Mixing
Finding Area for Related Rates Ladder Question [closed]
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A 52-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate of 8 feet per second, at what rate is the area of the triangle formed by the wall, the ground, and the ladder changing, in square feet per second, at the instant the bottom of the ladder is 48 feet from the wall?
How do I go about solving this one? I know A = 1/2bh but get a little confused on what the next step is. If anyone could help out I'd be really grateful. Thanks!
calculus derivatives word-problem
asked Jan 25 at 17:37
Jelly BiscuitJelly Biscuit
204
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closed as off-topic by José Carlos Santos, Leucippus, Shailesh, max_zorn, ancientmathematician Jan 26 at 7:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Leucippus, Shailesh, max_zorn, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
A 52-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate of 8 feet per second, at what rate is the area of the triangle formed by the wall, the ground, and the ladder changing, in square feet per second, at the instant the bottom of the ladder is 48 feet from the wall?
How do I go about solving this one? I know A = 1/2bh but get a little confused on what the next step is. If anyone could help out I'd be really grateful. Thanks!
calculus derivatives word-problem
asked Jan 25 at 17:37
Jelly BiscuitJelly Biscuit
204
$endgroup$
closed as off-topic by José Carlos Santos, Leucippus, Shailesh, max_zorn, ancientmathematician Jan 26 at 7:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Leucippus, Shailesh, max_zorn, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
A 52-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate of 8 feet per second, at what rate is the area of the triangle formed by the wall, the ground, and the ladder changing, in square feet per second, at the instant the bottom of the ladder is 48 feet from the wall?
How do I go about solving this one? I know A = 1/2bh but get a little confused on what the next step is. If anyone could help out I'd be really grateful. Thanks!
calculus derivatives word-problem
asked Jan 25 at 17:37
Jelly BiscuitJelly Biscuit
204
$endgroup$
A 52-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate of 8 feet per second, at what rate is the area of the triangle formed by the wall, the ground, and the ladder changing, in square feet per second, at the instant the bottom of the ladder is 48 feet from the wall?
How do I go about solving this one? I know A = 1/2bh but get a little confused on what the next step is. If anyone could help out I'd be really grateful. Thanks!
calculus derivatives word-problem
calculus derivatives word-problem
asked Jan 25 at 17:37
Jelly BiscuitJelly Biscuit
204
asked Jan 25 at 17:37
Jelly BiscuitJelly Biscuit
204
asked Jan 25 at 17:37
Jelly BiscuitJelly Biscuit
204
asked Jan 25 at 17:37
Jelly BiscuitJelly Biscuit
204
asked Jan 25 at 17:37
Jelly BiscuitJelly Biscuit
204
204
closed as off-topic by José Carlos Santos, Leucippus, Shailesh, max_zorn, ancientmathematician Jan 26 at 7:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Leucippus, Shailesh, max_zorn, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by José Carlos Santos, Leucippus, Shailesh, max_zorn, ancientmathematician Jan 26 at 7:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Leucippus, Shailesh, max_zorn, ancientmathematician
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
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oldest
votes
$begingroup$
For these kinds of problems, I suggest always drawing a picture. In this case, you can think of the positive $y$-axis as the wall and the positive $x$-axis as the floor on which the ladder stands. Then the ladder will be the line segment connecting the point on the $x$-axis where the base of ladder sits, and the point on the $y$-axis where the top of the ladder rests. You will then see that the ladder is the hypoteneuse of a right triangle.
Now we need to label our variables and obtain an equation relating them so that we can use the chain rule. Let's let $x$ be the the $x$-coordinate of the bottom of the ladder i.e. how far (horizontally) the bottom of the ladder is from the base of the wall. And let $y$ be the the $y$-coordinate of the top of the ladder i.e. how far (vertically) the top of the ladder is from the base of the wall. Then the area, as you pointed out, is
$$A=frac{1}{2}xy$$
We want the rate of change of the area i.e. $frac{dA}{dt}$. The key point here is that $x$ and $y$ are both functions of $t$ since they are changing with time. To emphasize this, we can write
$$A(t)=frac{1}{2}x(t)y(t)$$
Now we can obtain $frac{dA}{dt}$ using the product rule and chain rule:
$$frac{dA}{dt}=frac{1}{2}left[x(t)frac{dy}{dt}+y(t)frac{dx}{dt}right]$$
Most people won't use the extra notation here, so you'd probably see it as
$$frac{dA}{dt}=frac{1}{2}left[xfrac{dy}{dt}+yfrac{dx}{dt}right]$$
Now we plug in our information. We know that $x=48$ at the time of interest, and $frac{dx}{dt}=8$. But we still need to plug in $y$ and $frac{dy}{dt}$. To get $y$, we'll use the Pythagorean theorem $x^{2}+y^{2}=52^{2}$ with $x=8$ to get $y=sqrt{52^{2}-8^{2}}=sqrt{2630}$. But what about $frac{dy}{dt}$? Well, we can also use the Pythagorean theorem here (as well as chain rule):
$$x^{2}+y^{2}=52^{2}$$
$$2xfrac{dx}{dt}+2yfrac{dy}{dt}=0$$
$$2(48)(8)+2sqrt{2630}frac{dy}{dt}=0$$
$$frac{dy}{dt}=-frac{384}{sqrt{2630}}$$
Observe that $frac{dy}{dt}$ is negative (why?). I'll leave it to you to simplify the fraction. To finish the problem, return to our original equation
$$frac{dA}{dt}=frac{1}{2}left[xfrac{dy}{dt}+yfrac{dx}{dt}right]$$
and plug in the values of $x,y,frac{dx}{dt},$ and $frac{dy}{dt}$
answered Jan 25 at 17:51
pwerthpwerth
3,300417
$endgroup$
$begingroup$
Maybe easier to understand for some people would be to plug in $y=sqrt{52^2-x^2}$ into the first equation for the area, and just to a single derivative. It is the same answer (even some calculation steps might overlap), but it's more intuitive for beginners.
