The equation of curve is y=(5x-4)^1/2 I) calculate the gradient of the curve at the point where x=1 ii) a point with coordinates (x,y) moves along the curve in such a way the rate of increase of x has the constant value 0.03 units per second. Find the rate of increase of y at the instant when x=1

Question

kainui · Answer

So where are you stuck? Show me your best guess and I'll help you get through this interesting problem.

abmon98 · Answer

I will differentiate the given and substitute by x

undeadknight26 · Answer

so this is your equation?
y=(5x-4)^1/2

abmon98 · Answer

Yes

abmon98 · Answer

0.5(9)^-0.5 *5

kainui · Answer

Perfect, so you've got part 1 down.

abmon98 · Answer

Yep

abmon98 · Answer

Could anyone please help me in that problem

kainui · Answer

So it appears to be that the change in x with respect to time is .03 units per second. Can you represent this as a derivative?

abmon98 · Answer

No could you?

kainui · Answer

$$\frac{ dx }{ dt }=.03$$

You already know how much y changes with respect to x, but now you know that x changes with respect to t. So when you take the derivative you have to realize this and do the chain rule and take the derivative of your original problem with respect to t, knowing that y is a function of x and x is a function of t.

abmon98 · Answer

When I looked back to my textbook it appears to be 0.025

campbell_st · Answer

well the question uses related rates... the change in y with respect to time is 

$$\frac{dy}{dt} $$

and this can be found using 

$$\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$$

so using your value for dy/.dx when x = 1 you get 

$$\frac{dy}{dt} = 0.5 \times \frac{1}{\sqrt{9}} \times 5 \times 0.03$$

and you'll find you get your answer.