problem 1) Suppose that a hawk, whose initial position is (a,0)=(7000,0) on the x-axis, spots a pigeon at (0,-5000) on the y-axis. Suppose that the pigeon flies at a constant speed of 30 ft/sec in the direction of the -axis (oblivious to the hawk), while the hawk flies at a constant speed of 80 ft/sec, always in the direction of the pigeon.
The problem is to find an equation for the flight path of the hawk (the curve of pursuit) and to find the time and place where the hawk will catch the pigeon. Assume that in this problem all distances are measured in feet and all times measured in seconds. Leave out all dimensions from your answers.
Consider the diagram above (click on it for a better view) which represents the situation at an arbitrary time t during the pursuit. The points P and Q represent the positions of the hawk and pigeon respectively at that time instant t, with y=f(x) representing the flight path of the hawk.
The pigeon's position Q=(0, g(t))is given by the following function of time
g(t)=-5000+30t
The fact that the hawk is always headed in the direction of the pigeon means that the line PQ is tangent to the pursuit curve y=(f(x)). This tells us that (dy/dx)=h(x,y,t) where
H(x,y,t)=(y-(-5000+30t))/x
(Your answer must involve the three variables x, y, and t. Also note that different variables must be separated by either a space or a multiplication sign in a product: xy^2 must be denoted as x*y\^{}2 or x y\^{}2.)
If we solve the equation
P=h(x,y,t)
where p=dy/dx, for time we obtain that t=k(x,y,p) where
K(x,y,p) =(y-px+5000)/30
(Your answer must involve the three variables x, y, and p, where p stands for dy/dx. Also note that different variables must be separated by either a space or a multiplication sign in a product: xy^2 must be denoted as x*y\^{}2 or x y\^{}2.)
2) Again referring to the diagram above (click on it for a better view) we see that the distance that the hawk has flown in time t is given by the integral integral Fdx (top d, bottom c) where
C= x
d= 7000
(Hints: Note that this integral computes the length of a curve. Also recall that the hawk's initial position is at . )
and
F = (1+p^2)^(1/2)
(Use P to denote dy/dx in your last answer above.)
On the other hand the hawk is flying at a constant speed of 80 for time . Hence the total distance it has flown is 80t. If we equate this to the distance we just computed and solve for t we obtain
t= integral Gdx (top d, bottom c)
where G = ((1+p^2)^(1/2))/80
(Remember to use p to represent dy/dx.)
3) K(x,y,p)= integral G dx (range upper d, bottom c)
To get rid of the integral, we differentiate both sides of the equation with respect to . On the left hand side of the resulting equation we obtain the following expression (which might involve x, y, p and p=dy/dx q=dp/dx =dy^2/dx^2
-xq/30
(remember to separate different variables in a product with spaces or multiplication signs)
while on the right hand side (after applying the Fundamental Theorem of Calculus) we obtain
-(1+p^2)^(1/2)/80
The resulting differential equation we obtained above is a separable differential equation in the variables and . We can rewrite it in the form
K(p)dp=L(x)dx
(with all numerical factors moved to the right hand side of the equation so that L(1)=30/80.) where
K(p)=
L(x)=
4) Integrating the left hand side of the equation integral K(p)dp, using the methods of sections 8.3 and 8.2, we obtain
Integral K(p)dp=
while on the right hand side we obtain integral L(x)dx= +C
Plugging in the initial positions of the hawk and pigeon, and recalling that p=dy/dx is the slope of the tangent line, we find that
C=
Note that it is important to compute to at least 5 or 6 decimal places of precision. While computing C to less precision may not affect the correctness of your answer at this point, the roundoff error tends to get magnified in your subsequent calculations in problems 5 and 6 and may lead to your answers there being unexpectedly rejected as incorrect.
5) Solving the equation we obtained in Problem 4 for P in terms of x, we obtain
P=
(Hints for solving for P: Exponentiate to get rid of the logarithm. Then isolate the square root on one side of the equation and square both sides.)
Recalling that P=dy/dx and integrating. we obtain that
Y= +c
Plugging in the initial position of the hawk we obtain that the constant of integration is given by
C =
Note: if your answer is unexpectedly rejected as incorrect, go back to problem 4 part c, recompute your constant of integration there to more decimal places and recalculate your answers here with this more accurate value.
