How do you find the value of m in the equation y=k+asin(mx)?
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If BE−→−BE→ bisects ∠ABD and m∠ABD = 66°, find m∠ABE.
y=k+asin(mx)
asin(mx) = y-k
sin(mx) = (y-k)/a
mx = arcsin( (y-k)/a )
m = (1/x)( arcsin( (y-k)/a ) ) or (1/x)( sin^-1 ( (y-k)/a ) )
asin(mx) = y-k
sin(mx) = (y-k)/a
mx = arcsin( (y-k)/a )
m = (1/x)( arcsin( (y-k)/a ) ) or (1/x)( sin^-1 ( (y-k)/a ) )
what if you had a maximum point let's say (30,6) and a y-int of (2,0) and you had to find the equation of the graph? I get how to find the other variables just not m.
Ahhh, so that was the original question, ok ....
go back to your original:
y = k+asin(mx)
y' = acos(mx)*m = 0 for a max
amcos(mx) = 0
am = 0 or cos(mx) = 0 , forget about the am=0
cos(mx) = 0 , but x = 30 from the max of (30,6)
cos 30m = 0
30m = π/2
m = π/60
sub (30,6) into original
6 = k + asin((π/60)(30) )
6 = k + a sin π/2
6 = k + a(1) ----> a + k = 6 **
also (2,0) lies on it.
0 = k + a sin( 2π/60)
0 = k + asin π/30 --- k + .104528..a = 0 ***
subtract *** from **
.89547a = 6
a = appr 6.7
k = 6 - a = appr .7
so y = .7 + 6.7 sin((π/60)x) , check my work
go back to your original:
y = k+asin(mx)
y' = acos(mx)*m = 0 for a max
amcos(mx) = 0
am = 0 or cos(mx) = 0 , forget about the am=0
cos(mx) = 0 , but x = 30 from the max of (30,6)
cos 30m = 0
30m = π/2
m = π/60
sub (30,6) into original
6 = k + asin((π/60)(30) )
6 = k + a sin π/2
6 = k + a(1) ----> a + k = 6 **
also (2,0) lies on it.
0 = k + a sin( 2π/60)
0 = k + asin π/30 --- k + .104528..a = 0 ***
subtract *** from **
.89547a = 6
a = appr 6.7
k = 6 - a = appr .7
so y = .7 + 6.7 sin((π/60)x) , check my work
sorry my response is so late...
I tried solving it again and I got the equation y=2+4sin(3x). I got the two because the graph is shifted 2 units up and I got the four from the graph being stretched vertically 4 times. Then I plugged in the point (30,6) into the equation y=2+4sin(mx) and I got that m was 3(?). Is this right? I'm not sure...
I tried solving it again and I got the equation y=2+4sin(3x). I got the two because the graph is shifted 2 units up and I got the four from the graph being stretched vertically 4 times. Then I plugged in the point (30,6) into the equation y=2+4sin(mx) and I got that m was 3(?). Is this right? I'm not sure...
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