Calculus related rate question!

Question

I'm really stump at this question, asking for instantaneous rate of change!

A visitor of the Jurassic Park attraction lost his way and is now walking to the East from point $A$. At the same time, a Tyrranosaurus is moving from Point $B$ towards point $A$. Point $B$ is directly to the South from point $A$.

At some moment, the distance from Tyrranosaurus to point $A$ is $8$ km and Tyrranosaurus is moving at the speed of $34$ km /h. At the same time, the distance from the man to point $A$ is $5$ km and his speed is $6$ km / h.

Find the instantaneous rate of change of the distance from Tyrranosaurus to the man. Its sign will show whether it is increasing or decreasing. Enter your answer with four decimal places.

I tried placing them in grid, having the position of the man be on the positive $x$-axis, and the dinosaur to be on the negative $y$-axis. Finding the distance from the dinosaur to the man, I did this : $(-8y)^2+(5x)^2=D$, where $D$ is the distance apart. Is it correct to assign $$\frac{\mathrm{d}y}{\mathrm{d}t} = 34$$ and

$$\frac{\mathrm{d}x}{\mathrm{d}t} = 6$$

?

I tried doing implicit differentiation on the eqn but stuck at finding $D$. Please help!!!

Answer 1

Hint: By Pythagoras, we have $x^2 + y^2 = D^2$. Implicitly differentiating, we obtain: $$ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 2D \frac{dD}{dt} $$ So to solve for $\frac{dD}{dt}$ at the moment when $x = 5$ and $y = 8$, it remains to figure out the value of $D$ at this moment by substituting the above values into the original formula from Pythagoras.

Answer 2

Let $D$ be the distance from dinosaur to man, $x(t)$ the distance from $A$ to man, and $y(t)$ the distance from $A$ to dinosaur. So, with Pythagoras Theroem, $$x^2(t)+y^2(t)=D^2(t)$$

Then, you know that, at the instant $t_0$, where $x(t_0)=5$ and $y(t_0)=8$, then $$D(t_0)=\sqrt{5^2+8^2}=\sqrt{25+64}=\sqrt{89}$$

Also, with implicit differenciation, $$2x\frac{dx}{dt}+2y\frac{dy}{dt}=2D\frac{dD}{dt}$$

So $$\begin{array}{rcl} \frac{dD}{dt}&=&\frac{1}{2D}\bigg(2x\frac{dx}{dt}+2y\frac{dy}{dt}\bigg)\\ &=&\frac{1}{2\cdot\sqrt{89}}\big(2\cdot5\cdot6+2\cdot8\cdot34\big)\\ &=&\frac{302}{\sqrt{89}}=32.011935976192...=32.0119 \end{array}$$