Question

Find the points (if they exist) at which the following planes and curves intersect. $$8 x+y+z=60 ; \quad \mathbf{r}(t)=\left\langle t, t^{2}, 3 t^{2}\right\rangle, \text { for }-\infty

Short answer

Question: Find the points at which the plane \(8x + y + z = 60\) intersects the curve given by the vector function \(\mathbf{r}(t) = \langle t, t^2, 3t^2 \rangle\). Answer: The intersection points are: $$P_1 = \left( \frac{-8 + \sqrt{1184}}{8}, \left(\frac{-8 + \sqrt{1184}}{8}\right)^2, 3\left(\frac{-8 + \sqrt{1184}}{8}\right)^2 \right)$$ $$P_2 = \left( \frac{-8 - \sqrt{1184}}{8}, \left(\frac{-8 - \sqrt{1184}}{8}\right)^2, 3\left(\frac{-8 - \sqrt{1184}}{8}\right)^2 \right)$$

Step-by-step solution

Identify the equation of the plane

The given equation of the plane is: $$8x + y + z = 60$$

Parameterize the curve

The given parameterized curve is: $$\mathbf{r}(t) = \langle t, t^2, 3t^2 \rangle$$ This means that the function is defined in terms of the parameter \(t\) as follows: $$x=t, \quad y=t^2, \quad z=3t^2$$

Substitute the parameterized curve into the plane's equation

To find the intersection points, substitute the parameterized curve into the plane's equation: $$8(t) + (t^2) + (3t^2) = 60$$

Solve for t

Now, solve the equation to find the value(s) of \(t\): $$8t + t^2 + 3t^2 = 60$$ $$4t^2 + 8t - 60 = 0$$ This is a quadratic equation. To find the roots, we can use the quadratic formula: $$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where \(a = 4, b = 8,\) and \(c = -60\). Substitute the values of \(a, b, c\) to find \(t\): $$t = \frac{-8 \pm \sqrt{8^2 - 4(4)(-60)}}{2(4)}$$ $$t = \frac{-8 \pm \sqrt{1184}}{8}$$ Now, we have two values of \(t\): $$t_1 = \frac{-8 + \sqrt{1184}}{8}$$ $$t_2 = \frac{-8 - \sqrt{1184}}{8}$$

Find the intersection points

Use the parameterized curve to find the intersection points by substituting the values of \(t\): For \(t_1 = \frac{-8 + \sqrt{1184}}{8}\): $$x_1 = t_1, \quad y_1 = t_1^2, \quad z_1 = 3t_1^2$$ For \(t_2 = \frac{-8 - \sqrt{1184}}{8}\): $$x_2 = t_2, \quad y_2 = t_2^2, \quad z_2 = 3t_2^2$$ The intersection points are: $$P_1 = \left( \frac{-8 + \sqrt{1184}}{8}, \left(\frac{-8 + \sqrt{1184}}{8}\right)^2, 3\left(\frac{-8 + \sqrt{1184}}{8}\right)^2 \right)$$ $$P_2 = \left( \frac{-8 - \sqrt{1184}}{8}, \left(\frac{-8 - \sqrt{1184}}{8}\right)^2, 3\left(\frac{-8 - \sqrt{1184}}{8}\right)^2 \right)$$ These are the points of intersection between the given plane and curve.