$endgroup$
– Andrei
Jan 25 at 18:03
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For these kinds of problems, I suggest always drawing a picture. In this case, you can think of the positive $y$-axis as the wall and the positive $x$-axis as the floor on which the ladder stands. Then the ladder will be the line segment connecting the point on the $x$-axis where the base of ladder sits, and the point on the $y$-axis where the top of the ladder rests. You will then see that the ladder is the hypoteneuse of a right triangle.
Now we need to label our variables and obtain an equation relating them so that we can use the chain rule. Let's let $x$ be the the $x$-coordinate of the bottom of the ladder i.e. how far (horizontally) the bottom of the ladder is from the base of the wall. And let $y$ be the the $y$-coordinate of the top of the ladder i.e. how far (vertically) the top of the ladder is from the base of the wall. Then the area, as you pointed out, is
$$A=frac{1}{2}xy$$
We want the rate of change of the area i.e. $frac{dA}{dt}$. The key point here is that $x$ and $y$ are both functions of $t$ since they are changing with time. To emphasize this, we can write
$$A(t)=frac{1}{2}x(t)y(t)$$
Now we can obtain $frac{dA}{dt}$ using the product rule and chain rule:
$$frac{dA}{dt}=frac{1}{2}left[x(t)frac{dy}{dt}+y(t)frac{dx}{dt}right]$$
Most people won't use the extra notation here, so you'd probably see it as
$$frac{dA}{dt}=frac{1}{2}left[xfrac{dy}{dt}+yfrac{dx}{dt}right]$$
Now we plug in our information. We know that $x=48$ at the time of interest, and $frac{dx}{dt}=8$. But we still need to plug in $y$ and $frac{dy}{dt}$. To get $y$, we'll use the Pythagorean theorem $x^{2}+y^{2}=52^{2}$ with $x=8$ to get $y=sqrt{52^{2}-8^{2}}=sqrt{2630}$. But what about $frac{dy}{dt}$? Well, we can also use the Pythagorean theorem here (as well as chain rule):
$$x^{2}+y^{2}=52^{2}$$
$$2xfrac{dx}{dt}+2yfrac{dy}{dt}=0$$
$$2(48)(8)+2sqrt{2630}frac{dy}{dt}=0$$
$$frac{dy}{dt}=-frac{384}{sqrt{2630}}$$
Observe that $frac{dy}{dt}$ is negative (why?). I'll leave it to you to simplify the fraction. To finish the problem, return to our original equation
$$frac{dA}{dt}=frac{1}{2}left[xfrac{dy}{dt}+yfrac{dx}{dt}right]$$
and plug in the values of $x,y,frac{dx}{dt},$ and $frac{dy}{dt}$
answered Jan 25 at 17:51
pwerthpwerth
3,300417
$endgroup$
$begingroup$
Maybe easier to understand for some people would be to plug in $y=sqrt{52^2-x^2}$ into the first equation for the area, and just to a single derivative. It is the same answer (even some calculation steps might overlap), but it's more intuitive for beginners.
$endgroup$
– Andrei
Jan 25 at 18:03
add a comment |
$begingroup$
For these kinds of problems, I suggest always drawing a picture. In this case, you can think of the positive $y$-axis as the wall and the positive $x$-axis as the floor on which the ladder stands. Then the ladder will be the line segment connecting the point on the $x$-axis where the base of ladder sits, and the point on the $y$-axis where the top of the ladder rests. You will then see that the ladder is the hypoteneuse of a right triangle.
