Asked by Anonymous
this is a relate rates problem in calculus
in a right triangle, the hypotenuse is 5 and the vertical side is x and the angle is theta.
so if theta increases at a constant rate of 3 radians per minute, at what rate is x increasing in units per minute when x equals 3 units?
d(theta)/dt=3 rad/min
so if you draw the triangle, you get:
sin(theta)=x/5::::theta=u
take derivative:
cos(u(du/dt)=(1/5)dx/dt
i solve for dx/dt and i'm assuming the theta is 45 degrees because it's right triange. but the problem asks for when x=3 but the x is gone so what do i do?
no!!!
you had x = 5sin(theta), right?
now you have to differentiate that with respect to t, because it is a rate
dx/dt = 5cos(theta) d(theta)/dt
only now do you consider the case when x=3
when x=3, r=5, then the third side =4
and cos(theta)= 4/5
also you were given that d(theta)/dt = 3 rad/min
so.....
dx/dt = 5(4/5)(3) units/min
=15 units/min
5 x (4/5) x 3 = 12
in a right triangle, the hypotenuse is 5 and the vertical side is x and the angle is theta.
so if theta increases at a constant rate of 3 radians per minute, at what rate is x increasing in units per minute when x equals 3 units?
d(theta)/dt=3 rad/min
so if you draw the triangle, you get:
sin(theta)=x/5::::theta=u
take derivative:
cos(u(du/dt)=(1/5)dx/dt
i solve for dx/dt and i'm assuming the theta is 45 degrees because it's right triange. but the problem asks for when x=3 but the x is gone so what do i do?
no!!!
you had x = 5sin(theta), right?
now you have to differentiate that with respect to t, because it is a rate
dx/dt = 5cos(theta) d(theta)/dt
only now do you consider the case when x=3
when x=3, r=5, then the third side =4
and cos(theta)= 4/5
also you were given that d(theta)/dt = 3 rad/min
so.....
dx/dt = 5(4/5)(3) units/min
=15 units/min
5 x (4/5) x 3 = 12
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Meredith
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