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Consider a point moving along the curve $$f(x) = \sqrt x$$.

a). Find the position of the point on the curve where both coordinates of the point are changing at the same rate.

b). If $\dfrac{dx}{dt}$ is $2 \text{ m/sec}$ at the point $(4,f(4))$, how fast is the point moving away from the origin?


My attempt:

Find point where dy/dt = dx/dt,

Given $y = \sqrt x$ ==> $dy/dt = 1/2\sqrt x$* dx/dt

and this is where I'm still lost

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  • $\begingroup$ Where does $\frac{1}{2\sqrt{x}} = 1$? $\endgroup$
    – KM101
    Commented Nov 2, 2019 at 10:09
  • $\begingroup$ I know how to find the point, my main concern is how do you know the derivative/tangent slope is 1? $\endgroup$
    – harold232
    Commented Nov 2, 2019 at 10:10
  • $\begingroup$ What does it mean when $f'(x) = 1$? $\endgroup$
    – KM101
    Commented Nov 2, 2019 at 10:11
  • $\begingroup$ From the answer below it means the derivative which both coordinates change at the same rate. I am just unsure how to get this value for the derivative $\endgroup$
    – harold232
    Commented Nov 2, 2019 at 10:13
  • $\begingroup$ What did you do to get f'(x) = 1? $\endgroup$
    – harold232
    Commented Nov 2, 2019 at 10:14

4 Answers 4

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On your third attempt you have a solution in front of you, you just need to recognize it.

You found that at all points along the curve, $$\frac{dy}{dt} = \frac{1}{2\sqrt x} \frac{dx}{dt}.$$

(But don't write ${1}/{2\sqrt x}$; strictly interpreted it means $(1/2)\times\sqrt x$, which is not what you want.)

You're looking for a point at which $$\frac{dy}{dt} = \frac{dx}{dt}.$$

But if $\frac{dy}{dt} = \frac{1}{2\sqrt x} \frac{dx}{dt}$ (because it always is) and $\frac{dy}{dt} = \frac{dx}{dt}$ (because it is at your point) then $\frac{1}{2\sqrt x} \frac{dx}{dt} = \frac{dx}{dt}$ (because all equal things are equal) and the only way that can happen (provided that you don't let $\frac{dx}{dt}=0$) is if $$\frac{1}{2\sqrt x} = 1.$$

So you're looking for a point at which $\frac{1}{2\sqrt x} = 1.$

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a) No. You're not looking for the point where the x and y have the same value, you're looking for the point where the values are changing at the same rate. Whenever you see "change" in calculus, that is a free clue that you should be thinking about the derivative. So you are looking for the point where $\frac{dy}{dx}=1$.

b) For this, we are being asked about how quickly a different quantity is changing, so we will need a new function to take the derivative of. The distance from the origin to $(x,(f(x))$ is given by $$g(x)=\sqrt{x^2+(f(x))^2}=\sqrt{x^2+(\sqrt x)^2}=\sqrt{x^2+x}$$ The problem is asking you to calculate $\frac {dg}{dt}$ at $x=4$ given that $\frac{dx}{dt}=2$. That looks like a lot to work through, but remember that the Chain Rule tells us that $\frac{dg}{dt}=\frac{dg}{dx}\cdot\frac{dx}{dt}$, so you've got just enough clues to work it out.

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  • $\begingroup$ why is the derivative 1? Why can't it be any value? How do you know they change at a rate of 1 when changing at the same rate? $\endgroup$
    – harold232
    Commented Nov 2, 2019 at 9:24
  • $\begingroup$ @harold232 By the Chain Rule, $\frac{\mathrm dy}{\mathrm dt} = \frac{\mathrm dy}{\mathrm dx} \cdot \frac{\mathrm dx}{\mathrm dt}$, so $\frac{\mathrm dy}{\mathrm dt} = \frac{\mathrm dx}{\mathrm dt}$ when $\frac{\mathrm dy}{\mathrm dx} = 1$. $\endgroup$
    – KM101
    Commented Nov 2, 2019 at 9:27
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PART a) Same rate means Same derivative: $$x'_t=y'_t=(\sqrt x)'_t=\frac{x'_t}{2\sqrt x} \Longrightarrow x'_t=\frac{x'_t}{2\sqrt x} \Longrightarrow \sqrt x = \frac{1}{2} \Longrightarrow x = \frac{1}{4} \Longrightarrow (\frac{1}{4},\frac{1}{2})$$

