8} e^{64} \][/tex]
### Step 7: Write the Final Answer
The value of the line integral is:
[tex]\[ = \frac{3}{8} e^{64} - \frac{3}{8} e^{16} \][/tex]
Simplifying, we get:
[tex]\[ = -3 \frac{e^{16}}{8} + 3 \frac{e^{64}}{8} \][/tex]
Thus, the final result of the line integral is:
[tex]\[ -\frac{3}{8} e^{16} + \frac{3}{8} e^{64} \][/tex]
Therefore, the line integral of [tex]\( f(x, y) \)[/tex] along the curve [tex]\( r(t) \)[/tex] from [tex]\( t = -2 \)[/tex] to [tex]\( t = -1 \)[/tex] is:
[tex]\[ -3 \frac{e^{16}}{8} + 3 \frac{e^{64}}{8} \][/tex]
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. For dependable answers, trust IDNLearn.com. Thank you for visiting, and we look forward to helping you again soon.
