# Thread: Rate of Change

1. ## Rate of Change

I have a homework problem that I am unable to figure out.

Consider a triangle (not necessarily a right triangle!) with the length of the base increasing at a rate of 1/2 cm/s. At what rate must the height change so that the area remains constant at the instant the base is 10cm and the height is 18cm. Hint: What are the quantites that are changing?

2. ## Re: Rate of Change

the area of a triangle is calculated by the formula:

A = 1/2(b)(h), the area remains constant so you can calculate it's area for b = 10 and h = 18; it would be the same bc it's the height that is changing,

Next, having the Area and the rate of change of the base, db/dt = 1/2cm/s, take the derivative of A = 1/2(b)(h), and solve for dh/dt,

dokrbb

btw, can you say which sides of the triangle are changing? and in what direction (neg or pos), what that means?, cheers

3. ## Re: Rate of Change

Thank you so very much! This is great and very helpful.

I will think about those questions once I finish the problem. Thank you again for your help.

4. ## Re: Rate of Change

As I am trying to solve it I end up with 90=1/2(1/2)dh/dt and I have no idea how to continue. Also if the area remains the same and the base is increasing, the height must be decreasing which would make it neg.

5. ## Re: Rate of Change

Originally Posted by DreamingofWolves
I have a homework problem that I am unable to figure out.

Consider a triangle (not necessarily a right triangle!) with the length of the base increasing at a rate of 1/2 cm/s. At what rate must the height change so that the area remains constant at the instant the base is 10cm and the height is 18cm. Hint: What are the quantites that are changing?
Very important here is the 'per second' in the given 1/2 cm/s so you must differentiate the Aea function with respect to time ,

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

$\frac{d}{dt}(A) = \frac{d}{dt}(\frac{1}{2}bh)$

$\frac{dA}{dt} = \frac{1}{2} \frac{d}{dt}(bh)$

Now use the product rule on the right hand side of the equation ,

$\frac{dA}{dt} = \frac{1}{2}(b \frac{dh}{dt} + h \frac{db}{dt})$

Now you have...

$\frac{dA}{dt} = 0$ because area remains constant

$\frac{db}{dt} = \frac{1}{2}$ because the base is increasing at that rate

$b = 10 \ h = 18$

so when you plug these in your only unknown is $\frac{dh}{dt}$

Can you solve it now?

6. ## Re: Rate of Change

I got as far as plugging everything it, it is solving for dh/dt. Do I solve it as if it is another variable or am I going to have to take the derivative of something?

7. ## Re: Rate of Change

As far as you have got this formula, already differentiated,

$\frac{dA}{dt} = \frac{1}{2}(b \frac{dh}{dt} + h \frac{db}{dt})$

you solve for $\frac{dh}{dt}$ as it asks you - at what rate is the height changing?

dokrbb

8. ## Re: Rate of Change

Originally Posted by DreamingofWolves
I got as far as plugging everything it, it is solving for dh/dt. Do I solve it as if it is another variable or am I going to have to take the derivative of something?
You don't have to take any more derivatives, dh/dt is just like another variable now, isolate dh/dt.