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Estimating change with differentials?

The volume of a right circular cone is given by

V= \frac{1}{3} \pi r^2 h

where the

r

is the radius and

h

is the height. Find

\frac{dV}{dt},

if

r=4, h=5, \frac{dr}{dt}=6

and

\frac{dh}{dt}=-3

Im quite lost on what to do here.

You need to use the product rule and chain rule.

If I write

V = \frac{1}{3} \pi r(t)^2 h(t)

to emphasise that r and h are functions of t, does that help?

Sorry if you read my last post - I misread the question.

OP

DFranklin

You need to use the product rule and chain rule.

If I write

V = \frac{1}{3} \pi r(t)^2 h(t)

to emphasise that r and h are functions of t, does that help?

Is it sort of implicitly differentiating?- r's and h's in terms of t?

Do you know what the product rule and chain rule are?

OP

DFranklin

Do you know what the product rule and chain rule are?

Yes, I do.

OK. Suppose we have the formula

V = \frac{1}{3} F(t)G(t)

. What is dV/dt?

Edit: actually, maybe it is "sort of implicitly differentiating?- r's and h's in terms of t?". I'm not really sure what you meant - you might have meant the right thing.

OP

Wait - I think I got it- Ill have things in terms of r,r',h and h'

OP

DFranklin

OK. Suppose we have the formula

V = \frac{1}{3} F(t)G(t)

. What is dV/dt?

Edit: actually, maybe it is "sort of implicitly differentiating?- r's and h's in terms of t?". I'm not really sure what you meant - you might have meant the right thing.

Should I get something like

Unparseable latex formula:

\frac{dV}{dt}= 1/3 \pi r^2 \frac{dh}{dt} + h 2{\frac {dr}{dt}

nk9230

Should I get something like

Unparseable latex formula:

\frac{dV}{dt}= 1/3 \pi r^2 \frac{dh}{dt} + h 2{\frac {dr}{dt}

Almost.

Think about what the derivative of r^2 is with respect to t - it's not

Unparseable latex formula:

2{\frac {dr}{dt}

.

Letting

u=r^2

may help.

Also,

\frac{1}{3}\pi

should be in front of both terms.

Almost. The

\frac{1}{3}\pi

applies to both the dh/dt and dr/dt terms (which is probably a typo on your part). But you also haven't differentiated r^2 correctly.

OP

notnek

Almost.

Think about what the derivative of r^2 is with respect to t - it's not

Unparseable latex formula:

2{\frac {dr}{dt}

.

Letting

u=r^2

may help.

Also,

\frac{1}{3}\pi

should be in front of both terms.

I found that error- it should be

2r \frac {dr}{dt}

nk9230

I found that error- it should be

2r \frac {dr}{dt}

Yup.

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