Thank you for this video presentation, I can no longer be having doubts and hesitations about our recent topic during online classes. This video presentation is a good guide on students who do self-study more at this period of time.
.' if you used trigonometric function to solve problem 2 that 3/20 is the rate of how fast the angle theta is decreasing as the top of the ladder slides down the wall. That should be -3/20 rad/sec since its decreasing. Here it goes. Sin(theta) = y/25 Differentiating Cos(theta) d(theta)/dt = (1/25) dy/dt So we need to look for the value of d(theta)/dt since it is not given on the problem From the relationship cos(theta) = x/25 Theta = ArcCos(x/25) Differentiating d(theta)/dt =( -1/25)/[1-(x^2)/25^2]^(1/2) dx/dt When x= 15 and dx/dt = 3 ft/sec and cos(theta) = 15/25 We have d(theta)/dt = -3/20 Finally Cos(theta) d(theta)/dt = (1/25) dy/dt (15/25) (-3/20) = (1/25) dy/dt dy/dt = (15/25)(-3/20)(25) dy/dt = -9/4 ft/sec
Thanks for the video.Can you name a good book for learning this type of real life application based problems along with maxima , minima application problems ???
Thank you for this video presentation, I can no longer be having doubts and hesitations about our recent topic during online classes. This video presentation is a good guide on students who do self-study more at this period of time.
Absolutely true!
Problem 2. My answer is 3/20 ft/ sec. I used the cos(tetha) =x/25 in solving.
.' if you used trigonometric function to solve problem 2 that 3/20 is the rate of how fast the angle theta is decreasing as the top of the ladder slides down the wall. That should be -3/20 rad/sec since its decreasing.
Here it goes.
Sin(theta) = y/25
Differentiating
Cos(theta) d(theta)/dt = (1/25) dy/dt
So we need to look for the value of d(theta)/dt since it is not given on the problem
From the relationship
cos(theta) = x/25
Theta = ArcCos(x/25)
Differentiating
d(theta)/dt =( -1/25)/[1-(x^2)/25^2]^(1/2) dx/dt
When x= 15 and dx/dt = 3 ft/sec and cos(theta) = 15/25
We have d(theta)/dt = -3/20
Finally
Cos(theta) d(theta)/dt = (1/25) dy/dt
(15/25) (-3/20) = (1/25) dy/dt
dy/dt = (15/25)(-3/20)(25)
dy/dt = -9/4 ft/sec
Yasssss! Got it. So lutang lang dahil sa late night review. Thank you.
Thanks for the video.Can you name a good book for learning this type of real life application based problems along with maxima , minima application problems ???
.' you can consult the book TC7 (The Calculus 7) by Louie Leithold.
Thanks for the thumbs up.