Is my textbook wrong about this basic related rate question?

[The Student]

I am reviewing some of my high school calculus related rates questions, but I am not getting the same answer as the textbook. I get 519.6km/h, but the textbook gets 537km/h. Normally I would assume an error in the textbook, but it has always been right.

The question: A plane flys over over you at an altitude of 10km and with a horizontal speed of 600km/h. When it is 20km from you, how fast is it moving away from you?

Here's my work:

r = 20km ; y = 10km ; x = (r^2 - y^2)^(1/2) = 10(3)^(1/2) ; r' = 0 ; x' = 600km/h ; r' = ?

r^2 = x^2 + y^2

2rr' = 2xx' + 0

r' = 2xx'/(2r) = (2(10(3)^(1/2))(600))/(2(20)) = 519.6km/h is my answer. There answer is 537km/h.

Am I wrong anywhere?

 

[tkhunny]

First, r' = 0 seems a little odd. I'm guessing this is a typo and you meant y' = 0, meaning the altitude is not changing.

Second, Please simplfy your life. When you get 2rr' = 2xx' + 0, do not do anything with this until after you notice the common factors. Working with rr' = xx' is substantially simpler.

Third, is there a drawing? Perhaps it means x = 20 km

 

[Deleted member 4993]

I am reviewing some of my high school calculus related rates questions, but I am not getting the same answer as the textbook. I get 519.6km/h, but the textbook gets 537km/h. Normally I would assume an error in the textbook, but it has always been right.

The question: A plane flys over over you at an altitude of 10km and with a horizontal speed of 600km/h. When it is 20km from you, how fast is it moving away from you?

Here's my work:

r = 20km ; y = 10km ; x = (r^2 - y^2)^(1/2) = 10(3)^(1/2) ; r' = 0 ; x' = 600km/h ; r' = ?

r^2 = x^2 + y^2

2rr' = 2xx' + 0

r' = 2xx'/(2r) = (2(10(3)^(1/2))(600))/(2(20)) = 519.6km/h is my answer. There answer is 537km/h.

Am I wrong anywhere?

According to your assumption, r = 20, you have done the problem correctly.

If however, you assume x = 20 (and r =\(\displaystyle \sqrt{500}\)) according to TK's suggestion, you'll get book's answer.

 

[The Student]

First, r' = 0 seems a little odd. I'm guessing this is a typo and you meant y' = 0, meaning the altitude is not changing.

Second, Please simplfy your life. When you get 2rr' = 2xx' + 0, do not do anything with this until after you notice the common factors. Working with rr' = xx' is substantially simpler.

Third, is there a drawing? Perhaps it means x = 20 km

Yeah, I meant y' = 0.

With no picture in the textbook, it asks how fast is the plane travelling away from me when it is 20km away from me. So I take that to mean the shortest distance possible which would be the diagonal r distance.

 

[The Student]

According to your assumption, r = 20, you have done the problem correctly.

If however, you assume x = 20 (and r =\(\displaystyle \sqrt{500}\)) according to TK's suggestion, you'll get book's answer.

Oh great thanks, I didn't think to try that, but the question really does not seem to suggest that the plane has a horizontal distance of x = 20.

 

[The Student]

Thanks Subhotosh Khan and tkhunny, I am moving on.