Is This Answer Correct? Multiplying 10x and 12 by Opposite Values

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my question is... is this answer correct? for the last step. WOULDNT you multiply 10x by .5 and 12 by .2? the opposite of what this answer is? Itsays BASE is 10 in and BASE chagnges at .5 in per second
problem:
i45.tinypic.com/xknbci.png
answer:
i50.tinypic.com/2qcnsyv.jpg

this is what i get...

i50.tinypic.com/124cxgy.jpg
 

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indigo1 said:
my question is... is this answer correct? for the last step. WOULDN'T you multiply 10[STRIKE]x[/STRIKE] by .5 and 12 by .2? the opposite of what this answer is? It says BASE is 10 in and BASE changes at .5 in per second
problem:
i45.tinypic.com/xknbci.png
answer:
i50.tinypic.com/2qcnsyv.jpg

this is what i get...

i50.tinypic.com/124cxgy.jpg
Hello indigo1. Welcome to PF !

Why would you want to multiply the base by the rate of change of the base? Even your working has [itex]\displaystyle b\frac{dh}{dt}[/itex] and [itex]\displaystyle h\frac{db}{dt}\ .[/itex]

dh/dt is -0.2 in/sec and db/dt is 0.5 in/sec !

BTW: Below are your images, displayed so others can see them.

Printed problem text: a. The base of a triangle is 10 in and the height is 12 in. Find the rate of change in the area of the triangle if the base is changing at a rate of 0.5 in/sec and the height is changing at a rate of -0.2 in/sec.


Handwritten working on a related-rates problem: right triangle with vertices A, B, C, base b = 10, height h = 12, and the derivative of A(t) = (1/2)(b(t)h(t)) worked out, ending in 2 in/sec.


Handwritten calculus working on lined paper: da/dt = .5[b·dh/dt + h·db/dt], then da/dt = .5[10(.5) + 12(-.2)], then da/dt = 1.3 in²/sec circled.
 
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  • Printed problem text: a. The base of a triangle is 10 in and the height is 12 in. Find the rate of change in the area of the triangle if the base is changing at a rate of 0.5 in/sec and the height is changing at a rate of -0.2 in/sec.
    xknbci.png
    16 KB · Views: 916
  • Handwritten work on a triangle area rate problem: A(t) = (1/2)(b(t)h(t)), differentiated to dA/dt = (1/2)[b dh/dt + h db/dt], with b = 10, h = 12, db/dt = 0.5, dh/dt = -0.2, giving 2 in/sec. Beside it, a right triangle with vertex C at top, A and B at bottom, height 12 and base 10.
    2qcnsyv.jpg
    22.9 KB · Views: 2,953
  • Handwritten calculus working on lined paper: da/dt = .5[b · dh/dt + h · db/dt], then da/dt = .5[10(.5) + 12(-.2)], then da/dt = 1.3 in²/sec circled.
    124cxgy.jpg
    23.7 KB · Views: 932
This was really helpful
But i was wondering if this is the same for my HW problem i applied the same A' equation but instead looked for the db/dt and i got -3/2 idk if this is right.
"The base of an isosceles triangle is 6 feet. If the altitude is 4 feet and increasing at the rate of 2 inches per minute, at what rate is the vertex angle changing?"
I will really appreciate for your help.
 
seawolf23, instead of tacking your question onto the end of a question from almost two years ago, please start a new thread. Be sure to use the homework template, and include the problem statement and the work you have done.
 

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