im not too sure but i think you would start by

1. finding out what the radius is so r/h=tan30 so radium would be htan30 so you can replace the r in the formula with that

2. simplify the formula so it would be 1/9(pi)h^3 i think

3. find the full volume so find radius again using r/50 = tan 30 and also replace h with 50

not sure where to go from here but you might need to differentiate the first formula for dt/dh then multiply dv/dt by dt/dh to find dv/dh? not sure how to get time from here either sorry ive not revised rates of change much

1. finding out what the radius is so r/h=tan30 so radium would be htan30 so you can replace the r in the formula with that

2. simplify the formula so it would be 1/9(pi)h^3 i think

3. find the full volume so find radius again using r/50 = tan 30 and also replace h with 50

not sure where to go from here but you might need to differentiate the first formula for dt/dh then multiply dv/dt by dt/dh to find dv/dh? not sure how to get time from here either sorry ive not revised rates of change much

Original post by Matheen1

Can someone talk me through this question step by step please

Can someone talk me through this question step by step please

I got a weird answer and haven't done differential modelling in a while but here was my method

1 - Find the relationship between h and r (Triangles mean that r = htan(30), therefore r = (1/sqrt(3))h

2 - Use this relationship to find one between V and h by subbing r = (1/sqrt(3))h into V = (1/3)(pi)(r^2)(h) to get V = (1/9)(pi)(h^3)

3 - Now differentiate this with respect to h to get dV/dh = (1/3)(pi)(h^2)

4 - dV/dt = dV/dh x dh/dt, therefore we can now sub in dV/dh = (1/3)(pi)(h^2) to get dV/dt = (1/3)(pi)(h^2) x dh/dt

5 - Sub this into the equation the question gives us ( dV/dt = -2h ) to get (1/3)(pi)(h^2) x dh/dt = -2h

6 - divide both sides by h to get all of the h's on one side : (1/3)(pi)(h) x dh/dt = -2

7 - "Multiply" both sides by dt to get (1/3)(pi)(h) dh = -2 dt and add integration signs

8 - Integrate both sides with their respective variables to get (1/6)(pi)(h^2) = -2t + C

9 - Sub in the given initial value of h = 50 when t = 0 to get (1250/3)pi = C, now you have the complete equation (1/6)(pi)(h^2) = -2t + (1250/3)pi

10 - We want to find the time taken for the tank to become empty, so sub in h = 0 to get 0 = -2t + (1250/3)pi

11 - Add 2t to both sides to get 2t = (1250/3)pi

12 - Divide by 2 to solve for T and get T = (625/3)pi

I may have made a mistake somewhere and my explanation isn't the best but i'm pretty sure thats the answer

Original post by Throwaway1686

I got a weird answer and haven't done differential modelling in a while but here was my method

1 - Find the relationship between h and r (Triangles mean that r = htan(30), therefore r = (1/sqrt(3))h

2 - Use this relationship to find one between V and h by subbing r = (1/sqrt(3))h into V = (1/3)(pi)(r^2)(h) to get V = (1/9)(pi)(h^3)

3 - Now differentiate this with respect to h to get dV/dh = (1/3)(pi)(h^2)

4 - dV/dt = dV/dh x dh/dt, therefore we can now sub in dV/dh = (1/3)(pi)(h^2) to get dV/dt = (1/3)(pi)(h^2) x dh/dt

5 - Sub this into the equation the question gives us ( dV/dt = -2h ) to get (1/3)(pi)(h^2) x dh/dt = -2h

6 - divide both sides by h to get all of the h's on one side : (1/3)(pi)(h) x dh/dt = -2

7 - "Multiply" both sides by dt to get (1/3)(pi)(h) dh = -2 dt and add integration signs

8 - Integrate both sides with their respective variables to get (1/6)(pi)(h^2) = -2t + C

9 - Sub in the given initial value of h = 50 when t = 0 to get (1250/3)pi = C, now you have the complete equation (1/6)(pi)(h^2) = -2t + (1250/3)pi

10 - We want to find the time taken for the tank to become empty, so sub in h = 0 to get 0 = -2t + (1250/3)pi

11 - Add 2t to both sides to get 2t = (1250/3)pi

12 - Divide by 2 to solve for T and get T = (625/3)pi

I may have made a mistake somewhere and my explanation isn't the best but i'm pretty sure thats the answer

1 - Find the relationship between h and r (Triangles mean that r = htan(30), therefore r = (1/sqrt(3))h

2 - Use this relationship to find one between V and h by subbing r = (1/sqrt(3))h into V = (1/3)(pi)(r^2)(h) to get V = (1/9)(pi)(h^3)

3 - Now differentiate this with respect to h to get dV/dh = (1/3)(pi)(h^2)

4 - dV/dt = dV/dh x dh/dt, therefore we can now sub in dV/dh = (1/3)(pi)(h^2) to get dV/dt = (1/3)(pi)(h^2) x dh/dt

5 - Sub this into the equation the question gives us ( dV/dt = -2h ) to get (1/3)(pi)(h^2) x dh/dt = -2h

6 - divide both sides by h to get all of the h's on one side : (1/3)(pi)(h) x dh/dt = -2

7 - "Multiply" both sides by dt to get (1/3)(pi)(h) dh = -2 dt and add integration signs

8 - Integrate both sides with their respective variables to get (1/6)(pi)(h^2) = -2t + C

9 - Sub in the given initial value of h = 50 when t = 0 to get (1250/3)pi = C, now you have the complete equation (1/6)(pi)(h^2) = -2t + (1250/3)pi

10 - We want to find the time taken for the tank to become empty, so sub in h = 0 to get 0 = -2t + (1250/3)pi

11 - Add 2t to both sides to get 2t = (1250/3)pi

12 - Divide by 2 to solve for T and get T = (625/3)pi

I may have made a mistake somewhere and my explanation isn't the best but i'm pretty sure thats the answer

You are the best mate.

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