Math  /  Calculus

QuestionYou say goodbye to your friend at the intersection of two perpendicular roads. At time t=0t=0 you drive off North at a (constant) speed vv and your friend drives West at a (constant) speed ww. You badly want to know. how fast is the distance between you and your friend increasing at time tt ?
Enter here the derivative with respect to tt of the distance between you and your friend: \square Note: the next part will be MUCH easier if you simplify your answer to this part as much as possible.
Being scientifically minded you ask yourself how does the speed of separation change with time. In other words, what is the second derivative of the distance between you and your friend? \square Suppose that after your friend takes off (at time t=0t=0 ) you linger for an hour to contemplate the spot on which he or she was standing. After that hour you drive off too (to the North). How fast is the distance between you and your friend increasing at time tt (greater than one hour)? \square Again, you ask what is the second derivative of your separation: \square Notice how lingering can make things harder, mathematically speaking.

Studdy Solution

STEP 1

What is this asking? Two friends drive away from each other at a right angle, at different speeds.
How fast is the distance between them changing, and how is *that* rate of change, changing itself?
And what if one friend waits an hour before driving? Watch out! Don't mix up speed and acceleration!
Also, remember the Pythagorean theorem will be our best friend here!

STEP 2

1. Set up the distance formula
2. Find the first derivative
3. Find the second derivative
4. Adjust for the waiting period
5. Find the first derivative with the wait
6. Find the second derivative with the wait

STEP 3

Let's call the distance between the two friends DD.
After time tt, you've traveled a distance of vtv \cdot t North, and your friend has traveled wtw \cdot t West.
Since the roads are perpendicular, we have a right triangle!

STEP 4

Using the **Pythagorean theorem**, we get: D2=(vt)2+(wt)2 D^2 = (v \cdot t)^2 + (w \cdot t)^2 D=(vt)2+(wt)2=v2t2+w2t2 D = \sqrt{(v \cdot t)^2 + (w \cdot t)^2} = \sqrt{v^2 t^2 + w^2 t^2} D=t2(v2+w2)=tv2+w2 D = \sqrt{t^2(v^2 + w^2)} = t\sqrt{v^2 + w^2}

STEP 5

We want to find how fast DD is changing with respect to time, so we need to find dDdt\frac{dD}{dt}.
Since vv and ww are constants, v2+w2\sqrt{v^2 + w^2} is also a constant.

STEP 6

So, dDdt=ddt(tv2+w2)=v2+w2 \frac{dD}{dt} = \frac{d}{dt} \left(t\sqrt{v^2 + w^2}\right) = \sqrt{v^2 + w^2} This tells us the **rate of change of the distance** between the friends.

STEP 7

Now, we want to find how the *speed* of separation is changing, which means finding the second derivative, d2Ddt2\frac{d^2D}{dt^2}.

STEP 8

d2Ddt2=ddt(v2+w2)=0 \frac{d^2D}{dt^2} = \frac{d}{dt} \left(\sqrt{v^2 + w^2}\right) = 0 Since v2+w2\sqrt{v^2 + w^2} is a constant, its derivative is **zero**!
This means the rate at which the friends are separating is constant.

STEP 9

Now, you wait for an hour before driving.
This means your distance traveled is v(t1)v(t-1) for t>1t > 1.
Your friend's distance is still wtw \cdot t.

STEP 10

Our new distance formula is: D=(v(t1))2+(wt)2=v2(t1)2+w2t2 D = \sqrt{(v(t-1))^2 + (w \cdot t)^2} = \sqrt{v^2(t-1)^2 + w^2t^2}

STEP 11

Let's find dDdt\frac{dD}{dt} for this new scenario.
This will tell us how fast the distance between you and your friend is changing when you wait an hour.

STEP 12

dDdt=12v2(t1)2+w2t2(2v2(t1)+2w2t)=v2(t1)+w2tv2(t1)2+w2t2 \frac{dD}{dt} = \frac{1}{2\sqrt{v^2(t-1)^2 + w^2t^2}} \cdot (2v^2(t-1) + 2w^2t) = \frac{v^2(t-1) + w^2t}{\sqrt{v^2(t-1)^2 + w^2t^2}}

STEP 13

And finally, the second derivative!
This one is a bit more complicated, but we can do it!
This will tell us how the *speed* of separation is changing when you wait for an hour.

STEP 14

Using the quotient rule, we get: d2Ddt2=(v2+w2)(v2(t1)2+w2t2)(v2(t1)+w2t)v2(t1)+w2tv2(t1)2+w2t2v2(t1)2+w2t2 \frac{d^2D}{dt^2} = \frac{(v^2 + w^2)(\sqrt{v^2(t-1)^2 + w^2t^2}) - (v^2(t-1) + w^2t) \frac{v^2(t-1) + w^2t}{\sqrt{v^2(t-1)^2 + w^2t^2}}}{v^2(t-1)^2 + w^2t^2} =(v2+w2)(v2(t1)2+w2t2)(v2(t1)+w2t)2(v2(t1)2+w2t2)3/2 = \frac{(v^2 + w^2)(v^2(t-1)^2 + w^2t^2) - (v^2(t-1) + w^2t)^2}{(v^2(t-1)^2 + w^2t^2)^{3/2}} =v2w2(t(t1))2(v2(t1)2+w2t2)3/2=v2w2(v2(t1)2+w2t2)3/2 = \frac{v^2w^2(t-(t-1))^2}{(v^2(t-1)^2 + w^2t^2)^{3/2}} = \frac{v^2w^2}{(v^2(t-1)^2 + w^2t^2)^{3/2}}

STEP 15

The rate of change of the distance between the friends is v2+w2\sqrt{v^2 + w^2}.
The second derivative is 00.
With the one-hour wait, the rate of change of distance is v2(t1)+w2tv2(t1)2+w2t2\frac{v^2(t-1) + w^2t}{\sqrt{v^2(t-1)^2 + w^2t^2}}, and the second derivative is v2w2(v2(t1)2+w2t2)3/2\frac{v^2w^2}{(v^2(t-1)^2 + w^2t^2)^{3/2}}.

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