Question

Need help understanding this kind of problem. It is number 1. I will attach the problem

Answer 1

do u mind putting the answers on here, just want to make sure im on track

Answer 2

Ok here are the answers

Answer 3

cool im on track. :) Basically what is asking you to do is find the slope (gradient) from point P (15,250) to all the other points. we can use the formula y2-y1/x2-x1 to find the slope

Answer 4

I will do the first one. PQ when t=5
we are finding the gradient or slope of the secant line from point (15, 250) to (5, 695)
y2-y1/x2-x1
695-250/5-15
=-44.5

Answer 5

next lets to PQ where t=10 which are points (15, 250) to (10, 444)

i got -38.8 as my gradient...

then you would do PQ where t=20 which are points (15, 250) to (20, 111)


are u getting what im doing?

Answer 6

Yeah I get that so fare all u do is find the t that is given and put it in the y2-y1/x - x1 to get ur PQ.

Answer 7

Let m do those real quick and then will need help with b and c if you don't mind.

Answer 8

your finding the slope of the secant line PQ where P is(15,250) and Q are points on the graph where t=5, 10 .. etc .. be careful how you word it. cuz you are finding the gradient of lines

Answer 9

awesome, ill wait for your response

Answer 10

Ok for t=10 i got -38.5 then i got for t=20 it was -27.8, t=25 got -22.2 and t=30 i got -16.7

Answer 11

t=10 (444-250)/(10-15)=-38.8

yes other than t=10 i got the same ans as the answered u posted.

Answer 12

yea sorry i got that as well idk where i got the .5

Answer 13

all good. what bout B and C

Answer 14

how do i know which points to pick to find the average? do i just take the highest and lowest then multiply it by1/2?

Answer 15

we want the slope of the tangent line at point P (15,250) and the question tells us average the secant lines. so what this means I believe is average the slope of the secant lines we worked out PQ where t=10 and t=20

Answer 16

did you pick t=10 and t=20 because it is close to the point P given to us which is 15?

Answer 17

yes those two secant lines can provide us with a average slope line that can act as a tangent line at point P

Answer 18

oh ok so when ever they want the average of the tangent line at P, what ever point that was given from P we fine the 2 points closes to average the slopes of the two secant lines.

Answer 19

ok i got -33.3 for the average of the two slopes of the two secant lines

Answer 20

yes cuz we had coordinates in our question.

Answer 21

so at this point is is just asking us to work with slopes (gradient)

Answer 22

Question C is another way of also finding out the same ans (or near same ans) as Question B. I used calculus after fitting a quadratic equation to the points (using technology) and found the gradient at pointP

Answer 23

im sorry you lost me there so you found a quadratic equation and then plugged it into a calculator i assume and plugged in points to get the slope at point P

Answer 24

yes i plugged in the points on my calculator.

then i can choose what function i want it to fit.. like a quadratic model, a cubic, or quadratic model.

Answer 25

so example, i fitted a quadratic and got the following function (but a cubic is 'better')
y=1.11x^2-66.604285x+999.2

Answer 26

great i never really learned how to use the graphing calculator haha this is gonna be fun figuring it out haha

Answer 27

but come to think of it. my y intercept is at 999.2 for the QUad model i fitted, but the tank held 1000 ...
so again i mean, these models are not 100% accurate. however i got a r^2 value of 0.99 which is pretty good i guess

Answer 28

then i derived y=1.11x^2-66.604285x+999.2 and found the gradient at 15..in which i got -33.3 which is the same ans i got as question B