Applications of Derivatives using Pythagoras

In summary, a man is walking away from a lampost at a speed of 1.5 m/s. The lampost is 5m above the ground and the man is 2m tall. Using pythagoras and taking the derivative, it is found that the shadow is growing at a rate of 1m/s when the man is 10m away from the lampost. In another question, two ships at different speeds and directions are a certain distance apart at 1:00 p.m. and the question asks for the rate of change of their distance at 3:00 p.m. Finally, a water trough with a triangular cross section is being filled at a rate of 0.4 m^3/min
  • #1
erik05
50
0
Hello. Could anyone help me with a few questions that I'm stuck on? Any suggestions would be much appreciatd.

1) A man 2m tall walks away from a lampost whose light is 5m above the ground. If he walks at a speed of 1.5 m/s, at what rate is his shadow growing when he is 10m from the lampost?


Using pythagoras and taking the derivative I got:

[tex] \frac{dy}{dx}= \frac{-x}{y} \frac{dx}{dt}[/tex]

The derivative of x with respect to t(time) is 1.5 m/s so I think what I'm trying to find is the derivative of y with respect to t. Not too sure what to do here now but the answer is suppose to be 1m/s.

2) At 1:00 p.m. ship A was 80 km south of ship B. Ship A is sailing north at 30 km/h and ship B is sailing east at 40 km/h. How fast is the distance between them changing at 3:00 p.m.?

I think it's the times that is messing me up. Tried using the derivative of pythagoras again but where does the 80 km fit in? Answer: [tex] \frac{130}{\sqrt{17}} = 31.5 km/h [/tex]

3) A water trough is 10 m long and a cross section has the shape of an isosceles triangle that is 1m across at the top and is 50 cm high. The trough is being filled with water at a rate of 0.4 m^3/min. How fast will the water level rise when the water is 40 cm deep? Answer: 5 cm/min

Would differiantiating the equation for the area of a triangle fit into this?

Anyways, thank you in advance.
 
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  • #2
erik05 said:
Hello. Could anyone help me with a few questions that I'm stuck on? Any suggestions would be much appreciatd.

1) A man 2m tall walks away from a lampost whose light is 5m above the ground. If he walks at a speed of 1.5 m/s, at what rate is his shadow growing when he is 10m from the lampost?

Not too sure what to do here now but the answer is suppose to be 1m/s.

I'll do this one for you. Yes, the answer is 1m/s. Make a diagram of the lampost and the man (let base of lampost be at origin (0,0) and base of man at (10,0). Horizontal distance from man to lampost is 10 m. Draw a line from top of lampost passing through top of man and extend it till it hits x-axis. It will hit x-axis at (16.67,0), so length of shadow = 16.67-10 = 6.67.

Generalise: let x be the distance from base of post to base of man, let y be length of shadow. Then by properties of similar triangles:

y/2 = (x+y)/5, so y = 2x/3, hence dy/dx = 2/3. So if x is growing by 1.5 m/s, then y will be increasing by 1.5*(2/3) = 1 m/s. QED.
 
  • #3
erik05 said:
3) A water trough is 10 m long and a cross section has the shape of an isosceles triangle that is 1m across at the top and is 50 cm high. The trough is being filled with water at a rate of 0.4 m^3/min. How fast will the water level rise when the water is 40 cm deep? Answer: 5 cm/min

Would differiantiating the equation for the area of a triangle fit into this?

Anyways, thank you in advance.
Previous msg completed Problem #1. Solution to Prob #2 uses similar techniques to Prob #1. Here's Prob #3:
{Length of Trough} = (10 m)
{Base of Triangular Cross-section} = (1 m)
{Height of Triangular Cross-section} = (50 cm) = (0.5 m)
{Area of Triangular Cross-section} = (1/2)*{Base}*{Height} = (1/2)*(1)*(0.5) = (0.25 m^2)
{Volume of Trough} = {Length}*{Area} = (10)*(0.25) = (2.5 m^3)

{Height of Water in Trough} = H
{Base of Water Triangular Cross-section} = (1 - 0)*(H - 0)/(0.5 - 0) = 2*H
{Area of Water Triangular Cross-section} = (1/2)*{Base}*{Height} = (1/2)*(2*H)*(H) = H2
{Volume of Water in Trough} = V = {Length}*{Area} = (10)*(H2) = 10*H2
{Water Fill Rate} = (0.4 m^3/min)

Water Volume time rate of change is given by (dV/dt). Thus, from the above data:
(dV/dt) = (d/dt){10*H2} = (10)*(2)*(H)*(dH/dt) = (20)*H*(dH/dt)

It's given that {Water Fill Rate}=(0.4 m^3/min), so we place this value into the above equation and solve for (dH/dt) when H=(40 cm)=(0.4 m):
(0.4) = (20)*H*(dH/dt)
::: ⇒ (dH/dt) = (0.4)/{(20)*H} = (0.02)/H = (0.02)/(0.4)
::: ⇒ (dH/dt) = (0.05 m/min) = (5 cm/min)

~~
 
Last edited:
  • #4
Ah, you have to remember the property of similar triangles for number 1. I see. Any hints for question number two?
 
  • #5
erik05 said:
Ah, you have to remember the property of similar triangles for number 1. I see. Any hints for question number two?

Questions like #2 are solved as examples in most calculus texts. Please look them up. If you still can't get the solution by tomorrow evening, I'll do it for you.
 

1. How is Pythagoras' theorem used in calculus?

Pythagoras' theorem is used in calculus to find the derivative of a function involving two or more variables. It helps to determine the rate of change of a function with respect to one of its variables. This is done by using the Pythagorean identity, which states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides.

2. What are some real-life applications of derivatives using Pythagoras?

There are many real-life applications of derivatives using Pythagoras, such as finding the maximum or minimum values of a function, calculating rates of change in physics and engineering, and optimizing routes in transportation and logistics. It is also used in economics to determine the marginal cost and revenue of a product.

3. How does the Pythagorean theorem relate to optimization problems?

The Pythagorean theorem is used in optimization problems to find the shortest distance between two points. In these problems, the derivative of a function is used to find the critical points, which are then plugged into the Pythagorean theorem to determine the shortest distance. This is useful in various fields, including engineering, economics, and physics.

4. Can Pythagoras' theorem be used to find the slope of a curve?

Yes, Pythagoras' theorem can be used to find the slope of a curve. This is done by taking the derivative of the function and then using the Pythagorean identity to calculate the slope at a specific point. This is useful in determining the steepness of a curve, which is important in many applications, such as optimizing a roller coaster design or predicting the path of a projectile.

5. How is the concept of derivatives using Pythagoras applied in engineering?

In engineering, derivatives using Pythagoras are used in various applications, such as calculating the velocity and acceleration of a moving object, determining the stress and strain on a material, and optimizing designs for maximum efficiency. It is also used in electrical engineering to calculate the rate of change of current and voltage in circuits.

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