Differntiation Bloody Confusing

[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 doesn't 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 traveling east towards an intersection at a speed of 80km/hr while a blue car simultaneously traveling 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

 

[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.

 

[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

 

[whozum]

Type in (4-t)^(5t) into http://www.calc101.com/webMathematica/derivatives.jsp and change the variable from 'x' to 't'

 

[HallsofIvy]

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

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.

 

[OlderDan]

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.

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

 

[HallsofIvy]

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