AP Precalculus Functions Involving Parameters, Vectors, and Matrices — Worked Answer Explanations

Unit 4 · 12 questions explained

1. Question 1 · Easy

   A particle moves along the parametric curve $x(t)=t+1$, $y(t)=2t-3$. What is the position at $t=2$?

   • A
     $(2,-3)$

     Why not A: Evaluated $x(0)$ and $y(0)$ instead of $t=2$.
   • B
     $(3,1)$Correct
   • C
     $(2,1)$

     Why not C: Used $t=2$ for $x$ but $t=2$ for $y$ as if $y(t)=t-1$.
   • D
     $(1,2)$

     Why not D: Reversed which coordinate is $x$ and which is $y$.
   Explanation

   Substitute $t=2$ into both equations: $x(2)=2+1=3$ and $y(2)=2(2)-3=1$. Position: $(3,1)$.

   Key takeaway

   Evaluating a parametric curve at $t$ gives the position $(x(t), y(t))$ at that parameter value.

2. Question 2 · Easy

   Find the magnitude of the vector $\mathbf{v}=⟨3,4⟩$.

   • A
     $3$

     Why not A: Used only the $x$-component.
   • B
     $4$

     Why not B: Used only the $y$-component.
   • C
     $5$Correct
   • D
     $7$

     Why not D: Added the components instead of using the Pythagorean formula.
   Explanation

   Magnitude: $|\mathbf{v}|=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$.

   Key takeaway

   Magnitude of $\langle a, b \rangle$ is $\sqrt{a^2 + b^2}$ — Pythagorean theorem in disguise.

3. Question 3 · Easy

   Eliminate the parameter from $x=t+1$, $y=t^2$ to express $y$ as a function of $x$.

   • A
     $y=x^2$

     Why not A: Forgot to substitute $t=x-1$.
   • B
     $y=(x-1{)}^2$Correct
   • C
     $y=(x+1{)}^2$

     Why not C: Solved $t=x+1$ instead of $t=x-1$.
   • D
     $y=x^2-1$

     Why not D: Distributed incorrectly: $(x-1{)}^2\ne x^2-1$.
   Explanation

   Solve the first equation for $t$: $t=x-1$. Substitute into $y=t^2$: $y=(x-1{)}^2$.

   Key takeaway

   Eliminate the parameter by solving one equation for $t$ and substituting into the other.

4. Question 4 · Easy

   Compute $\mathbf{u}+\mathbf{v}$ where $\mathbf{u}=⟨2,-1⟩$ and $\mathbf{v}=⟨-3,4⟩$.

   • A
     $⟨-1,3⟩$Correct
   • B
     $⟨5,-5⟩$

     Why not B: Subtracted instead of added componentwise.
   • C
     $⟨-6,-4⟩$

     Why not C: Multiplied componentwise instead of adding.
   • D
     $⟨1,5⟩$

     Why not D: Arithmetic slip on the $x$-component.
   Explanation

   Add componentwise: $\mathbf{u}+\mathbf{v}=⟨2+(-3),-1+4⟩=⟨-1,3⟩$.

   Key takeaway

   Vectors add componentwise: $\langle a, b \rangle + \langle c, d \rangle = \langle a + c, b + d \rangle$.

5. Question 5 · Medium

   A parametric curve is given by $x(t)=2\cos (t)$, $y(t)=2\sin (t)$ for $t\in [0,2\pi )$. Which curve does it trace?

   • A
     A circle of radius $1$ centered at the origin.

     Why not A: Forgot the factor of $2$ on both components — that gives radius $2$, not $1$.
   • B
     A circle of radius $2$ centered at the origin.Correct
   • C
     An ellipse with semi-axes $2$ and $1$.

     Why not C: Both components are scaled by the same $2$, so the curve is a circle, not an ellipse.
   • D
     A line of slope $1$ through the origin.

     Why not D: Sine and cosine produce periodic, not linear, motion.
   Explanation

   Compute $x^2+y^2=4{\cos }^{2}t+4{\sin }^{2}t=4({\cos }^{2}t+{\sin }^{2}t)=4$. This is a circle of radius $\sqrt{4}=2$ centered at the origin.

