How Fast Is the Shadow's Height Changing as the Dog Approaches the Wall?

In summary, this problem involves related rates and finding the rate of change of the height of a dog's shadow on a building as the dog moves closer to the building. Using similar triangles, the equation 24 = Lx can be derived, where L is the length of the shadow and x is the ground covered by the spotlight. Differentiating this equation with respect to time, we get dL/dt = -24(12-8)^-2 = 9/32.
  • #1
venger
58
0

Homework Statement


A spotlight on the ground shines on a wall 14m away. If a dog, 0.5m tall, runs from the spotlight towars the building at a speed of 1 m/s, how fast is the height of the animal's shadow on the building decreasing when the dog is 5 meters from the building?

Wrt = with respect to

Homework Equations


Pythagoreon theorem, implicit differentiation, you name it.


The Attempt at a Solution



Knowing that Dc/dt is canine wrt time is 1 m/s and i am trying to find dh/dt 'h' for height wrt time

and I am stuck...
.
..
...
14^2 + wall^2 = Hyp^2
 
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  • #2
This is called a "related rates" problem, not "implicit differentiation."

dc/dt? if you're using the pythagorean theorem, that implies (at least to me) that the dog is running on the hypotenuse??

At a glance, I'm having a tough time getting the pythagorean theorem to really work for me on this problem. Have you thought about the angle made by the flashlight between the ground and the beam to the top of the dog? Or, have you thought about similar triangles?
 
  • #3
You should have put "similar triangles", not "Pythagorean theorem", in "Relevant Equations". Can you draw a picture and see why?
 
  • #4
Here i go furthering my development in related rates
let l be length of shadow and let x be ground covered with light
2/L = X/12
24=LX
L=24X^-1
dL/dt(1)=-24X^-2
dL/dt=-24(12-8)^-2
dL/dt=9/32
 

FAQ: How Fast Is the Shadow's Height Changing as the Dog Approaches the Wall?

1. What is implicit differentiation?

Implicit differentiation is a mathematical technique used to find the derivative of a function that is not expressed explicitly in terms of its independent variable. It is used when the dependent variable cannot be easily isolated on one side of the equation.

2. Why is implicit differentiation useful?

Implicit differentiation allows us to find the derivative of a function without having to solve for the dependent variable. This is especially useful for more complex functions where isolating the dependent variable may be difficult or impossible.

3. How do I use implicit differentiation?

To use implicit differentiation, you first need to differentiate both sides of the equation with respect to the independent variable. Then, you solve for the derivative of the dependent variable in terms of the derivative of the independent variable.

4. What are some common applications of implicit differentiation?

Implicit differentiation is commonly used in physics and engineering to find rates of change and in optimization problems. It is also used in economics and finance to analyze marginal effects and elasticity.

5. Are there any limitations to implicit differentiation?

Implicit differentiation can only be used for functions that are differentiable and continuous. It also requires a good understanding of the chain rule and other basic differentiation rules. Additionally, it may not always be the most efficient or accurate method for finding derivatives.

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