Help on Related Rates implicit differentiation

[riri]

Hi!

I recently came upon this problem : the height of a right angled triangle is increasing at a rate of 5cm/min while the area is constant. How fast must the base be decreasing at the moment when the height is 5 times the base?

I drew a picture of the triangle, labelled the height (h) and base (b)... area of a triangle = 1/2bh correct?
And it also states I should use PRODUCT RULE.

Now what I'm confused about is how to proceed. I have to find \d{b}{dt} and was wondering if there's a simple way to do this? dh/dt = 5cm/min and I'm a bit confused on how to write the next part of the equation and how to solve this step by step.

Thank you! :)

 

[MarkFL]

I would begin, as you did, by stating the formula for the area $A$ of a triangle which involves the height $h$ and the base $b$ (but in a slightly different form):

\(\displaystyle 2A=bh\)

Now, we know that both $b$ and $h$ are changing over time, and so will be represented as functions of time, and so what do we get when we differentiate both sides of the equation w.r.t time $t$ (keeping in mind that $A$ is a constant)?

 

[riri]

Hi! :)

Okay so if A is a constant and you differentiate, would I get : 2= (\d{db}{dt})h+(\d{dh}{dt})b ??
And since h= 5b
I just insert h into the h in the equation above??

 

[MarkFL]

Hi! :)

Okay so if A is a constant and you differentiate, would I get : 2= (\d{db}{dt})h+(\d{dh}{dt})b ??
And since h= 5b
I just insert h into the h in the equation above??

Hello! (Wave)

First, you need to enclose your $\LaTeX$ codes within tags, and the easiest way is to click the $\Sigma$ button on the editor toolbar, and then put your code in between the generated [MATH][/MATH] tags.

Next, when using the \d command, you only need to put the variables within the braces, for example:

\d{y}{x}

results in:

\(\displaystyle \d{y}{x}\)

Now, when something remains constant with respect to a certain variable, in this case time $t$, then it does not change, and so its derivative with respect to that variable will be zero, and so you would have:

\(\displaystyle 0=b\d{h}{t}+\d{b}{t}h\)

Since we are asked about the time rate of change of $b$, we want to solve this for \(\displaystyle \d{b}{t}\)...what do we get?

 

[riri]

\(\displaystyle \d{b}{t}\)h+5b=0

Is what I got! (I hope I'm doing the tags correctly this time:) )
and then I isolated \(\displaystyle \d{b}{t}\) and ended up with... -1 !
Is this the correct process?

Thank you!

 

[MarkFL]

\(\displaystyle \d{b}{t}\)h+5b=0

Is what I got! (I hope I'm doing the tags correctly this time:) )
and then I isolated \(\displaystyle \d{b}{t}\) and ended up with... -1 !
Is this the correct process?

Thank you!

You should enclose the entire equation within the tags. :)

I would write (solving for \(\displaystyle \d{b}{t}\)):

\(\displaystyle \d{b}{t}=-\frac{b}{h}\d{h}{t}\)

This was we have a formula that we can now plug in the given values...

And so (using $h=5b$):

\(\displaystyle \left.\d{b}{t}\right|_{h=5b}=-\frac{b}{5b}\d{h}{t}=-\frac{1}{5}\d{h}{t}\)

Now, plug in the given value \(\displaystyle \d{h}{t}=5\frac{\text{cm}}{\text{min}}\):

\(\displaystyle \left.\d{b}{t}\right|_{h=5b}=-\frac{1}{5}\left(5\frac{\text{cm}}{\text{min}}\right)=-1\frac{\text{cm}}{\text{min}}\)

So, you got the correct answer, I just wanted to demonstrate how to preserve the units. :)