# Difference between revisions of "009A Sample Final A, Problem 9"

9. A bug is crawling along the ${\displaystyle x}$-axis at a constant speed of   ${\displaystyle {\frac {dx}{dt}}=30}$. How fast is the distance between the bug and the point ${\displaystyle (3,4)}$ changing when the bug is at the origin? (Note that if the distance is decreasing, then you should have a negative answer).

Foundations:
Like most geometric word problems, you should start with a picture. This will help you declare variables and write meaningful equation(s). In this case, we will have to use implicit differentiation to arrive at our related rate. In particular, we need to choose variables to describe the distance between the bug and the point ${\displaystyle (3,4)}$, which we can call ${\displaystyle z}$. By the given information, we can consider the position on the ${\displaystyle x}$-axis simply as ${\displaystyle x}$.

Solution:

Step 1:
Write the Basic Equation: From the picture, we can see there is a right triangle involving both the bug and the point ${\displaystyle (3,4)}$. From this, we can see that ${\displaystyle z^{2}=x^{2}+4^{2}}$.
Step 2:
Use Implicit Differentiation: We take the derivative of the equation from Step 1 to find
${\displaystyle 2z{\frac {dz}{dt}}=2x{\frac {dx}{dt}},}$
or
${\displaystyle {\frac {dz}{dt}}={\frac {x}{z}}\cdot {\frac {dx}{dt}},}$
Step 3:
Evaluate and Solve: When the bug is at the origin, we have ${\displaystyle x=3}$. By the Pythagorean Theorem, ${\displaystyle z=5}$. Based on our drawing, our variable ${\displaystyle x}$ is actually decreasing at a rate of ${\displaystyle 30}$, so we should really write ${\displaystyle dx/dt=-30}$. We now simply plug in to the result of our implicit differentiation to find
${\displaystyle {\frac {dz}{dt}}={\frac {3}{5}}\cdot (-30)=-18.}$