1995 cssa trial Q 4a (1 Viewer)

[lyounamu]

http://www.boredofstudies.org/courses/maths/2u/1995_Maths2_T_CSSA_q.pdf

For this question, is it important to be exactly right? This is really an easy question but I am not sure if mine would get me full mark.

I would still get full mark if I can show all the relevant points and the shape being roughly right, right?

As in the relevant points, I did horizontal point of inflexion, and maximum local point where the gradient graph touches y=0, x=0 and y=0 and x=directly below the local maximum turning point.

I also placed the maximum turning point of the gradient graph where I presumed to be the point of inflexion.

And I am sure the shape is roughly right.

By the way, for the gradient graph, all graph is above the x-axis except the part where the graph starts to move down from the local maximum turning point.

 

[henry08]

It would only have to have the correct shape and be roughly in the right position. Certainly wouldn't need to be 'perfect'.

 

[tommykins]

sketches only look for shape and a decent scale of critical points.

 

[pLuvia]

You just need critical points i.e. x/y intercepts, any relevant points that the old graph touches you mark on the new graph and sketch it so it shows that you know what kind of feature it is. Like if at one of the intercepts the point is a maximum turning point, draw that bit so it shows the marker that you know it is a max point

 

[lyounamu]

You just need critical points i.e. x/y intercepts, any relevant points that the old graph touches you mark on the new graph and sketch it so it shows that you know what kind of feature it is. Like if at one of the intercepts the point is a maximum turning point, draw that bit so it shows the marker that you know it is a max point

Thanks! That helps a lot~

Thanks to other members who helped me here.

By the way, what about 4(c). It's another easy question but I don't really know how to explain. Should I just say that the total amount of rain is representated by the area under the curve? (and this area under the curve is represented by the integral?)

 

[pLuvia]

Since the 31.8t is the rate at which the water is being released hence it is something like dV/dt, and so if you integrate this function with respect to t, then you are essentially deriving the function for the amount of water that was released at time a specific time hence why you integrate it. But seeing as though you want the amount of water released at time k and so the limits are k and 0

 

[lyounamu]

Since the 31.8t is the rate at which the water is being released hence it is something like dV/dt, and so if you integrate this function with respect to t, then you are essentially deriving the function for the amount of water that was released at time a specific time

Thanks for your help.