Asked by Matt
I had this question answered before,but I'm totally confused with the answer and need a clearer answer,as I'm totally lost. This is composite functions by the way.
Answers
Answered by
jolly rancher
which question? - i'll give it a shot
Answered by
Matt
Suppose a ship is sailing at a rate of 35km/h parellel to a straight shoreline. The ship is 10km from shore when it passes a lighthouse at 11am.
Question 6 options:
a)
Let k be the distance between the lighthouse and the ship. Let d be the distance from the ship has travelled since 11am. Express k as a function of d. Please include a diagram.
b)
Express d as a function of t, the time elapsed since 11am.
c)
Find k∘d . What does this function represent?
Question 6 options:
a)
Let k be the distance between the lighthouse and the ship. Let d be the distance from the ship has travelled since 11am. Express k as a function of d. Please include a diagram.
b)
Express d as a function of t, the time elapsed since 11am.
c)
Find k∘d . What does this function represent?
Answered by
jolly rancher
"ship passes lighthouse" tells me that the ship (at that instant) is on a straight line that is perpendicular to the shoreline and 10km from the lighthouse, moving parallel with the shoreline.
a) initially (at time t=0, the instant of the ship's passing at 11am) k = 10km and d = 0.
The ship continues to move parallel with the shoreline at 35km/h.
You can see that (for example) after 1 hour the ship will have traveled 35km from its earlier point (now d = 35km). The distance from the lighthouse, k, will now be the hypotenuse of the right triangle that has one of its two shorter sides equal to d=35km and the other equal to 10km. (sorry I can't draw a diagram with this tool)
After 1 hour: k = sqrt[(35km)^2 + (10km)^2] = 36.4km at 12noon (here d = 35km)
After 2 hours: k = sqrt[(70km)^2 + (10km)^2] = 70.7km at 1pm (here d = 70km)
In general: k (as a function of d) = sqrt[d^2 + (10km)^2], where d is in km
b) d as a function of t:
d = (35km/h) * t, where t is in hours
c) recall that (k o d) just means k(d(t))
you’re just putting one function into another function as the independent variable
you end up with k as a function of t
so just embed the function d(t) above into the function k(d) above
it’s as though t becomes the independent variable inside the function k so you get
k(d(t)) = k(t) = sqrt[(35km/h * t)^2 + (10km)^2], where t is in hours
the function is now k as a function of t instead of k as a function of d
-questions let me know
a) initially (at time t=0, the instant of the ship's passing at 11am) k = 10km and d = 0.
The ship continues to move parallel with the shoreline at 35km/h.
You can see that (for example) after 1 hour the ship will have traveled 35km from its earlier point (now d = 35km). The distance from the lighthouse, k, will now be the hypotenuse of the right triangle that has one of its two shorter sides equal to d=35km and the other equal to 10km. (sorry I can't draw a diagram with this tool)
After 1 hour: k = sqrt[(35km)^2 + (10km)^2] = 36.4km at 12noon (here d = 35km)
After 2 hours: k = sqrt[(70km)^2 + (10km)^2] = 70.7km at 1pm (here d = 70km)
In general: k (as a function of d) = sqrt[d^2 + (10km)^2], where d is in km
b) d as a function of t:
d = (35km/h) * t, where t is in hours
c) recall that (k o d) just means k(d(t))
you’re just putting one function into another function as the independent variable
you end up with k as a function of t
so just embed the function d(t) above into the function k(d) above
it’s as though t becomes the independent variable inside the function k so you get
k(d(t)) = k(t) = sqrt[(35km/h * t)^2 + (10km)^2], where t is in hours
the function is now k as a function of t instead of k as a function of d
-questions let me know
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