PART b) By euclidean metric we have: $$l=\sqrt{x^2+y^2}=\sqrt{x^2+x} \Longrightarrow l'_t=\frac{2x'_tx_t+x'_t}{2\sqrt {x^2+x}} \overset{x'_t=2}{\underset{x=4}{===}} \frac{9}{2\sqrt{5}}\frac{m}{s}$$

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For the first part, you want to compare $\frac{\mathrm dx}{\mathrm dt}$ and $\frac{\mathrm dy}{\mathrm dt}$, not $x$ and $y$. Remember, the problem is referred to as "related rates" for a reason. Using the Chain Rule, you get

$$\frac{\mathrm dy}{\mathrm dt} = \frac{\mathrm dy}{\mathrm dx} \cdot \frac{\mathrm dx}{\mathrm dt} \tag{1}$$

Hence, $\frac{\mathrm dx}{\mathrm dt} = \frac{\mathrm dy}{\mathrm dt}$ when $\frac{\mathrm dy}{\mathrm dx} = 1$. You have $y = \sqrt{x}$, so where does this occur?

For the second part, you have distance too. Whenever you have distance, you're dealing with $r^2 = (\Delta x)^2 + (\Delta y)^2$, by the Pythagorean Theorem. Since you start from the origin, this just becomes $r^2 = x^2+y^2$. Differentiating implicitly with respect to time gives

$$\frac{\mathrm d}{\mathrm dt}r^2 = \frac{\mathrm d}{\mathrm dt} \left(x^2+y^2\right)$$

$$2r\frac{\mathrm dr}{\mathrm dt} = 2x\frac{\mathrm dx}{\mathrm dt} + 2y\frac{\mathrm dy}{\mathrm dt}$$

$$r\frac{\mathrm dr}{\mathrm dt} = x\frac{\mathrm dx}{\mathrm dt} + y\frac{\mathrm dy}{\mathrm dt}$$

Well, you're given $x = 4$, so you can find $y$ by $y = \sqrt{x}$, meaning you can find $r$ as well. You're also given $\frac{\mathrm dx}{\mathrm dt}$. This just leaves $\frac{\mathrm dy}{\mathrm dt}$, which you can find by using $(1)$. Finally, plug in the values and find $\frac{\mathrm dr}{\mathrm dt}$.

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  • $\begingroup$ the point where they change at the same rate would be (1, 1/2) , but how are related rates even used in that question? $\endgroup$
    – harold232
    Commented Nov 2, 2019 at 9:27
  • $\begingroup$ Because the rate of change and $x$ and $y$ are related, by some factor $y'$. That's the core idea of related rates problems. $\endgroup$
    – KM101
    Commented Nov 2, 2019 at 9:28
  • $\begingroup$ I meant for implicit differentiation. When you do a related rates problem, you use implicit differentiation for the first step or two, but for this part it wasn't even used $\endgroup$
    – harold232
    Commented Nov 2, 2019 at 9:29
  • $\begingroup$ When differentiating implicitly, you're actually using the Chain Rule. So, it actually was used to relate the rate of change of $x$ and $y$. $\endgroup$
    – KM101
    Commented Nov 2, 2019 at 9:30
  • $\begingroup$ how did you relate x and y for a)? y=x then dy/dt = dx/dt? $\endgroup$
    – harold232
    Commented Nov 2, 2019 at 9:39

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