Rate of Change of Area of Triangle with Changing Base and Height

[fghtffyrdmns]

Homework Statement

Question: the base b and height h change with a time t in such a way that b= $$(t+1)^{2}$$ and h= $$t^{2}+1$$
Determine the rate of change of the area of the triangle when t=3

Homework Equations

$$A=\frac {1}{2}(b)(h)$$

The Attempt at a Solution

I am not sure what to do here.

Do you just multiply it out using the area of a triangle? Or should I take the derivative of both then use the area formula?

 

[lanedance]

find the area in terms of t, then differentiate w.r.t t

 

[LCKurtz]

You want to calculate dA/dt. Can you do that somehow?

 

[fghtffyrdmns]

I was thinking of doing something like this:
$$A=\frac {1}{2}(b)(h)$$
$$A=\frac {1}{2}((t+1)^{2})((t^{2}+1))$$

and continue from here.

thus then I can take dA/dt?

 

[LCKurtz]

Are you asking for permission? If nobody's looking, go ahead and do it.

 

[fghtffyrdmns]

I was just wondering if it is correct :D

 

[lanedance]

sounds good to me

 

[fghtffyrdmns]

I've been wondering, is it possible to use the product rule on 3 functions? Because I could split the (t+1)^2 into (t+1)(t+1) then just 0.5(t+1)(t+1)(t^2+1)

I'm not sure if this is allowed though since I've only see the product rule for two functions.

 

[LCKurtz]

I've been wondering, is it possible to use the product rule on 3 functions? Because I could split the (t+1)^2 into (t+1)(t+1) then just 0.5(t+1)(t+1)(t^2+1)

I'm not sure if this is allowed though since I've only see the product rule for two functions.

Yes there is a three factor product rule:

(fgh)' = f'gh + fg'h + fgh'

But in your example the easiest way is to leave it as (1/2)(t+1)²(t²+1) and just use the ordinary product rule, which involves using the chain rule on one of the factors.

 

[lanedance]

or just multiply the expression out before differentiating

 

[fghtffyrdmns]

$$\frac {d}{dx} A =\frac {1}{2}$$$$(t+1)$$$$(t+1)$$$$(t^{2}+1)$$

$$\frac {d}{dx} A = f'(t)g(t)h(t) + f(t)g'(t)h(t) + f(t)g(t)h'(t)$$

$$\frac {d}{dx} A = (t+1)$$$$(t^{2}+1)$$ +$$(t+1)$$$$(t^{2}+1)$$ +$$(t+1)$$$$(t+1)$$$$(2t)$$

the d/dx for t+1 is just 1.

$$\frac {d}{dx} A = t^{3}+t+t^{2}+1+t^{3}+t+t^{2}+1+2t^{2}+4t+2t$$

$$\frac {d}{dx} A = 2t^{3}+4t^{2}+8t+2$$

Now, to multiply the $$\frac {1}{2}$$ in.

$$\frac {d}{dx} A = t^{3}+2t^{2}+4t+1$$

Is this correct? Something doesn't seem right.

 

[fghtffyrdmns]

If I multiply it out, I get this as the derivative:

$$2t^{3}+3t-2t$$

which seems to be a more reasonable answer.

 

[LCKurtz]

You both have it wrong. If you multiply it out you should get:

2t³ + 3t² + 2t + 1

 

[fghtffyrdmns]

You both have it wrong. If you multiply it out you should get:

2t³ + 3t² + 2t + 1

But why is that answer different if I used the product rule?

 

[slider142]

$$\frac {d}{dx} A = (t+1)$$$$(t^{2}+1)$$ +$$(t+1)$$$$(t^{2}+1)$$ +$$(t+1)$$$$(t+1)$$$$(2t)$$

the d/dx for t+1 is just 1.

$$\frac {d}{dx} A = t^{3}+t+t^{2}+1+t^{3}+t+t^{2}+1+2t^{2}+4t+2t$$

Two problems.
1) You are not taking the derivative with respect to x; there is no x in the problem which means your derivative would be 0. You want $$\frac{d}{dt}$$.
2) The last product in the sum was not distributed properly. You should have gotten 2t³ + 4t² + 2t, not 2t² + 4t + 2t.