College Algebra

1. To find a parametric representation of the curve \( x^3 + 2x^2 + y^2 = 3 \) with \( y \geq 0 \), we can use the following approach: Let's set \( x(t) = t \), where \( t \) is a parameter. Then we can solve for \( y(t) \) in terms of \( t \) using the given equation: \( t^3 + 2t^2 + y^2 = 3 \) Solving for \( y \), we get: \( y^2 = 3 - t^3 - 2t^2 \) Since \( y \geq 0 \), we take the positive square root: \( y(t) = \sqrt{3 - t^3 - 2t^2} \) The relevant range of \( t \) is determined by the condition that the expression under the square root must be non-negative: \( 3 - t^3 - 2t^2 \geq 0 \) This inequality can be solved to find the interval for \( t \). However, without specific methods to solve this cubic inequality, we can't provide the exact range for \( t \). In general, one would look for real roots of the cubic equation \( t^3 + 2t^2 - 3 = 0 \) and analyze the sign of the expression within the relevant intervals. 2. For the curve \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), a common parametric representation is: \( x(t) = a\cos(t) \) \( y(t) = b\sin(t) \) where \( t \) is the parameter, typically ranging from \( 0 \) to \( 2\pi \) for a complete traversal of the ellipse. The velocity \( \vec{v}(t) \) is the derivative of the position vector \( \vec{r}(t) = (x(t), y(t)) \): \( \vec{v}(t) = \left( \frac{dx}{dt}, \frac{dy}{dt} \right) = (-a\sin(t), b\cos(t)) \) The acceleration \( \vec{a}(t) \) is the derivative of the velocity vector: \( \vec{a}(t) = \left( \frac{d^2x}{dt^2}, \frac{d^2y}{dt^2} \right) = (-a\cos(t), -b\sin(t)) \) In the special case where \( a = b \), the ellipse becomes a circle, and the parametric equations simplify to: \( x(t) = a\cos(t) \) \( y(t) = a\sin(t) \) The speed \( v \) is the magnitude of the velocity vector: \( v = \sqrt{(-a\sin(t))^2 + (a\cos(t))^2} = \sqrt{a^2\sin^2(t) + a^2\cos^2(t)} = a \) The relation between the velocity and acceleration in this case is that the acceleration vector is always directed towards the center of the circle and is perpendicular to the velocity vector. 3. For the curve given by \( x = \sin^2(t), y = 1 - \cos(t), z = 2t \), we first find the derivatives of each component with respect to \( t \): \( \frac{dx}{dt} = 2\sin(t)\cos(t) \) \( \frac{dy}{dt} = \sin(t) \) \( \frac{dz}{dt} = 2 \) At \( t = t_0 = \frac{\pi}{2} \), we have: \( x(t_0) = \sin^2\left(\frac{\pi}{2}\right) = 1 \) \( y(t_0) = 1 - \cos\left(\frac{\pi}{2}\right) = 1 \) \( z(t_0) = 2\left(\frac{\pi}{2}\right) = \pi \) The tangent vector at \( t_0 \) is: \( \vec{T}(t_0) = \left( \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \right)_{t=t_0} = (0, 1, 2) \) The equation of the tangent line at \( M_0 \) is: \( \frac{x - 1}{0} = \frac{y - 1}{1} = \frac{z - \pi}{2} \) Since the denominator for \( x \) is zero, the tangent line is parallel to the \( x \)-axis, and we can write the equations as: \( x = 1 \) \( y = 1 + s \) \( z = \pi + 2s \) where \( s \) is a parameter along the tangent line. The normal plane is perpendicular to the tangent vector and passes through the point \( M_0 \). Its equation is: \( 0(x - 1) + 1(y - 1) + 2(z - \pi) = 0 \) Simplifying, we get: \( y - 1 + 2z - 2\pi = 0 \) \( y + 2z = 1 + 2\pi \) 4. To find the directional derivative of \( f(x,y) = xy^2 + x^3y \) at the point \( (1,2) \) in the direction of vector \( \vec{a} = \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \), we first compute the gradient of \( f \): \( \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (y^2 + 3x^2y, 2xy + x^3) \) Evaluating the gradient at \( (1,2) \): \( \nabla f(1,2) = (2^2 + 3 \cdot 1^2 \cdot 2, 2 \cdot 1 \cdot 2 + 1^3) = (4 + 6, 4 + 1) = (10, 5) \) The directional derivative in the direction of \( \vec{a} \) is the dot product of the gradient and the unit vector in the direction of \( \vec{a} \): \( D_{\vec{a}}f = \nabla f \cdot \vec{a} = (10, 5) \cdot \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) = \frac{10}{\sqrt{2}} + \frac{5}{\sqrt{2}} = \frac{15}{\sqrt{2}} \) 5. The temperature \( T \) of a solid is given by \( T(x,y,z) = e^{-x} + e^{2y} + e^{4z} \). The direction of the steepest descent is opposite to the gradient of \( T \). First, we compute the gradient: \( \nabla T = \left( \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}, \frac{\partial T}{\partial z} \right) = (-e^{-x}, 2e^{2y}, 4e^{4z}) \) Evaluating the gradient at \( (1,1,1) \): \( \nabla T(1,1,1) = (-e^{-1}, 2e^{2}, 4e^{4}) \) The direction of the steepest descent is the opposite of the gradient: \( -\nabla T(1,1,1) = (e^{-1}, -2e^{2}, -4e^{4}) \) This vector points in the direction where the temperature decreases the fastest from the point \( (1,1,1) \).

Math

1. Find a parametric representation of the curve defined as \( x^3 + 2x^2 + y^2 = 3, y \geq 0 \), i.e., describe it in the form \( x = x(t), y = y(t) \). Find the relevant range of \( t \). 2. Find a parametric representation of the curve defined as \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Consider this parametrization as the equations of motion of a particle and calculate the velocity and acceleration. Consider the special case \( a = b \). What is the value of the speed in this case? What is the relation between the velocity and acceleration? 3. Find the equations of the tangent line and of the normal plane to the curve described by the following parametrization: \( x = \sin^2(t), y = 1 - \cos(t), z = 2t \), at the point \( M_0 \) corresponding to \( t = t_0 = \frac{\pi}{2} \). 4. Find the directional derivative of \( f(x,y) = xy^2 + x^3y \) at the point \( (1,2) \) in the direction of vector \( a = \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \). 5. The temperature \( T \) of a solid is given by the function \( T(x,y,z) = e^{-x} + e^{2y} + e^{4z} \), where \( x, y, \) and \( z \) are the coordinates relative to the center of the solid. In which direction from the point \( (1,1,1) \) will the temperature decrease the fastest?

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