Question

A man is walking at the rate of 4.5 km/hr towards the foot of the tower 120m high. At what rate is he approaching the top of the tower when he is 50m away from the tower?

Answer 1

Hint: we will use the concept of derivative as rate measures. we used Pythagoras theorem to form an equation and hypotenuse, which we need to find here, will be the unknown factor. we will differentiate the equation with respect to time since we need to find the speed.

Complete step-by-step answer:
First, let Draw a rough sketch which will give us a rough idea of what exactly is happening in the sums.

So from the sketch, we get the idea that the base, that is, the distance between the man and foot of the tower is in taken as \[x\] and we have also been told according to the question that the man covers this distance with a speed of \[4.5\,\text{km/hr}\] thus that will be \[\dfrac{dx}{dt}\] which means that the man is walking a certain amount of distance per unit time.
The hypotenuse will be taken as \[y\] which is the distance between the man and the top of the tower. We also need to find the rate at which the man is passing this hypotenuse distance which will be \[\dfrac{dy}{dt}\] which means the distance of y with respect to per unit time.
We have been given the height of the tower which is also the perpendicular of this triangle.
Thus given to us in expression is;
\[ y=\text{hypotenuse} \]
\[x=base \]
\[\text{perpendicular}=120\text{m} \] & 

\[ \text{hypotenus}{{\text{e}}^{2}}=\text{b}as{{e}^{2}}+\text{perpendicula}{{\text{r}}^{2}}\text{ }[\text{ pythagoras theorem }\]
\[{{y}^{2}}={{x}^{2}}+{{(120)}^{2}}-(1) \]
Using the required data we will use...
\[{{y}^{2}}={{x}^{2}}+{{(120)}^{2}}-(1)\]
We will differentiate both sides;
\[ 2y\dfrac{dy}{dt}=2x\dfrac{dx}{dt} \]
\[ \dfrac{dy}{dt}=\dfrac{x}{y}\dfrac{dx}{dt}-(2) \]
putting \[ \dfrac{dx}{dt}=-4.5\text{ km/hr in (2)} \]
\[ \dfrac{dy}{dt}=-4.5\times \dfrac{x}{y}.........(3) \]
The negative sign in \[\dfrac{dx}{dt}\] shows that the man is moving towards the tower thus the distance is decreasing. If the man was going away from the tower it would have been positive since the distance would have been increasing.
Further in the sum, we have been told that the distance between the foot of the tower and the man is \[x=50\] so we will use this parameter in the equation (1)...
\[\text{putting }x=50\text{ in (1) we will get: }y=\sqrt{{{50}^{2}}+{{120}^{2}}}=130\]
Thus using the parameter we found the value of \[y=130\,\] which is precisely the distance between the man and the top of the tower.
Now using this value of \[y\] we will get the rate at which the man is reaching the top of the tower;
putting x=50, y=130 in (3), we will get
\[ \dfrac{dy}{dt}=-\dfrac{4.5\times 50}{130}=-1.73\,\text{km/hr}\] 

x and y has same unit of meter thus on dividing & they get canceled leaving only km/hr from the speed x.
The negative sign shows that the man is moving towards the top of the tower.
Thus the man is approaching the top of the tower at the rate \[1.73\,\text{km/hr}\text{.}\,\]

Note: The \[\text{km}\] is distance in kilometre and \[\text{m}\] is distance in meter. \[\text{hr}\] is the hours in time. Differentiation gives us the per unit. That means at a very small unit. Here we took per unit time which means every hour or second.
They took negative signs for\[\dfrac{dx}{dt}\] because the man was moving toward the base of the tower. If the man would have moved away from the tower we would have taken a positive sign.