The hypotenuse AB of a right triangle ABC is 5 ft, and one leg, AC, is decreasing at the rate of 2 ft/sec. The rate, in square feet per second, at which the area is changing when AC = 3 is?
This is a related rates question, Calculus. I keep -4 ft^2/s, but the answers are: A. 25/4 B. 7/4 C. -3/2 D. -7/2 E. -7/4 Am I doing something wrong, or was there a typo in the answers?
im not taking calculus but would u mind posting exactly what you are doing step by step so you can actually see and evaluate d work?
ok...we have two relevant equations here:\[c^2=a^2+b^2\]\[A=\frac{1}{2}bh\]
hang on...i have to put my kids to bed...be back...
no problem, thanks for helping
ok. we have c=5 and we know that the length of side a is decreasingat a rate of 2 ft/s. In math speak this looks like:\[\frac{d}{dt}a=-2\]we want:\[\frac{d}{dt}A=?\]when a=3. Ok, first we'll re-write the area equation as\[A=\frac{1}{2}ab\]since that's what we called the sides in the pythagorus equation. Notice that the hypotenuse is not changing in length, it remains constant at c=5. Differentiating c^2=a^2+b^2 with respect to time gives us:\[2c(\frac{d}{dt}c)=2a(\frac{d}{dt}a)+2b(\frac{d}{dt}b)\]Since c is not changing it's time derivative is zero:\[0=2(3)(-2)+2b(\frac{d}{dt}b)\]we get:\[12=2b(\frac{d}{dt}b)\]now we can get the value of b by c^2=a^2+b^2, where c=5 and a=3. Thus b=4. We get:\[\frac{d}{dt}b=\frac{3}{2}\]So the rate of change of side b is 3/2. Differentiating the area equation we get:\[\frac{dA}{dt}=\frac{1}{2}(a'b+ab')\]We have everything we need to get dA/dt:\[\frac{dA}{dt}=\frac{1}{2}(-2*4+3*\frac{3}{2})\]We get:\[\frac{dA}{dt}=-4+\frac{9}{2}=\frac{1}{2}\]which isn't one of your answers....hmmmm.....let me check for an error.
ooopps my error is in the last step...my arithmetic is off:\[\frac{dA}{dt}=\frac{1}{2}(-8+\frac{9}{2})=-4+\frac{9}{4}=\frac{-16+9}{4}=\frac{-7}{4}\]there you go :)
I love you, thank you so much! :D
lol...your welcome. Off to watch Thurs night football (go Rams!). good luck.
Thank you again-- have funzies! :D
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