Related Rates word problem.

Question

A spotlight is placed on the ground, and shines on a wall 12 metres away. Spot is a 75 cm tall dog who walks at a speed of 1.6 m/s. Spot walks from the spotlight directly towards the wall, projecting a shadow onto the wall. When Spot is 9 metres from the wall: (a) how high is his shadow on the wall?
y = height of Spot's shadow on the wall
y/12 = 0.75/1.6
1.6y = 9
Spot's shadow on the wall = 5.625m .

(b) how fast is the length of his shadow decreasing?
x = distance Spot is from the wall
y = height of Spot's shadow on the wall
12-x/0.75 = 12/y
12y - xy = 9
Took the derivative of both sides and rearranged to make dy/dx the subject

dy/dx (12 - x) = d/dx y
dy/dx = (y * d/dx) / (12 - x)
Subbed in d/dx = -1.6, y = 3 --> (y = 9 / (12 - x)), x = 9
dy/dx = -1.6m/s

Can someone please verify whether the followed the correct procedure and arrived at the correct answer? Thank you

Answer 1

• For part $(a)$, you shouldn't involve the speed that Spot is walking.

The right formula appears in part $(b)$ where

$$\frac{12-x}{0.75}=\frac{12}y$$

You just have to use the right value for $x$ and solve for $y$.

• For part $(b)$, your goal is to compute $\frac{dy}{dt}$ and not $\frac{dy}{dx}$.

From $$y=\frac{12(0.75)}{12-x},$$

You might like to simplify it and then use chain rule to compute the derivative with respect to $t$. Note that you have access to $\frac{dx}{dt}$.