Week 4 tutorial 8

[aravind.s Pai]

8. An advertiser is analysing the growth of likes for their new ad on YouTube. She analyzed

that the increase in likes in a given second is equal to 4tav where tav is midpoint of the

time interval. For example, the increase in likes from 3 seconds to 4 seconds is equal to

4 × 3.5. Answer the following questions.

(a) If the total likes follow the path as l(t) = at2 + bt + c then what is the value of b?

(b) Find the total likes at 60 seconds.

(c) If the domain of the function l is [k, inf), what will be the value of k?

How does the l(t ) function change in only square term and constant in video lectures ??

[Mathematics 1 Support 2]

Hi,

As you can see the increase in likes is 4t only, which is nothing but the slope of a quadratic function. It is also clear that the slope of the quadratic function is 2ax+b. If t is treated as x then we will get 4x. From here it is very clear that there is no constant term in slope that is b=0. And it is common sense that the likes when someone just uploads the picture will be zero which means the curve represented by the quadratic function passes through (0,0), which only be possible if and only if c =0.

Best wishes,

Rahul

[shyam sundar]

Hello Rahul,

Would you please consider that increase in likes is not 4t it is 4tav which is 4(2t+1)/2 which gives 4t + 2. In this case b term is very much possible.

Please correct me if my understanding is incorrect.

Thanks!

[subhajit POD]

Hi Sundar, 

It is given that,  the total likes follow the path as l(t) = at2 + bt + c. 

So I(t+1) -I(t)= 2at +a +b

Now it is given that the increase in likes in a given second is equal to 4tav where tav is midpoint of the time interval.

Hence the increase from t to t+1 will be= 4t+2 

Now If you compare these two you will get, b=0 

Observe that what we are calculating here is not the slope. We are just comparing the differences in like at time = t and time = t+1, which is only the difference of the functional value. 

Hope this clarifies your doubt. 

Best

Subhajit

[shyam sundar]

Thank you very much Subhajit sir!

Yes I understood that. However, my understanding was slope is equal to change in y for unit change in x which is in this case change in likes to unit change in time from 3 to 4 seconds or 0 to 1 second. Hence I was expecting slope calculation also to yield the result. Could you please point out to me why slope is not fetching the correct result in this case? This is for my understanding only. Thank you so much!

Regards,

Shyam.

[shyam sundar]

Hello Subhajit sir/All,

After much deliberation I figured the error. The understanding of slope of correct. However the difference in likes between 3 and 4 seconds is the slope at 3.5. Similarly difference in likes between 0 and 1 second is equal to slope at 0.5. This is because slope of parabola at a point is equal to slope at tangent at that point. This tangent is parallel to line connecting 2 points and touches the slope at midpoint.

So, slope is 2at + b.

At 3.5 slope is 7a +b = 14 and at 0.5 slope is a +b = 2. Solving gives a =2 and b=0.

Thank you all and happy Deepavali.