Exam (elaborations)
calculus, Cambridge, past papers, tenth edition
- Course
- Institution
Calculus 10th edition Ex 2.8 .A complete solution with geometrical figure to explain the concept
[Show more]
Seller
Follow
shahbazahmed
Content preview
CalculusAnton Bivens Davis
10 TH Edition
Shahbaz Ahmed
email: shahbazpunjabuniversity@gmail.com
Ex 2.8
Implicit Differentiation
If the reader find any answer uncomfortable ,he/she may email me and i shall
immediately revise the same answer up to the entire satisfaction of the learner.To
satisfy the learner is my professional commitment and ultimate goal
Q1
dx dy
Equation : y = 3x + 5. (a) Given that dt = 2, find dt when x = 1.
dy dx
(b) Given that dt = −1 , find dt when x = 0
Solution
y = 3x + 5
Taking derivative with respect to t
1
,dy d(3x+5) d(3x) d(5)
dt = dt = dt + dt
dy d(3x) dx d(5) dx
dt = dx dt + dx dt
dy
dt = 3 dx dx
dx dt + 0 ×
dx
dt
dy
dt = 3 dx dx
dt + 0 = 3 dt .................. eq (1)
dx
(a)Putting dt = 2 in equation (1)
dy
dt = 3(2) = 6 when x=1
dy
(b) Putting dt = −1 in equation (1)
−1 = 3 dx
dt
dx
dt = − 31 when x=0
Q2
Equation: x + 4y = 3.
dx dy
(a) Given that dt = 1, find dt when x = 2.
dy dx
(b) Given that dt = 4 , find dt dx/dt when x = 3.
Solution
x + 4y = 3
Taking derivative with respect to t
2
,d(x+4y) d(3)
dt = dt
dx d(4y)
dt + dt =0
dx
dt + 4 dy
dt = 0 ....................... eq (1)
dx
(a) Putting dt = 1 in equation (1)
1 + 4 dy
dt = 0 Or
4 dy
dt = −1 Or
dy
dt = − 41
dy
(b)Putting dt = 4 in equation (1)
dx
dt +4×4=0
dx
dt = −16
Q3
Equation: 4x2 + 9y 2 = 1
dx dy
(a) Given that dt = 3, find dt when
1 1
(x, y) = ( 2√ , √
2 3 2
)
dy dx
(b) Given that dt = 8, find dt when
√
5
(x, y) = ( 13 , − 9 ) Solution
3
, 4x2 + 9y 2 = 1
Taking derivative with respect to t
d(4x2 +9y 2 )) d(1)
dt = dt
d(4x2 ) d(9y 2 )
dt + dt =0
d(4x2 ) dx d(9y 2 ) dy
dx dt + dy dt =0
2 2
4 d(x ) dx d(y ) dy
dx dt + 9 dy dt = 0
dy
4 × 2x dx
dt + 9 × 2y dt = 0
dy
8x dx
dt + 18y dt = 0 ........................ eq (1)
dx 1 1
(a)Putting dt = 3, x = √
2 2
,y = √
3 2
in equation (1)
1 1 dy
8× √
2 2
× 3 + 18 × √
3 2 dt
=0
1 dy 1
18 × √
3 2 dt
= −8 × √
2 2
×3
dy 1
√ 1
dt = −8 × √
2 2
×3×3 2× 18 = −2
√
dy 5
(b) Putting dt = 8, x = 13 , y = − 9 in equation (1)
√
1 dx 5
8× 3 dt − 18 × 9 ×8=0
dx
√
5 1
√
dt = 18 × 9 ×8×3× 8 =6 5
Q4
4
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller shahbazahmed. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $7.99. You're not tied to anything after your purchase.
Can Stuvia be trusted?
4.6 stars on Google & Trustpilot (+1000 reviews)
79064 documents were sold in the last 30 days
Founded in 2010, the go-to place to buy study notes for 14 years now