Question

Linear approximation word problem (Calculus):

Using Newton's equation:
s = (1/32)v^2 x sin(Θ)

A player located 18.1 ft from the basket launches a successful jump shot from a height of 10 ft (level with the rim of the basket), at an angle Θ = 34° and initial velocity v = 25ft/s.

a) Show that Δs ≈ 0.255ΔΘ ft for a small change of ΔΘ.

b) Is it likely that the shot would have been successful if the angle had been off by 2°?


Thanks! :)

Answer 1

a linear approximation uses a line or plane at a given point

Answer 2

yeah that might not be what it's called, sorry haha. but it's in the same section as that in the book. Still don't understand this question

Answer 3

using other formulas:\[h=-\frac12gt^2+25sin(34)t\]
\[d=25cos(34)t=18.1~:~t=\frac{18.1}{25cos(34)}\]
with any luck, this "t"ime gives us 0 for height
\[-\frac12(32)(\frac{18.1}{25cos(34)})^2+25sin(34)(\frac{18.1}{25cos(34)})\]
\[-16(\frac{18.1}{25cos(34)})^2+18.1tan(34)=0~??\]

http://www.wolframalpha.com/input/?i=-16%2818.1%2F%2825cos%2834%29%29%29%5E2%2B18.1tan%2834%29

its prolly close enough :) all we need to do to confirm if a 2 degree error fits is to try out 36 and 32 degrees

Answer 4

its prolly not what the question is asking as far as a method goes ...

Answer 5

at 32 degrees that works out to .35 ft under the basket

and at 36 degrees that works out to .34 ft over

so about 4 inches off

Answer 6

It's okay, that works for me! Thanks you are a life saver, got a test in a couple hours on these!

Answer 7

s = k sin(a)

\[\frac{\triangle s}{\triangle v}=k\frac{sin(a+h)-sin(a)}{h}\]
\[\frac{\triangle s}{\triangle v}=k\frac{~sin(a)cos(h)+sin(h)cos(a)- sin(a)}{h}\]
\[\frac{\triangle s}{\triangle v}=k\frac{sin(a)[cos(h)-1]+sin(h)cos(a)}{h}\]
\[\frac{\triangle s}{\triangle v}=ksin(a)\frac{cos(h)-1}{h}+kcos(a)\frac{sin(h)}{h}\]
\[\frac{\triangle s}{\triangle v}=\frac{25^2}{32}sin(34)\frac{cos(2)-1}{2}+\frac{25^2}{32}cos(34)\frac{sin(2)}{2}\]
\[\frac{\triangle s}{\triangle v}=0.279222\]
\[\triangle s=0.279222~\triangle v\]

im not really sure how they are getting the number they did, unless im missing something about Newtons Linear approximations

Answer 8

Yeah, that was confusing me too. we have a review before our test so I will just check with the professor. But anyways you helped a lot, thank you!!

Answer 9

your posts is off:

Answer 10

sin(2a) not sin(a)

Answer 11

Ohhh oops! haha my bad, that clears things up a bit

Answer 12

\[h=-\frac12gt^2+v~sin(a)~t\]
\[0=-\frac12gt^2+v~sin(a)~t\]
\[t=\frac{-v~sin(a)\pm\sqrt{v^2sin^2(a)}}{-g}\]
\[t=\frac{-v~sin(a)\pm v~sin(a)}{-g}\]
\[t=0,~or~\frac{2v~sin(a)}{g}\]
horizontal distance is:\[s = vt~cos(a)\]
\[s = v\frac{2v~sin(a)}{g}~cos(a)\]
\[s = \frac{v^2~}{g}~2sin(a)~cos(a)\]
\[s = \frac{v^2~}{g}~sin(2a)\]
to which they start with this equation
\[s+\Delta s \approx \frac{v^2~}{g}~2sin(a+\Delta a)cos(a+\Delta a)\]
\[\Delta s \approx \frac{v^2~}{g}~2sin(a+\Delta a)cos(a+\Delta a)-s\]
\[\Delta s \approx \frac{v^2~}{g}~2sin(a+\Delta a)cos(a+\Delta a)-\frac{v^2}{g}2sin(a)cos(a)\]
just need to determine a way to get the left side worked out