   Key takeaway

   $x = r\cos t$, $y = r\sin t$ traces a circle of radius $r$ — use $\sin^2 + \cos^2 = 1$ to eliminate the parameter.

6. Question 6 · Medium

   What is the unit vector in the direction of $\mathbf{w}=⟨6,-8⟩$?

   • A
     $⟨6,-8⟩$

     Why not A: Original vector has magnitude $10$, not $1$.
   • B
     $⟨\frac{3}{5},-\frac{4}{5}⟩$Correct
   • C
     $⟨\frac{6}{10},\frac{-8}{10}⟩\cdot 2$

     Why not C: Doubled the unit vector — magnitude would be $2$, not $1$.
   • D
     $⟨\frac{1}{6},-\frac{1}{8}⟩$

     Why not D: Reciprocated each component instead of dividing by the magnitude.
   Explanation

   Magnitude: $|\mathbf{w}|=\sqrt{36+64}=\sqrt{100}=10$. Unit vector: $\frac{1}{10}⟨6,-8⟩=⟨3/5,-4/5⟩$.

   Key takeaway

   Unit vector: $\hat{\mathbf{v}} = \dfrac{\mathbf{v}}{|\mathbf{v}|}$ — divide each component by the magnitude.

7. Question 7 · Medium

   If $A=\left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right]$ acts on a vector by left multiplication, what is $A⟨1,4{⟩}^T$?

   • A
     $⟨2,4{⟩}^T$

     Why not A: Multiplied only the first component by $2$ — forgot to scale $y$.
   • B
     $⟨2,12{⟩}^T$Correct
   • C
     $⟨1,12{⟩}^T$

     Why not C: Multiplied only the second component — missed scaling $x$.
   • D
     $⟨5,7{⟩}^T$

     Why not D: Added matrix entries to vector entries instead of multiplying.
   Explanation

   $A\left[\begin{array}{c}1\\ 4\end{array}\right]=\left[\begin{array}{c}2\cdot 1+0\cdot 4\\ 0\cdot 1+3\cdot 4\end{array}\right]=\left[\begin{array}{c}2\\ 12\end{array}\right]$.

   Key takeaway

   Diagonal matrices stretch each axis independently: $\text{diag}(a, b)$ sends $\langle x, y \rangle$ to $\langle ax, by \rangle$.

8. Question 8 · Medium

   Compute the dot product of $\mathbf{u}=⟨2,3⟩$ and $\mathbf{v}=⟨4,-1⟩$.

   • A
     $5$Correct
   • B
     $11$

     Why not B: Multiplied the $y$-components as $+3\cdot +1$ — sign error on the $-1$.
   • C
     $⟨8,-3⟩$

     Why not C: Componentwise multiplied — that's the Hadamard product, not the dot product.
   • D
     $2$

     Why not D: Subtracted the two products instead of adding them.
   Explanation

   Dot product: $\mathbf{u}\cdot \mathbf{v}=(2)(4)+(3)(-1)=8-3=5$.

   Key takeaway

   Dot product = sum of componentwise products; result is a scalar, not a vector.

9. Question 9 · Hard

   Find the inverse of the matrix $A=\left[\begin{array}{cc}2& 1\\ 1& 1\end{array}\right]$.

   • A
     $\left[\begin{array}{cc}1& -1\\ -1& 2\end{array}\right]$Correct
   • B
     $\left[\begin{array}{cc}-1& 1\\ 1& -2\end{array}\right]$

     Why not B: Sign error throughout — forgot the determinant sign.
   • C
     $\left[\begin{array}{cc}2& -1\\ -1& 1\end{array}\right]$

     Why not C: Only flipped signs on the off-diagonal; should also have swapped the diagonal entries.
   • D
     $\left[\begin{array}{cc}1& 2\\ 1& 1\end{array}\right]$

     Why not D: Did not apply the inverse formula at all — just rearranged the matrix.
   Explanation

   For a $2\times 2$ matrix $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$, the inverse is $\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$. Determinant: $2(1)-1(1)=1$. So $A^{-1}=\left[\begin{array}{cc}1& -1\\ -1& 2\end{array}\right]$. Verify: $A\cdot A^{-1}=I$.

