Thread: Need a little help from gifted math people...

Hi all,

I've tried solving the following example, I think I'm on the right track but I need reassurance and help simplifying the result, because my ever so strict professor picks random people to show their solution at the blackboard, and I absolutely dread this.

Here it is:

You have the function f(x,y,z)= x^2+2yz+y^2
Imagine that you travel in the xyz-room among the following lines

x(t)= 0.5 - (sqrt(3)/2)t
y(t)=sqrt(3)/2+0.5t
z(t)= 1

During your travels you continously get the values of the function f.

a) State the function that shows how the functionvalues of f depend on t

I think that means I should put the equations of x(t), y(t) and z(t) into f(x,y,z)= x^2+2yz+y^2

Am I correct? This gives me:

f(x,y,z)=(0.5 - (sqrt(3)/2)t)^2 + 2(sqrt(3)/2 + 0.5t) + (sqrt(3)/2 + 0.5t)^2=

t^2+ t+ 1+ sqrt(3)

Is this right..?

In the next question he asks how fast the function value increases/decreases when we pass the point that is prevailing when t=0

So, when t = 0: x= .05, y=sqrt(3)/2, z=1
They're looking for the "direction-derivative", how do I find the answer to how fast the function value increases/decreases when we pass the point that is prevailing when t=0..?



Please help me out, I dont want to make a fool out of myself at the blackboard!!

Thanks so much

Originally Posted by tinyone

Hi all,

I've tried solving the following example, I think I'm on the right track but I need reassurance and help simplifying the result, because my ever so strict professor picks random people to show their solution at the blackboard, and I absolutely dread this.

Here it is:

You have the function f(x,y,z)= x^2+2yz+y^2
Imagine that you travel in the xyz-room among the following lines

x(t)= 0.5 - (sqrt(3)/2)t
y(t)=sqrt(3)/2+0.5t
z(t)= 1

During your travels you continously get the values of the function f.

a) State the function that shows how the functionvalues of f depend on t

I think that means I should put the equations of x(t), y(t) and z(t) into f(x,y,z)= x^2+2yz+y^2

Am I correct? This gives me:

f(x,y,z)=(0.5 - (sqrt(3)/2)t)^2 + 2(sqrt(3)/2 + 0.5t) + (sqrt(3)/2 + 0.5t)^2=

t^2+ t+ 1+ sqrt(3)

Is this right..?

Yes, that is what it means and you are completely correct!

In the next question he asks how fast the function value increases/decreases when we pass the point that is prevailing when t=0

So, when t = 0: x= .05, y=sqrt(3)/2, z=1
They're looking for the "direction-derivative", how do I find the answer to how fast the function value increases/decreases when we pass the point that is prevailing when t=0..?

"How fast the function value is changing" or "rate of change" is the derivative. Find the derivative of when t= 0.

Please help me out, I dont want to make a fool out of myself at the blackboard!!

Thanks so much

The very first time I did a problem before the class in graduate school I completely misunderstood the question! So I have been there!

Thanks!!

Here is my solution now, the derivative is:
h '(t)= 2t+ 1
So h'(0)= 2*0+1= 1

I guess this was the easy part because the question builds on to state that "you have calculated the directional derivative of the function f in a certain point and direction", which point, which I think is: 0.5, sqrt(3)/2, 1 (because that is what x, y and z are when t=0)

Now the prof asks me to calculate the directional derivative with the help from the gradient to f..Does this mean I should find the derivative to (x,y,z)= x^2+2yz+y^2 with respect to x, y and z? This is it:

grad f = (2x, 2z+2y, 2y)...Putting in the points: 0.5, sqrt(3)/2, 1 gives me : (1, 2+sqrt(3), sqrt(3))

Now how do I find the direction-derivative? I know I'm asking a lot but I really need help since I failed the last seminar