# Rate of change

• June 12th 2010, 09:27 AM
Glitch
Rate of change
Sorry if this should be in the Calculus section, it still shows up as having malware.

The question:

A 5m ladder is leaning against a vertical wall. Suppose that the bottom of the ladder is being pulled away from the wall at a rate of 1m/s. How fast is the area of the triangle underneath the ladder changing at the instant that the top of the ladder is 4m from the floor?

My attempt:

A = (1/2)xy

dA/dt = dA/dx + dx/dt

dA/dt = (1/2)(dx/dt)y + (1/2)(dy/dt)x

Sub in y = 4, x = 3 (using trig),

dA/dt = 3.5(dy/dt)

I'm not sure if I'm doing this correctly, it doesn't seem right that I'd have to solve for dy/dx. The answer is supposed to be 7/8.

Any assistance would be great!
• June 12th 2010, 09:42 AM
skeeter
Quote:

Originally Posted by Glitch
The question:

A 5m ladder is leaning against a vertical wall. Suppose that the bottom of the ladder is being pulled away from the wall at a rate of 1m/s. How fast is the area of the triangle underneath the ladder changing at the instant that the top of the ladder is 4m from the floor?

you need two equations ...

$x^2 + y^2 = 5^2$

and

$A = \frac{1}{2}xy$

take the time derivative of the first ...

$2x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt} = 0$

$\frac{dy}{dt} = -\frac{x}{y} \cdot \frac{dx}{dt}
$

you were given $\frac{dx}{dt} = 1$ and $y = 4$ , and you should be able to easily calculate the value of $x$ when $y = 4$

using these values, calculate the value of $\frac{dy}{dt}$

take the time derivative of the second equation ...

$\frac{dA}{dt} = \frac{1}{2}\left(x \cdot \frac{dy}{dt} + y \cdot \frac{dx}{dt}\right)
$

... sub in all your values to find $\frac{dA}{dt}$
• June 12th 2010, 09:46 AM
Mukilab
Since you didn't mention, I assume it is the base of the ladder that is sliding away from the wall at a rate of 1m/s and not the whole ladder.

Assuming the above is true the first triangle formed would have an area approximate to 6cm^2 (this is easily known without using trig because the triangle is made of lengths of which are phythagorean triplets)

If the ladder slips 1m in one second the next area would still be 6cm^2

Not sure if my answer is correct.

Assuming the ladder slips again 1m away the hypotenuse would be 5 and the base would also be 5 which is not possible I believe therefore the ladder would be lying on the ground.
• June 12th 2010, 09:54 AM
Glitch
Cheers skeeter, I believe our answers are the same, apart from calculating dy/dt.

Mukilab, the base slides away at 1m/s. The area gets smaller, so it can't remain at 6cm^2.
• June 12th 2010, 10:10 AM
Unknown008
A = (xy)/2

$\frac{dA}{dt} = \frac{dA}{dx} \times \frac{dx}{dt}$

Now, dx/dt = 1.

$y = \sqrt{5^2 - x^2} = \sqrt{25-x^2}$

So, $A = \frac{x}{2}\sqrt{25-x^2}$

$\frac{dA}{dx} = \frac{x}{2}.\frac12 (25-x^2)^{-\frac12} + \frac12 (25-x^2)^{\frac12}.1 = \frac{x}{4}(25-x^2)^{-\frac12} + \frac12 (25-x^2)^{\frac12}$

Use it up there.

$\frac{dA}{dt} = \frac{x}{4}(25-x^2)^{-\frac12} + \frac12 (25-x^2)^{\frac12}$

$\frac{dA}{dt} = \frac{x + 2(25-x^2)}{4\sqrt{25-x^2}}$

Use y = 4, so x = 3

$\frac{dA}{dt} = \frac{3 + 2(25-3^2)}{4\sqrt{25-3^2}} = \frac{35}{16} m^2/s$

Geesh!!! :(
• June 12th 2010, 11:47 AM
skeeter
Quote:

Originally Posted by Glitch
Cheers skeeter, I believe our answers are the same, apart from calculating dy/dt.

no, they're not.

$\frac{dy}{dt} = -\frac{3}{4} \, m/s
$

$\frac{dA}{dt} = \frac{1}{2}\left(3 \cdot -\frac{3}{4} + 4 \cdot 1\right) = \frac{7}{8} \, m^2/s
$