Brain Gains (Rapid Recall, Jivin' Generation)

1. For the curve $\vec r(t) = (t^2+2, -3t+4)$, note that $\frac{d\vec r}{dt} =(2t,-3)$. Give a vector equation of the tangent line to $\vec r(t)$ at $t=2$. In other words, give a vector equation of a line that passes through $\vec r(2) = (6,-2)$ and is parallel to $\frac{d\vec r}{dt}(2) = (4,-3)$.

Solution

Passing through $(6,-2)$ and parallel to $(4,-3)$ means an equation is $$(x,y) = (4,-3)t+(6,-2)\quad \text{or}\quad \vec r(t) = (4,-3)t+(6,-2).$$ This is the same as $$(x,y) = (4t+6,-3t-2).$$

2. The curve above represents the path of an object. Find the object's velocity and speed at any time $t$, and then at $t=1$.

Solution

The derivative $\frac{d\vec r}{dt} =(2t,-3)$ gives us the velocity at any time $t$. So at $t=1$ we have the velocity as $$\vec v = (2,-3).$$ The speed is the magnitude of the velocity, which gives the speed as $$v=|\vec v|=||\vec v|| = \sqrt{4t^2+9}.$$ At $t=1$ we have $v(1) = \sqrt{13}$. Note that we may use the same letter, without a vector symbol, to represent the magnitude of the corresponding vector, and you'll find that some people prefer double bars $||\vec v||$ instead of single bards $|\vec v|$.

3. The (vector) projection of $\vec P$ onto $\vec Q$ is given by the formula $$\ds \text{proj}_\vec Q\vec P = \frac{\vec P\cdot \vec Q}{\vec Q\cdot \vec Q}\vec Q.$$ Compute the projection of $\vec P$ onto $\vec Q$ using $P=(3,4)$ and $Q=(-2,5)$.

Solution

We compute $$\ds \text{proj}_\vec Q\vec P = \frac{\vec P\cdot \vec Q}{\vec Q\cdot \vec Q}\vec Q = \frac{(3)(-2)+(4)(5)}{(-2)^2+5^2}(-2,5) = \frac{14}{29}(-2,5) .$$ Note that this is a little less than half $\vec Q$.

Group Problems

Remember to pass the chalk after each problem (or part of a problem).

  1. Give a vector equation of the line that passes through the point $(1,2,3)$ and $(-2,4,9)$.
  2. An object starts at $P=(1,2,3)$ and each unit of time its displacement is $\vec v=(-4,5,1)$. Give an equation for the position $(x,y,z)$ at any time $t$.
    • What is the speed of an object that follows the path described above?
  3. Draw the parametric curve $\vec r(t) = (1+t^2,3t-2)$ for $-2\leq t\leq 3$.
    • Give $\frac{d\vec r}{dt}$. Then state $\frac{dx}{dt}$, $\frac{dy}{dt}$, and $\frac{dy}{dx}$.
    • Give a vector equation of the tangent line to the curve at $t=2$.
  4. Let $P=(3,4)$ and $Q=(2,0)$.
    • Using the formula $$\ds \text{proj}_\vec Q\vec P = \frac{\vec P\cdot \vec Q}{\vec Q\cdot \vec Q}\vec Q,$$ compute $\text{proj}_\vec Q\vec P$, the project of $\vec P$ onto $\vec Q$. We may also write this as $\vec P _{\parallel \vec Q}$, which we read as, "The vector component of $\vec P$ that is parallel to $\vec Q$."
    • Draw $\vec P$, $\vec Q$ and $\text{proj}_\vec Q\vec P $ on the same grid with their base at the origin.
    • Add to your picture the vector difference $\vec P_{\perp \vec Q} = \vec P - \text{proj}_{\vec Q}\vec P$, which we call "The vector component of $\vec P$ that is orthogonal to $\vec Q$." How is $\vec P_{\perp \vec Q}$ related to the other vectors?
    • Now compute the projection of $\vec Q$ onto $\vec P$ (so swap which vector is projected onto the other).
    • Draw $\vec P$, $\vec Q$ and $\text{proj}_\vec P\vec Q $ on the same grid with their base at the origin, and add to your picture the vector difference $\vec Q_{\perp \vec P} = \vec Q - \text{proj}_{\vec P}\vec Q $.
  5. Repeat the previous problem with $P=(3,4)$ and $Q=(-1,1)$.