MA005: Calculus I

1. (a) –3/4 (b) 1/2 (c) 0 (d) 2 (e) undefined

3. (a) $\dfrac{4}{3}$ (b) $\dfrac{ –9}{5}$ (c) $x + 2$ (if $x ≠ 2$) (d) $4 + h$ (if $h ≠ 0$) (e) $a + x$ (if $a ≠ x$)

5. (a) $t = 5: \dfrac{5000}{1500} = \dfrac{10}{3} , t = 10: \dfrac{5000}{3000} =\dfrac{5}{3} , t = 20: \dfrac{5000}{6000} =\dfrac{5}{6}$

(b) any $t > 0: \dfrac{5000}{300t} = \dfrac{50}{3t}$

(c) decreasing, since the numerator remains constant at 5000 while the denominator increases.

7. The restaurant is 4 blocks south and 2 blocks east. The distance is $\sqrt{4^2 + 2^2} =\sqrt{20} ≈ 4.47$ blocks.

9. $y = \sqrt{20^2 – 4^2} = \sqrt{384} ≈ 19.6$ feet, $m = \dfrac{\sqrt{384}}{4} ≈ 4.9 . tan( q ) = \dfrac{\sqrt{384}}{4} ≈ 4.9$ so $q ≈ 1.37 (≈ 78.5^o)$.

11. The equation of the line through $P = (2,3)$ and $Q = (8,11)$ is $y – 3 =\dfrac{8}{6} (x – 2)$ or $6y – 8x = 2$. Substituting $x = 2a + 8(1–a) = 8 – 6a$ and $y = 3a + 11(1–a) = 11 – 8a$ into the equation for the line, we get $6(11 – 8a) –8(8 – 6a) = 66 – 48a –64 + 48a$ which equals 2 for every value of a, so the point with $x = 2a + 8(1–a)$ and $y = 3a + 11(1–a)$ is on the line through P and Q for every value of a.

$The Dist(P,Q) =\sqrt{6^2 + 8^2} = 10$. $Dist(P,R) = \sqrt{(8–6a – 2)^2 + (11–8a – 3)^2}$

$=\sqrt{(6 – 6a)^2 + (8 – 8a)^2} = \sqrt{6^2(1–a)^2 + 8^2(1–a)^2} = \sqrt{100(1–a)^2} = 10 . |1–a| = |1–a| . Dist(P,Q)$.

13. (a) $m_1 \cdot m_2 = (1)(-1) = -1$ so the lines are perpendicular.

(b) Because 20 units of x-values are physically wider on the screen than 20 units of y-values.

(c) Set the window so (xmax - xmin) ≈ 1.7 (ymax - ymin).

15. (a) $y – 5 = 3(x – 2)$ or $y = 3x – 1$.

(b) $y – 2 = –2(x – 3)$ or $y = 8 – 2x$

(c) $y – 4 = –\dfrac{1}{2}(x – 1)$ or $y = –\dfrac{1}{2} x + \dfrac{9}{2}$

17. (a) $y – 5 = \dfrac{3}{2}(x–2)$ or $y = \dfrac{3}{2} x + 2$

(b) $y – 2 = \dfrac{3}{2}(x+1)$ or $y = \dfrac{3}{2} x + \dfrac{7}{2}$

(c) $x = 3$.

19. The distance between the centers is $\sqrt{6^2 + 8^2} = 10$.

(a) $10–2–4 = 4$

(b) $10–2–7 = 1$

(c) 0 (they intersect)

(d) $15–10–3 = 2$

(e) $12–10–1 = 1$.

21. Find $Dist( P,C ) = \sqrt{ (x–h)^2 + (y–k)^2}$, and compare the value to r: P is

inside the circle              if $Dist( P,C ) < r$

on the circle                   if $Dist( P,C ) = r$

outside the circle          if $Dist( P,C ) > r$


23. A point $P =(x,y)$ lies on the circle if and only if its distance from $C = (h,k)$ is $r : Dist( P,C ) = r$. So P is on the circle if and only if $\sqrt{(x–h)^2 + (y–k)^2} = r$ or $(x–h)^2 + (y–k)^2 = r^2$.

25. (a) $slope is –\dfrac{5}{12}$

(b) undefined (vertical line)

(c) $\dfrac{12}{5}$

(d) 0 (horizontal line)

27. (a) distance ≈ 2.22.

(b) Distance ≈ 2.24.

(c) (by inspection) 3 units which occurs at the point (5, 3).

29. (a) If $B ≠ 0$, we may solve for y: $y = – \dfrac{A}{B} x + \dfrac{C}{B}$. The slope is the coefficient of x: $m = –\dfrac{A}{B}$.

(b) The required slope is B/A (the negative reciprocal of –A/B) so the equation is $y = \dfrac{B}{A} x$ or $Bx–Ay = 0$.

(c) Solve ${ Ax + By = C, Bx – Ay = 0 }$ to get $x = \dfrac{AC}{A^2 + B^2}$ and $y = \dfrac{BC}{A^2 + B^2}$.

(d) Distance = $\sqrt { (\dfrac{AC}{(A^2 + B^2)})^2 + (\dfrac{BC}{A^2 + B^2})^2}$ = $\sqrt{\dfrac{A^2C^2}{(A^2 + B^2)^2} + \dfrac{B^2C^2}{(A^2 + B^2)^2}}$

$= \sqrt{\dfrac{(A^2 + B^2)C^2}{(A^2 + B^2)^2}} = \sqrt{\dfrac{C^2}{A^2 + B^2}} = \dfrac{| C |}{\sqrt{A^2 + B^2}}$