# Related rates visualization (Increasing distances)

"A bridge goes straight across a river at a height of $60$ m. A car on the bridge travelling at $40$ m/sec passes directly over a boat travelling up the river at $15$ m/sec. Find the rate at which the distance between the car and the boat is increasing 3 seconds later."

I don't think I need help on the actual problem itself but I'm a very hard time visualizing this situation. Is is possible for someone to draw me a picture of the situation? I just need to be able to visualize what is going on here.

Thanks!

• I'm no good at drawing pictures, but maybe this will help. Visualize a 3-dimensional coordinate system. The $y$-axis is the river. The bridge is parallel to the $x$-axis, but up at $z=60$. At time zero, the boat is at the origin, while the car is on the bridge, directly above the boat, at $(0,0,60)$. The car is moving to the right along the bridge at 40 m/sec, while the boat is heading up the $y$-axis at 15 m/sec. – Gerry Myerson Nov 5 '17 at 23:15
• That's the thing. I'm very skeptical this would be a 3D problem since we're only working with 2 dimensions at most. This isn't multivariable calculus or anything so I am not sure if 3D is required here. – Future Math person Nov 5 '17 at 23:17
• The action is taking place in 3D, but you don't need multivariable calculus to solve it. Everything is a function of a single variable, namely, the time. You can work out the position of the car at time $t$, and the position of the boat at time $t$, and then you need a little bit of 3D geometry (actually, just two applications of 2D Pythagoras) to get a formula for the distance between them at time $t$. – Gerry Myerson Nov 5 '17 at 23:20
• The position of the bridge doesn't change, but you can't ignore the fact that the bridge is 60 meters up from the river. – Gerry Myerson Nov 5 '17 at 23:38
• Imagine the bridge being a million miles high. Then the distance between car and boat will barely change at all as they move – it will stay very close to a million miles. So the height comes into the formula for the rate of change of the distance. But, look – just write down the formula for the distance, and see for yourself! – Gerry Myerson Nov 5 '17 at 23:44

Here is a rendering of the scene:

The car travels at $$r_c(t) = (v_c t, 0, 60)$$ (grey line), the boat at $$r_b(t) = (0, v_b t, 0)$$ (light blue line). The distance (three difference vectors shown as black arrows) is $$d(t) = \lVert r_c(t) - r_b(t) \rVert$$ and your task is to find $\dot{d}(t)$ for $t=3\text{s}$.

Hint: $\lVert (x,y,z) \rVert = \sqrt{x^2 + y^2 + z^2}$

Update:

It seems the dot on the $d$ did not get enough attention.

Find the rate at which the distance between the car and the boat is increasing 3 seconds later

A rate is a measure of change. In these context it is often the time derivative.

For convenience I had used $t=0$ as the time when the car is above the boat. But for completeness let us shift that to a time $t_0$ to be a bit more general, though it should lead to the same result.

Then we have $$r_c(t) = (v_c(t-t_0), 0, 60) \\ r_b(t) = (0, v_b(t-t_0), 0)$$ Then the difference vector is $$\Delta r(t) = r_c(t) - r_b(t) = (v_c (t-t_0), -v_b(t-t_0), 60)$$ and the length of this vector is the distance between car and boat: $$d(t) = \lVert \Delta r(t) \rVert$$ Asked is for rate of change of the distance, this is the time derivative of the distance $$\dot{d}(t) = d'(t) = \frac{d}{dt} d(t)$$ depending on your preference of notation.

And it is wanted 3 seconds later than the meeting time, thus $$\dot{d}(t_0 + 3)$$ where I left out the physical units for the 3.

Putting everything together we get \begin{align} \dot{d}(t) &= \frac{d}{dt} \sqrt{(v_c (t-t_0))^2 + (-v_b(t-t_0))^2 + 60^2} \\ &= \frac{1}{2 \sqrt{(v_c (t-t_0))^2 + (-v_b(t-t_0))^2 + 60^2}} (2 v_c + 2 v_b) \\ &= \frac{v_c + v_b}{\sqrt{(v_c (t-t_0))^2 + (-v_b(t-t_0))^2 + 60^2}} \end{align} Inserting $t = t_0 + 3$ gives $$\dot{d}(t_0 + 3) = \frac{v_c + v_b}{\sqrt{9(v_c^2 + v_b^2) + 60^2}}$$

The values for $v_c$ and $v_b$ were given in the task, so there is only one calculation left to do to get the desired rate.

• Ohh boy. Well this would essentially collapse down to a 2D problem though no? The height of the bridge is fixed and the only variables I care about is the ship and the car right? – Future Math person Nov 5 '17 at 23:27
• @FutureMathperson Correct – gen-z ready to perish Nov 5 '17 at 23:33
• It's not a 2D problem – you can't ignore the height of the bridge when calculating the distance, and then you will find that it also comes into the rate of change of the distance. – Gerry Myerson Nov 5 '17 at 23:41
• But then is question is quite simple then. It would simply be $d(3) = \sqrt{120^2 + 45^2 + 60^2} = 15\sqrt{89}$ . I would then differentiate $d(t)$ and plug in my values for $x,y,z,x',y',z'$. Since $z' = 0$, that vanishes but everything else stays. Is that the correct intuition? – Future Math person Nov 5 '17 at 23:57
• You don't want $d(3)$, you want $d(t)$. – Gerry Myerson Nov 6 '17 at 11:54