1) Two cars are traveling along perpendicular roads, car A at 40 mph, car B at 60 mph. At noon, when car A reaches the intersection, car B is 90 mi away, and moving toward it. At 1 pm, the rate, in miles per hour, at which the distance between the cars is changing is:

A. -40 | B. 68 | C. 4 | D. -4 | E. 40

2) If t is the number of hours of travel after noon, then the cars are closest together when t is:

A. 0 | B. $\frac{27}{26}$ | C. $\frac{9}{5}$ | D. $\frac{3}{2}$ | E. $\frac{14}{13}$



To begin, note down the "givens": $\frac{da}{dt} = 40$, $\frac{db}{dt} = 60$. At 12pm, cars A and B are at (0,0) and (0, -90) respectively. Because A is traveling along the x-axis and B along the y-axis for the purposes of this solution, position $a$ represent's car A's x value, and $b$ is car B's y value.

Next, evaluate the cars' positions at 1pm, keeping in mind they are traveling perpendicular to each other. Cars A and B will be at (40, 0) and (0, -30) respectively, because of the velocities provided. Simply put, car A has moved "right" along the x-axis by 40, and car B has moved "up" along the y-axis by 60, because exactly one hour passed since 12pm, hence 1 * the velocity.

The first question asks us to find the rate at which the distance is changing, or $\frac{dd}{dt}$, where $d$ is the distance between cars A and B. To do this, we use the distance formula, $d = \sqrt{(a_1 - b_1)^2 + (a_2 - b_2)^2}$, where $a_1$ and $b_1$ represent car A and B's x coords, and $a_2$ and $b_2$ represent car A and B's y coords. But because car A always travels along the x-axis, and car B along the y-axis, $a_2$ (car A's y value) is always zero and $b_1$ (car B's x value) is always zero. $$d = \sqrt{(a_1 - 0)^2 + (0 - b_2)^2}$$ $$d = \sqrt{(a_1)^2 + (-b_2)^2}$$ $$d^2 = (a_1)^2 + (-b_2)^2$$ $$d^2 = (a_1)^2 + (b_2)^2$$ And finally, $d^2 = a^2 + b^2$, because using the axes simplifies the problem, as explained in the first paragraph.

Implicitly derive $$2d\frac{dd}{dt} = 2a\frac{da}{dt} + 2b\frac{db}{dt}$$ $$\frac{dd}{dt} = \frac{a\frac{da}{dt} + b\frac{db}{dt}}{d}$$

Plug in $40$ for $a$, $-30$ for $b$, $40$ for $\frac{da}{dt}$, $60$ for $\frac{db}{dt}$, and $50$ for $d$ (using Pythagorean Theorem, $30^2 + 40^2 = 50^2$). $$\frac{dd}{dt} = \frac{40(40) + -30(60)}{50}$$ $$\frac{dd}{dt} = \frac{1600 -1800}{50}$$ $$\frac{dd}{dt} = \frac{-200}{50}$$ $$\frac{dd}{dt} = -4$$

Answer 1 is D

Question two concerns optimization; finding the minimum value of the distance. We know the initial positions of the cars at 12pm, and their velocities. The initial point plus the distance traveled per hour for $t$ hours tells us the positions of the cars at $t$. We can find that car A at time $t$ is at $0 + 40t$ or $40t$ and car B is at $-90 + 60t$.

Using the rate of change of the distance computed above, $$\frac{dd}{dt} = \frac{a\frac{da}{dt} + b\frac{db}{dt}}{d}$$

Plug in the values and set the derivative to zero to optimize.

$$\frac{dd}{dt} = \frac{40t(40) + (-90 + 60t)60}{d} = 0$$ $$40t(40) + (-90 + 60t)60 = 0$$ $$5200t - 5400 = 0$$ $$5200t = 5400$$ $$t=27/26$$

And perform a simple check to verify the minimum. Plugging in $t = 1$ into the numerator portion, for simplicity, yields $-200$; $t = 2$ yields $5000$. The derivative is 0 and is increasing at $27/26$, so the distance is indeed minimized at that time.

Answer for 2 is B

Feel free to offer things to clarify, add, or remove, as this is my first lengthy and sophisticated solution.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.