Now we need to label our variables and obtain an equation relating them so that we can use the chain rule. Let's let $x$ be the the $x$-coordinate of the bottom of the ladder i.e. how far (horizontally) the bottom of the ladder is from the base of the wall. And let $y$ be the the $y$-coordinate of the top of the ladder i.e. how far (vertically) the top of the ladder is from the base of the wall. Then the area, as you pointed out, is
$$A=frac{1}{2}xy$$
We want the rate of change of the area i.e. $frac{dA}{dt}$. The key point here is that $x$ and $y$ are both functions of $t$ since they are changing with time. To emphasize this, we can write
$$A(t)=frac{1}{2}x(t)y(t)$$
Now we can obtain $frac{dA}{dt}$ using the product rule and chain rule:
$$frac{dA}{dt}=frac{1}{2}left[x(t)frac{dy}{dt}+y(t)frac{dx}{dt}right]$$
Most people won't use the extra notation here, so you'd probably see it as
$$frac{dA}{dt}=frac{1}{2}left[xfrac{dy}{dt}+yfrac{dx}{dt}right]$$
Now we plug in our information. We know that $x=48$ at the time of interest, and $frac{dx}{dt}=8$. But we still need to plug in $y$ and $frac{dy}{dt}$. To get $y$, we'll use the Pythagorean theorem $x^{2}+y^{2}=52^{2}$ with $x=8$ to get $y=sqrt{52^{2}-8^{2}}=sqrt{2630}$. But what about $frac{dy}{dt}$? Well, we can also use the Pythagorean theorem here (as well as chain rule):
$$x^{2}+y^{2}=52^{2}$$
$$2xfrac{dx}{dt}+2yfrac{dy}{dt}=0$$
$$2(48)(8)+2sqrt{2630}frac{dy}{dt}=0$$
$$frac{dy}{dt}=-frac{384}{sqrt{2630}}$$
Observe that $frac{dy}{dt}$ is negative (why?). I'll leave it to you to simplify the fraction. To finish the problem, return to our original equation
$$frac{dA}{dt}=frac{1}{2}left[xfrac{dy}{dt}+yfrac{dx}{dt}right]$$
and plug in the values of $x,y,frac{dx}{dt},$ and $frac{dy}{dt}$
answered Jan 25 at 17:51
pwerthpwerth
3,300417
$endgroup$
$begingroup$
Maybe easier to understand for some people would be to plug in $y=sqrt{52^2-x^2}$ into the first equation for the area, and just to a single derivative. It is the same answer (even some calculation steps might overlap), but it's more intuitive for beginners.
$endgroup$
– Andrei
Jan 25 at 18:03
add a comment |
$begingroup$
For these kinds of problems, I suggest always drawing a picture. In this case, you can think of the positive $y$-axis as the wall and the positive $x$-axis as the floor on which the ladder stands. Then the ladder will be the line segment connecting the point on the $x$-axis where the base of ladder sits, and the point on the $y$-axis where the top of the ladder rests. You will then see that the ladder is the hypoteneuse of a right triangle.
Now we need to label our variables and obtain an equation relating them so that we can use the chain rule. Let's let $x$ be the the $x$-coordinate of the bottom of the ladder i.e. how far (horizontally) the bottom of the ladder is from the base of the wall. And let $y$ be the the $y$-coordinate of the top of the ladder i.e. how far (vertically) the top of the ladder is from the base of the wall. Then the area, as you pointed out, is
$$A=frac{1}{2}xy$$
We want the rate of change of the area i.e. $frac{dA}{dt}$. The key point here is that $x$ and $y$ are both functions of $t$ since they are changing with time. To emphasize this, we can write
$$A(t)=frac{1}{2}x(t)y(t)$$
Now we can obtain $frac{dA}{dt}$ using the product rule and chain rule:
$$frac{dA}{dt}=frac{1}{2}left[x(t)frac{dy}{dt}+y(t)frac{dx}{dt}right]$$
Most people won't use the extra notation here, so you'd probably see it as
$$frac{dA}{dt}=frac{1}{2}left[xfrac{dy}{dt}+yfrac{dx}{dt}right]$$
Now we plug in our information. We know that $x=48$ at the time of interest, and $frac{dx}{dt}=8$. But we still need to plug in $y$ and $frac{dy}{dt}$. To get $y$, we'll use the Pythagorean theorem $x^{2}+y^{2}=52^{2}$ with $x=8$ to get $y=sqrt{52^{2}-8^{2}}=sqrt{2630}$. But what about $frac{dy}{dt}$? Well, we can also use the Pythagorean theorem here (as well as chain rule):
$$x^{2}+y^{2}=52^{2}$$
$$2xfrac{dx}{dt}+2yfrac{dy}{dt}=0$$
$$2(48)(8)+2sqrt{2630}frac{dy}{dt}=0$$
$$frac{dy}{dt}=-frac{384}{sqrt{2630}}$$
Observe that $frac{dy}{dt}$ is negative (why?). I'll leave it to you to simplify the fraction. To finish the problem, return to our original equation
$$frac{dA}{dt}=frac{1}{2}left[xfrac{dy}{dt}+yfrac{dx}{dt}right]$$
and plug in the values of $x,y,frac{dx}{dt},$ and $frac{dy}{dt}$
answered Jan 25 at 17:51
pwerthpwerth
3,300417
$endgroup$
For these kinds of problems, I suggest always drawing a picture. In this case, you can think of the positive $y$-axis as the wall and the positive $x$-axis as the floor on which the ladder stands. Then the ladder will be the line segment connecting the point on the $x$-axis where the base of ladder sits, and the point on the $y$-axis where the top of the ladder rests. You will then see that the ladder is the hypoteneuse of a right triangle.