6) Hence the hawk catches the pigeon at the point (0,c) where
C=
at time t =
I solve problem 1,2 and 3 (half) but i do not know how to solve 3(half) 4 5 6 help me
The problem is to find an equation for the flight path of the hawk (the curve of pursuit) and to find the time and place where the hawk will catch the pigeon. Assume that in this problem all distances are measured in feet and all times measured in seconds. Leave out all dimensions from your answers.
Consider the diagram above (click on it for a better view) which represents the situation at an arbitrary time t during the pursuit. The points P and Q represent the positions of the hawk and pigeon respectively at that time instant t, with y=f(x) representing the flight path of the hawk.
The pigeon's position Q=(0, g(t))is given by the following function of time
g(t)=-5000+30t
The fact that the hawk is always headed in the direction of the pigeon means that the line PQ is tangent to the pursuit curve y=(f(x)). This tells us that (dy/dx)=h(x,y,t) where
H(x,y,t)=(y-(-5000+30t))/x
(Your answer must involve the three variables x, y, and t. Also note that different variables must be separated by either a space or a multiplication sign in a product: xy^2 must be denoted as x*y\^{}2 or x y\^{}2.)
If we solve the equation
P=h(x,y,t)
where p=dy/dx, for time we obtain that t=k(x,y,p) where
K(x,y,p) =(y-px+5000)/30
(Your answer must involve the three variables x, y, and p, where p stands for dy/dx. Also note that different variables must be separated by either a space or a multiplication sign in a product: xy^2 must be denoted as x*y\^{}2 or x y\^{}2.)
2) Again referring to the diagram above (click on it for a better view) we see that the distance that the hawk has flown in time t is given by the integral integral Fdx (top d, bottom c) where
C= x
d= 7000
(Hints: Note that this integral computes the length of a curve. Also recall that the hawk's initial position is at . )
and
F = (1+p^2)^(1/2)
(Use P to denote dy/dx in your last answer above.)
On the other hand the hawk is flying at a constant speed of 80 for time . Hence the total distance it has flown is 80t. If we equate this to the distance we just computed and solve for t we obtain
t= integral Gdx (top d, bottom c)
where G = ((1+p^2)^(1/2))/80
(Remember to use p to represent dy/dx.)
3) K(x,y,p)= integral G dx (range upper d, bottom c)
To get rid of the integral, we differentiate both sides of the equation with respect to . On the left hand side of the resulting equation we obtain the following expression (which might involve x, y, p and p=dy/dx q=dp/dx =dy^2/dx^2
-xq/30
(remember to separate different variables in a product with spaces or multiplication signs)
while on the right hand side (after applying the Fundamental Theorem of Calculus) we obtain
-(1+p^2)^(1/2)/80
The resulting differential equation we obtained above is a separable differential equation in the variables and . We can rewrite it in the form
K(p)dp=L(x)dx
(with all numerical factors moved to the right hand side of the equation so that L(1)=30/80.) where
K(p)=
L(x)=
4) Integrating the left hand side of the equation integral K(p)dp, using the methods of sections 8.3 and 8.2, we obtain
Integral K(p)dp=
while on the right hand side we obtain integral L(x)dx= +C
Plugging in the initial positions of the hawk and pigeon, and recalling that p=dy/dx is the slope of the tangent line, we find that
C=
Note that it is important to compute to at least 5 or 6 decimal places of precision. While computing C to less precision may not affect the correctness of your answer at this point, the roundoff error tends to get magnified in your subsequent calculations in problems 5 and 6 and may lead to your answers there being unexpectedly rejected as incorrect.
5) Solving the equation we obtained in Problem 4 for P in terms of x, we obtain
P=
(Hints for solving for P: Exponentiate to get rid of the logarithm. Then isolate the square root on one side of the equation and square both sides.)
Recalling that P=dy/dx and integrating. we obtain that
Y= +c
Plugging in the initial position of the hawk we obtain that the constant of integration is given by
C =
Note: if your answer is unexpectedly rejected as incorrect, go back to problem 4 part c, recompute your constant of integration there to more decimal places and recalculate your answers here with this more accurate value.
6) Hence the hawk catches the pigeon at the point (0,c) where
C=
at time t =
I solve problem 1,2 and 3 (half) but i do not know how to solve 3(half) 4 5 6 help me