Question

The height of a triangle is increasing at a
rate of 5 cm/min while its area is increasing
at a rate of 4 sq. cms/min.
At what speed is the base of the triangle
changing when the height of the triangle is 6
cms and its area is 24 sq. cms?

Answer 1

The area, base, and height are functions of time t. We can write h(t) and A(t) from the problem description:\[h(t) = h _{0} + 5t\]\[A(t) = A_{0} + 4t\]They also tell us that at some point in time h(t) = 6 and A(t) = 24. Let's just call that time t=0. It's arbitrary and makes our work easier. So we can see that h0 = 6 and A0 = 24.

In a triangle A=(1/2)bh, so we can write an expression for b(t) using h(t) and A(t)\[A(t) = \frac{1}{2}b(t)h(t)\]\[b(t) = 2\frac{A(t)}{h(t)} = 2\frac{48+8t}{6+5t}\]The problem is asking what is the change in b(t) at time t=0, or what's b'(0)? So we just find b'(t) and plut in t=0. Use the quotient rule:\[\frac{d}{dx}\frac{f(x)}{g(x)} = \frac{gf' - fg'}{g^{2}}\]So b'(t) becomes\[b'(t) = \frac{(48+8t)(5)-(6+5t)(8)}{(6+5t)^{2}} = \frac{192}{(6+5t)^{2}}\]Plut in t=0 and we get b'(0) = 192/36 = 16/3.

Answer 2

dmancine seems to answer it correctly but b'(0) should be negative.

Answer 3

I did type a spurious 2 in my formula for b(t), but I didn't carry it through the calculations. It should say\[b(t) = \frac{48+8t}{6+5t}\]Can you find another mistake that might change my final answer?

Answer 4

compare last third and last second equations

Answer 5

Ah, yes. I did hiDlo-loDhi. Good catch. This question was reposted, anyway, with better answers. Please ignore my work completely.