VCE Specialist 3/4 Question Thread!

[Phy124]

A flexible beam is supported at its ends, which are the same horizontal level and at a distance L apart. The deflection y of the beam, measured downwards from the horizontal, satisfies the differential equation: 16 d^2y/dx^2 = L-3x, 0≤x≤L
The deflection of the beam and its inclination to the horizontal are both zero at x=0
Use calculus to find an expression for y in terms if x.

Is deflection d^2y/dx^2 and inclination dy/dx?

Thanks

Yep, is the deflection and is the gradient of the beam.

Thanks, a little confused here... how do I solve for a specific solution with the DE and find k?

You have to integrate and then use the conditions given (when , , when , ) to find and or whatever you choose to label your constants after integration.

When I get home, I'll type up the solution if you still need help

Also, not sure what you mean by "k"?

[#1procrastinator]

How do you prove the a the tangent to a point on any circle is always perpendicular to the radius? I can do it fine if it's not a translated circle, multiplying the derivative and the slope of the radius just gives me -1 but when I tried differentiating a translated circle, I get . I'm not sure how to deal with that....probably made an error when I differentiated but if I did, I can't spot it

[kamil9876]

Well for me at least, I think it's enough to prove it for a circle with centre at the origin, and then just notice that translations preserve angles.

But oh well if you want to do it directly: You've already done the hard part (I assume you used implicit differentiation or something of the sort). Then just notice that the radial segment between and has gradient .

I think another neat way of doing it is to parametrize the circle: and notice that then just take the dot product and notice it is zero. Calculating then also shows you that the acceleration is towards the centre - something that is often mentioned but not proved in VCE physics

[#1procrastinator]

When using definite integrals to find areas below the x-axis, could you change the terminals around to make it positive rather than use the modulus?

And is there a geometric interpretation of that (do you count the sums backwards or something?)

EDIT: Oh and thanks kamil, forgot I asked that! :p

[Phy124]

When using definite integrals to find areas below the x-axis, could you change the terminals around to make it positive rather than use the modulus?

And is there a geometric interpretation of that (do you count the sums backwards or something?)

Yes, if you are asked to represent the area between the curve and the x-axis by an integral and the graph is below the axis, you either alter the terminals or place a negative in front.

e.g. Use a definite integral to represent the area between the curve and the x-axis.



or



I'm not sure I can answer your second question with extreme certainty, so I'll leave that for someone else.

[Bhootnike]

so
and

how would i go about finding cartesian equation ?

[b^3]

Using the double angle formuals for cos

[Bhootnike]

Using the double angle formuals for cos

legend, thanks mate!

[cassettekid]

Could someone please help me with this differential equation question?
(sorry I don't know how to make all the math symbols come up!)


Many chemical reactions follow the law which states that the velocity of the reaction is proportional to the different between the initial concentration of the reagent and the amount transformed at any time.

dx/dt= k(a-x)   where, x is an element [0,a]

Where a is the original concentration and x is the amount transformed at time t. If a = 10 and x = 4 after 2 minutes, find the amount left after 5 minutes.

[soccerboi]

I'm stuck with this:

From a balloon ascending with a velocity of 10m/s a stone was dropped and reached the ground in 12 seconds. Given that the gravitational acceleration is 9.8m/s². Find the height of the balloon when the stone was dropped.

Thanks

[b^3]

Firstly I will take up as positive.
(positive as it is moving upwards initially)

(negative as acceleration due to gravity is downwards)



i.e. it dropped (downwards) by
So the height of the balloon when the stone was dropped was

[soccerboi]

Firstly I will take up as positive.
(positive as it is moving upwards initially)

I dont understand why the initial velocity is 10? It ascends at a velocity of 10m/s but then when the stone drops, wouldn't the velocity be instantaneously at rest for it to switch directions and fall downwards?

[b^3]

Firstly I will take up as positive.
(positive as it is moving upwards initially)

I dont understand why the initial velocity is 10? It ascends at a velocity of 10m/s but then when the stone drops, wouldn't the velocity be instantaneously at rest for it to switch directions and fall downwards?

When it is initially released, the balloon and the stone are moving upwards with a velocity of 10 m/s, the stone starts to accelerate towards the ground, but for a short period of time, it is still moving upwards, as it will take time for the stones velocity to change from 10 m/s upwards to 0 m/s. Here it is instantaneously at rest, but it's position will be above that of where it started. Then its velocity will increase in the negative direction (i.e. downwards) until it hits the ground.

We know that the total time for it to reach the ground is 12 seconds, that is when it travels upwards initially, then travels downwards and hits the ground.

It will be instantaneously at rest at some point, but as this point will be above the starting position (the position we want) we can't use it, and the time taken from that point at which it is instantaneously at rest to hit the ground, won't be the 12 seconds.

Draw a diagram of the situation, it should help.

Hopefully the above helps too

[paulsterio]

Just think of a ball being thrown into the air, by your logic, soccerboi, it would stop the moment it falls out of your hand, but that is not true, it continues going upwards, decreasing in speed until it reaches zero, then it continues back down. Velocity and acceleration can be in two different directions, in this case, up and down, this will cause the velocity to decrease.

[soccerboi]

Thanks a lot guys, i finally got my head around it