Find the radius of curvature of the following curve at the given point. Then write the equation of the circle of curvature at the point. The radius of curvature at a point \( P \) is given by \( \frac{1}{\mathrm{k}} \), whese k is the curvature at \( P \). \( r(t)=\left\langle t,-5 t^{2}\right\rangle \) at \( t=0 \) The radius of curvature at \( t=0 \) is The equation of the circle of curvature at \( t=0 \) is \( (x-(\square))^{2}+(y-(\square))^{2}=\square \).

To find the radius of curvature of the curve given by \( r(t) = \langle t, -5t^2 \rangle \) at the point corresponding to \( t = 0 \), we first need to compute the curvature \( k \) at that point. ### Step 1: Compute the first and second derivatives of \( r(t) \) 1. **First derivative** \( r'(t) \): \[ r'(t) = \left\langle \frac{d}{dt}(t), \frac{d}{dt}(-5t^2) \right\rangle = \langle 1, -10t \rangle \] At \( t = 0 \): \[ r'(0) = \langle 1, 0 \rangle \] 2. **Second derivative** \( r''(t) \): \[ r''(t) = \left\langle \frac{d^2}{dt^2}(t), \frac{d^2}{dt^2}(-5t^2) \right\rangle = \langle 0, -10 \rangle \] At \( t = 0 \): \[ r''(0) = \langle 0, -10 \rangle \] ### Step 2: Compute the curvature \( k \) The curvature \( k \) is given by the formula: \[ k = \frac{\| r'(t) \times r''(t) \|}{\| r'(t) \|^3} \] In two dimensions, the cross product can be computed as: \[ \| r'(t) \times r''(t) \| = \left| \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 0 \\ 0 & -10 & 0 \end{vmatrix} \right| = |(1)(-10) - (0)(0)| = 10 \] Now, we compute \( \| r'(0) \| \): \[ \| r'(0) \| = \sqrt{1^2 + 0^2} = 1 \] Now we can find the curvature \( k \): \[ k = \frac{10}{1^3} = 10 \] ### Step 3: Find the radius of curvature The radius of curvature \( R \) is given by: \[ R = \frac{1}{k} = \frac{1}{10} \] ### Step 4: Find the center of the circle of curvature To find the center of the circle of curvature, we need to find the unit normal vector at \( t = 0 \). The unit tangent vector \( T \) is: \[ T(0) = \frac{r'(0)}{\| r'(0) \|} = \langle 1, 0 \rangle \] The unit normal vector \( N \) can be obtained by rotating \( T \) by \( 90^\circ \): \[ N(0) = \langle 0, -1 \rangle \] The center of the circle of curvature is located at: \[ \text{Center} = r(0) + R \cdot N(0) \] Calculating \( r(0) \): \[ r(0) = \langle 0, 0 \rangle \] Thus, the center is: \[ \text{Center} = \langle 0, 0 \rangle + \frac{1}{10} \langle 0, -1 \rangle = \langle 0, -\frac{1}{10} \rangle \] ### Step 5: Write the equation of the circle of curvature The equation of the circle of curvature with center \( (0, -\frac{1}{10}) \) and radius \( \frac{1}{10} \) is: \[ (x - 0)^2 + \left(y + \frac{1}{10}\right)^2 = \left(\frac{1}{10}\right)^2 \] This simplifies to: \[ x^2 + \left(y + \frac{1}{10}\right)^2 = \frac{1}{100} \] ### Final Answers The radius of curvature at \( t = 0 \) is: \[ \frac{1}{10} \] The equation of the circle of curvature at \( t = 0 \