[Guest]

It's the third down, ten yards to go for a first. The quarterback calls a play that involves a roll out to the right and a square out to his tight end. On the snap from center, the tight end runs straight ahed for ten yards then makes a sharp right towards the sidelines. The quarterback rolls to the right and stops directly 6 yards behind where the tight end began. The quarterback doesn't make the pass until the tight end makes the break toward the sidelines. The tight end is running towards the sideline at a speed of 8 yards per second. The quarterback tracks the receiver deciding wen to throw the pass and the flight path of the ball. If the tight end makes the catch 12 yards after the break, how far does the quarterback throw the pass, in a straight line (he's not that good at curves), and at what rate is the distance between the receiver and quarterback changing?

Here is the best diagram that I could make:

[Faizaan]

you lost me at "ten yards to go for a first".

even though everyone around me is addicted to football, i have 0 IQ of it! but this might be a little different.

Do you have to know a lot about foot ball to answer this?

[Guest]

you lost me at "ten yards to go for a first".

even though everyone around me is addicted to football, i have 0 IQ of it! but this might be a little different.

Do you have to know a lot about foot ball to answer this?

no, that's why I drew the diagram.

other than understanding what the heck is going on, it's all math.

[Faizaan]

no, that's why I drew the diagram.

other than understanding what the heck is going on, it's all math.

math? ok, maybe later. my head hurt!

[Prof. Templeton]

It's a 20 yard pass. 16² + 12² = (Pass length)².

It's not constant, so I have to get back to you.

[Guest]

It's not constant, so I have to get back to you.

It's a 20 yard pass. 16² + 12² = (Pass length)².

nice use of the theorem

[HoustonHokie]

A generalized form of the Professor's equation is H = sqrt(X² + Y²), where H is the hypotenuse and X & Y are the horizontal and vertical distances betweeen the two players.

Y is fixed at 16; hence Y² = 256. X varies with time, but could be written as X = 12 + 8T, where 12 is the starting distance between the two and T is the time in seconds. Positive values of T are after the catch, negative values are before the catch.

Substituting X and Y into the equation for H gives H = sqrt[(12 + 8T)² + 256] = sqrt(400 + 192T + 64T²).

Taking the derivative of H w.r.t. T gives the rate of change of H dH/dT = [16T + 24]/sqrt(4T² + 12T + 25).

To find the specific rate of change when the ball is caught, plug in T=0 and you get dH/dT = 24/sqrt(25) = 24/5 = 4.8 yards/sec.

An alternative version of this answer could be obtained by using vectors. Maybe I'll show that later.

[Guest]

using a different way to get same answer.

first one.....

it's a 3,4,5 triangle, ergo 20.

second one....

x^2 + y^2= z^2

derivative

1/2x dx/dt + 1/2y dy/dt = 1/2z dz/dt

y is constant so dy/dt = 0

divide by 2

end up with x dx/dt = z dz/dt

x=12

dx/dt = 8 ft/s

z = 20

dz/dt = 4.8 ft/sec

[Guest]

14 yards, first down and ten to go now.

rate is faster then anyone you know can catch the ball, even as it slows down in that 14 yards.

By the way i don't remeber the equation

but if i have an unequal triagle

one side is 12 yards, another is 16 yards, the third side = x. What does x equation equal? There is a standard for this we learned in highschool either in geometry or trig, whatever you took. What is it?

[Guest]

using a different way to get same answer.

first one.....

it's a 3,4,5 triangle, ergo 20.

second one....

x^2 + y^2= z^2

derivative

1/2x dx/dt + 1/2y dy/dt = 1/2z dz/dt

y is constant so dy/dt = 0

divide by 2

end up with x dx/dt = z dz/dt

x=12

dx/dt = 8 ft/s

z = 20

dz/dt = 4.8 ft/sec

owe you made my head hurt, i haven't used derivitives since college and even then it was against my will! NIce though

[HoustonHokie]

owe you made my head hurt, i haven't used derivitives since college and even then it was against my will! NIce though

as promised above, the answer using vectors:

To get the rate of change in distance, remember that a speed is really a vector. In this case, the vector has a magnitude of 8 and is headed in the positive x direction. What we need is the component of that vector headed in the direction from the quarterback to the tight end at the time of the catch.

To do that, you can employ similar triangles. The 12-16-20 relationship of the distance triangle corresponds to a 4.8-6.4-8.0 relationship of the speed vectors. Simple, really. And no calculus