   Key takeaway

   $2 \times 2$ inverse: swap the diagonal, negate the off-diagonal, divide by the determinant.

10. Question 10 · Hard

    A projectile is launched with initial velocity ${\mathbf{v}}_0=⟨20,30⟩$ m/s. Ignoring gravity, what is its position vector at $t=4$ seconds, starting from the origin?

    • A
      $⟨24,34⟩$ m

      Why not A: Added $t=4$ to each component instead of multiplying.
    • B
      $⟨80,120⟩$ mCorrect
    • C
      $⟨20,30⟩$ m

      Why not C: Ignored time — position equals velocity here, which is only true at $t=1$.
    • D
      $⟨5,7.5⟩$ m

      Why not D: Divided velocity by time instead of multiplying.
    Explanation

    Position vector under constant velocity (no gravity): $\mathbf{r}(t)={\mathbf{r}}_0+t{\mathbf{v}}_0$. Here ${\mathbf{r}}_0=0$ and $t=4$, so $\mathbf{r}(4)=4⟨20,30⟩=⟨80,120⟩$.

    Key takeaway

    Under constant velocity, position scales linearly with time: $\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}_0$.

11. Question 11 · Hard

    The parametric curve $x(t)=\cos (t)$, $y(t)=\sin (2t)$ has what behavior at $t=0$?

    • A
      Passes through $(1,0)$ moving in the $-y$ direction.

      Why not A: Right position but wrong sign on the vertical motion; $y^{\prime }(0)=2>0$, not negative.
    • B
      Passes through $(1,0)$ moving in the $+y$ direction.Correct
    • C
      Passes through $(0,0)$ moving in the $+x$ direction.

      Why not C: $x(0)=\cos 0=1$, not $0$.
    • D
      Passes through $(1,0)$ stationary.

      Why not D: $y^{\prime }(0)=2\cos (0)=2\ne 0$, so it's moving vertically.
    Explanation

    Position: $x(0)=\cos 0=1$, $y(0)=\sin 0=0$, so the curve is at $(1,0)$. Velocity: $x^{\prime }(t)=-\sin t$, $y^{\prime }(t)=2\cos (2t)$. At $t=0$: $x^{\prime }(0)=0$, $y^{\prime }(0)=2$, so the motion is purely upward.

    Key takeaway

    Velocity vector at $t$: $\langle x'(t), y'(t) \rangle$ — its direction tells you which way the curve is moving.

12. Question 12 · Hard

    Solve the system using matrix methods: $2x+y=5$ and $x-y=1$.

    • A
      $x=1$, $y=2$

      Why not A: Doesn't satisfy the first equation: $2(1)+2=4\ne 5$.
    • B
      $x=2$, $y=1$Correct
    • C
      $x=3$, $y=-1$

      Why not C: Doesn't satisfy the second equation: $3-(-1)=4\ne 1$.
    • D
      $x=0$, $y=5$

      Why not D: Solved only the first equation; the second is not satisfied.
    Explanation

    Write as $A\mathbf{x}=\mathbf{b}$ with $A=\left[\begin{array}{cc}2& 1\\ 1& -1\end{array}\right]$ and $\mathbf{b}=⟨5,1{⟩}^T$. Determinant: $2(-1)-1(1)=-3$. Inverse: $\frac{1}{-3}\left[\begin{array}{cc}-1& -1\\ -1& 2\end{array}\right]=\left[\begin{array}{cc}1/3& 1/3\\ 1/3& -2/3\end{array}\right]$. Solution: $A^{-1}\mathbf{b}=⟨(5+1)/3,(5-2)/3⟩=⟨2,1⟩$. So $x=2$, $y=1$. Verify: $2(2)+1=5$ ✓, $2-1=1$ ✓.

    Key takeaway

    Linear systems can be solved as $\mathbf{x} = A^{-1}\mathbf{b}$ when $A$ is invertible.