Question

A​ hot-air balloon is 160 ftabove the ground when a motorcycle​ (traveling in a straight line on a horizontal​ road) passes directly beneath it going 45 mi divided by hr​(66 ft divided by s​).If the balloon rises vertically at a rate of 14 ft divided by s​,what is the rate of change of the distance between the motorcycle and the balloon 7 seconds​later?

Answer 1

73.77 ft/s

Explanation:

Let's imagine this situation as a right triangle where the distance between the two points is the hypotenuse.

-Applying Pythagoras theorem:

`z^2=x^2+y^2\\\\z=√(x^2+y^2)\\\\x=dx*t+x_i, x_i=0\\\\y=dy*t+y_i\\\\y_i=160\ ft`

#Take the derivative of the first equation and solve for
`dz`:

`d(z^2=x^2+y^2)\\\\2z*dz=2x*dx+2y*dy\\\\dz=(x*dx+y*dy)/(z)\\\\dz=(x*dx+y*dy)/(√(x^2+y^2))\\\\\\dz=(dx^2*t+dy(dy*t+160))/(√((dx*t)^2+(dy*t)))`

#We then substitute the values given in the question to solve for
`dz`:

`dz=(7*66^2+14(14*7+160))/(√((66*7)^2+7*14+160))\\\\=73.78`

Hence, the rate of change of the distance between the motorcycle and the balloon 7 seconds ​later is 73.77 ft/s