Now we need to label our variables and obtain an equation relating them so that we can use the chain rule. Let's let $x$ be the the $x$-coordinate of the bottom of the ladder i.e. how far (horizontally) the bottom of the ladder is from the base of the wall. And let $y$ be the the $y$-coordinate of the top of the ladder i.e. how far (vertically) the top of the ladder is from the base of the wall. Then the area, as you pointed out, is
$$A=frac{1}{2}xy$$
We want the rate of change of the area i.e. $frac{dA}{dt}$. The key point here is that $x$ and $y$ are both functions of $t$ since they are changing with time. To emphasize this, we can write
$$A(t)=frac{1}{2}x(t)y(t)$$
Now we can obtain $frac{dA}{dt}$ using the product rule and chain rule:
$$frac{dA}{dt}=frac{1}{2}left[x(t)frac{dy}{dt}+y(t)frac{dx}{dt}right]$$
Most people won't use the extra notation here, so you'd probably see it as
$$frac{dA}{dt}=frac{1}{2}left[xfrac{dy}{dt}+yfrac{dx}{dt}right]$$
Now we plug in our information. We know that $x=48$ at the time of interest, and $frac{dx}{dt}=8$. But we still need to plug in $y$ and $frac{dy}{dt}$. To get $y$, we'll use the Pythagorean theorem $x^{2}+y^{2}=52^{2}$ with $x=8$ to get $y=sqrt{52^{2}-8^{2}}=sqrt{2630}$. But what about $frac{dy}{dt}$? Well, we can also use the Pythagorean theorem here (as well as chain rule):
$$x^{2}+y^{2}=52^{2}$$
$$2xfrac{dx}{dt}+2yfrac{dy}{dt}=0$$
$$2(48)(8)+2sqrt{2630}frac{dy}{dt}=0$$
$$frac{dy}{dt}=-frac{384}{sqrt{2630}}$$
Observe that $frac{dy}{dt}$ is negative (why?). I'll leave it to you to simplify the fraction. To finish the problem, return to our original equation
$$frac{dA}{dt}=frac{1}{2}left[xfrac{dy}{dt}+yfrac{dx}{dt}right]$$
and plug in the values of $x,y,frac{dx}{dt},$ and $frac{dy}{dt}$
answered Jan 25 at 17:51
pwerthpwerth
3,300417
answered Jan 25 at 17:51
pwerthpwerth
3,300417
answered Jan 25 at 17:51
pwerthpwerth
3,300417
answered Jan 25 at 17:51
pwerthpwerth
3,300417
3,300417
$begingroup$
Maybe easier to understand for some people would be to plug in $y=sqrt{52^2-x^2}$ into the first equation for the area, and just to a single derivative. It is the same answer (even some calculation steps might overlap), but it's more intuitive for beginners.
$endgroup$
– Andrei
Jan 25 at 18:03
add a comment |
$begingroup$
Maybe easier to understand for some people would be to plug in $y=sqrt{52^2-x^2}$ into the first equation for the area, and just to a single derivative. It is the same answer (even some calculation steps might overlap), but it's more intuitive for beginners.
$endgroup$
– Andrei
Jan 25 at 18:03
$begingroup$
Maybe easier to understand for some people would be to plug in $y=sqrt{52^2-x^2}$ into the first equation for the area, and just to a single derivative. It is the same answer (even some calculation steps might overlap), but it's more intuitive for beginners.
$endgroup$
– Andrei
Jan 25 at 18:03
$begingroup$
Maybe easier to understand for some people would be to plug in $y=sqrt{52^2-x^2}$ into the first equation for the area, and just to a single derivative. It is the same answer (even some calculation steps might overlap), but it's more intuitive for beginners.
$endgroup$
– Andrei
Jan 25 at 18:03
add a comment |
