Hello Ms. Havrot! I love your videos, I have completed gr12 and am going to take math in university and I was to do some practice over the summer to get me prepared! at 13:54, how did you know we have to derive that formula? because i would have thought that we would use the y= 20/x and substitute it into the Pythagorean theorem or something.
We are trying to find the rate of change of the shadow so we need to have a function that describes the shadow at any given time. Then we can find the rate of change in the length of the shadow at a specific time. Hope that helps!
For #8 you want to know what value of k makes the line and the quadratic intersect at one point. That means that you can set the line equal to the quadratic, then simplify so that the equation is set to 0 You should end up with 2x^2 - 8x +3 -k =0 Now for there to only be ONE point of intersection that means that the discriminant must be equal to zero. D = b^2 - 4ac The rest is just plugging in a = 2, b = -8 and c = 3 - k and you should get k = -5 #12 is similar in that you will start by setting the line = the quadratic h(t) = g(t) in this case. Get everything to one side of the equation and then use the quadratic formula to solve for t You will get two solutions but in the context of the problem only one is admissible (the larger number is inadmissible because it would mean that g(t) would be negative). Then you substitute the smaller value (t = 0.18) into g(t) and you will find the height that the blocker's hands must be.
![How to Solve ANY Related Rates Problem [Calc 1]](http://i.ytimg.com/vi/gBHIZlF0TX8/mqdefault.jpg)
Hello Ms. Havrot! I love your videos, I have completed gr12 and am going to take math in university and I was to do some practice over the summer to get me prepared! at 13:54, how did you know we have to derive that formula? because i would have thought that we would use the y= 20/x and substitute it into the Pythagorean theorem or something.
We are trying to find the rate of change of the shadow so we need to have a function that describes the shadow at any given time. Then we can find the rate of change in the length of the shadow at a specific time. Hope that helps!
Year 1 civil engineering here. Thank you that was really useful
Thank you! 😊
Why do you multiply by dz/dt, dx/dt etc when taking the derivative at 19:00? Thanks
The reason is that you are taking the derivative with respect to time of each of the variables.
Hi mrs, I’m having trouble with page 199 question 8,12 . Grade 11 academic quadratic functions
For #8 you want to know what value of k makes the line and the quadratic intersect at one point. That means that you can set the line equal to the quadratic, then simplify so that the equation is set to 0
You should end up with 2x^2 - 8x +3 -k =0 Now for there to only be ONE point of intersection that means that the discriminant must be equal to zero. D = b^2 - 4ac The rest is just plugging in a = 2, b = -8 and c = 3 - k
and you should get k = -5
#12 is similar in that you will start by setting the line = the quadratic h(t) = g(t) in this case. Get everything to one side of the equation and then use the quadratic formula to solve for t You will get two solutions but in the context of the problem only one is admissible (the larger number is inadmissible because it would mean that g(t) would be negative). Then you substitute the smaller value (t = 0.18) into g(t) and you will find the height that the blocker's hands must be.
Ms Havrot's Canadian University Math Prerequisites thanks a lot mrs appreciate it
Ms Havrot's Canadian University Math Prerequisites thank you so much mrs, I appreciate it
Hello mam please Leibniz rule video
I haven't worked my way up to this level of calculus yet, so here's a very good explanation for you to check th-cam.com/video/zbWihK9ibhc/w-d-xo.html