Homework Help: Differntiation Bloody Confusing!

1. May 12, 2005

dagg3r

Differntiation!!! Bloody Confusing!

HI guys help me out here with some differentitaion problems i'll post here what i have done and show me the way to go thanks
1. find dx/dt where x=(4-t)^5t

i tried using the chain rule but it doesnt work though cos you got that t constant do i use the logarithic way of differntiation so log X=5t*log(4-t) then i diff this?

2. A red car is travelling east towards an intersection at a speed of 80km/hr while a blue car simultaneously travelling north away from the intersection at a speed of 60 km/hr. If the red car is 4km from the intersection and the blue car is 3km from the intersection what is the rate of change the cars are changing?

i drew pictures of this then started to think to try and use pythagoras and possibly use some differentiation there applying the direction changes of negatives and positive but got lost and somebody point me the steps on to solve this problem thanks

2. May 12, 2005

whozum

1. Try implicit differentiation
2. Set up a system of equations. I assume you meant RoC of the distance between the cars. The distance formula will probably come in handy.

Eq of motion for the cars are
x1 = 80t + 4
x2 = 60t - 3

The distance btwn the two functions as a function of time is the distance between the coordinates of x1 and x2 at any given time.

3. May 13, 2005

dagg3r

hmm how can you do implicit differentiation when its to the power of 5t? i tried to the log way and got this

lnx=5tloge(4-t)
then used product rule with u=5t v=loge(4-t)

dy/dx=(4-t)^5t * [ 5t(-1/4-t) + 5ln(4-t) ]

realling long and ugly i think i did it wrong reckon somebody can show me how to apply
implicit differentiation usually i can do it but got confused with the power of 5t

4. May 13, 2005

whozum

5. May 13, 2005

HallsofIvy

Of course, you DON'T mean "dy/dx"

$$\frac{1}{ln x} \frac{dx}{dt}= 5 ln(4-t)- \frac{5t}{4-t}$$
so
$$\frac{dx}{dt}= (4-t)^{5t}(5 ln(4-t)- \frac{5t}{4-t})$$

looks like just what you have.

6. May 13, 2005

OlderDan

And of course you don't mean $$\frac{1}{ln x} \frac{dx}{dt}$$

You mean

$$\frac{d}{dt} ln x =\frac{1}{x} \frac{dx}{dt}= 5 ln(4-t)- \frac{5t}{4-t}$$
so
$$\frac{dx}{dt}= (4-t)^{5t}(5 ln(4-t)- \frac{5t}{4-t})$$

7. May 14, 2005

HallsofIvy

Oops: $$\frac{1}{x}\frac{dx}{dt}$$