# Rates of Change Question

#### ISITIEIW

##### New member
A Tightrope is 40ft above ground between two buildings that are 60 feet apart. A tightrope walker starts along the rope and walks from building A to building B at a rate of 2 ft per second. 80 feet above the starting point of the tightrope walker on building A is a spotlight that is illuminating the tightrope walker as the tightrope walker is crossing between two buildings.
a)How fast is the shadow of the tightrope walker's feet moving along the ground when the tightrope walker is midway between the buildings?
b)How fast is the shadow of the tightrope walker's feet moving up the wall of building B when the tightrope walker is twelve feet away from building B?

OK ! I posted a picture of what i've done so far…
I got the answer for a) but I'm having troubles with b) as you can see in the picture.. Thanks #### MarkFL

Staff member
The image you posted is sideways and is hard to read even when I enlarge and rotate it, but sometimes images posted from a mobile device display correctly on the device but not for those of us still using PCs. Interestingly, this problem was posted by another user earlier today, and your IP addresses are very close. Anyway...I agree with what you have:

$$\displaystyle \frac{80}{x}=\frac{b}{60-x}$$

You then cross-multiplied

$$\displaystyle bx=80(60-x)=480-80x$$

I would next implicitly differentiate with respect to time $t$, bearing in mind that $b$ is a function of $t$ and so you need to use the product rule on the left side. What do you find?

#### ISITIEIW

##### New member
Thanks,

I got b=4800/x - 80

then db/dt = -4800x^-2 (dx/dt)

and we know the values of x which equals 60-12=48 and (dx/dt)= 2

so db/dt = -25/6 ft/s

Thanks !

#### MarkFL

Staff member
Thanks,

I got b=4800/x - 80

then db/dt = -4800x^-2 (dx/dt)

and we know the values of x which equals 60-12=48 and (dx/dt)= 2

so db/dt = -25/6 ft/s

Thanks !
If you choose (which is actually simpler) to write $b$ as a function of $x$, then you have:

$$\displaystyle b=480x^{-1}-80$$

and so:

$$\displaystyle \frac{db}{dt}=-480x^{-2}\frac{dx}{dt}$$

Your result is too large by a factor of 10, since you used 4800 instead of 480.

#### ISITIEIW

##### New member
Um, not too sure that 480 is correct because bx= 80(60-x)
and 80 times 60 is 4800 not 480
Thanks again! #### MarkFL

Thanks again!  You are absolutely right! Sorry for the confusion. I guess I am having one of those days... I will go to the corner for 5 minutes. 