UrbanPro

Take Class 12 Tuition from the Best Tutors

  • Affordable fees
  • 1-1 or Group class
  • Flexible Timings
  • Verified Tutors

Search in

Find the equations of the tangent and normal to the given curves at the indicated points:

(i) y = x4 − 6x3 + 13x2 − 10x + 5 at (0, 5)

(ii) y = x4 − 6x3 + 13x2 − 10x + 5 at (1, 3)

(iii) y = x3 at (1, 1)

(iv) y = x2 at (0, 0)

(v) x = cos ty = sin t at 

Asked by Last Modified  

Follow 1
Answer

Please enter your answer

(i) The equation of the curve is y = x4 − 6x3 + 13x2 − 10x + 5. On differentiating with respect to x, we get: Thus, the slope of the tangent at (0, 5) is −10. The equation of the tangent is given as: y − 5 = − 10(x − 0) ⇒ y − 5 = − 10x ⇒ 10x...
read more

(i) The equation of the curve is y = x4 − 6x3 + 13x2 − 10x + 5.

On differentiating with respect to x, we get:

Thus, the slope of the tangent at (0, 5) is −10. The equation of the tangent is given as:

y − 5 = − 10(x − 0)

y − 5 = − 10x

⇒ 10x + y = 5

The slope of the normal at (0, 5) is

Therefore, the equation of the normal at (0, 5) is given as:

(ii) The equation of the curve is y = x4 − 6x3 + 13x2 − 10x + 5.

On differentiating with respect to x, we get:

Thus, the slope of the tangent at (1, 3) is 2. The equation of the tangent is given as:

The slope of the normal at (1, 3) is

Therefore, the equation of the normal at (1, 3) is given as:

(iii) The equation of the curve is y = x3.

On differentiating with respect to x, we get:

Thus, the slope of the tangent at (1, 1) is 3 and the equation of the tangent is given as:

The slope of the normal at (1, 1) is

Therefore, the equation of the normal at (1, 1) is given as:

(iv) The equation of the curve is y = x2.

On differentiating with respect to x, we get:

Thus, the slope of the tangent at (0, 0) is 0 and the equation of the tangent is given as:

y − 0 = 0 (x − 0)

y = 0

The slope of the normal at (0, 0) is , which is not defined.

Therefore, the equation of the normal at (x0, y0) = (0, 0) is given by

(v) The equation of the curve is x = cos t, y = sin t.

∴The slope of the tangent atis −1.

When

Thus, the equation of the tangent to the given curve at is

The slope of the normal atis

Therefore, the equation of the normal to the given curve at is

read less
Comments

Related Questions

Find the slope of the normal to the curve x = 1 − sin θy = cos2θ at .

x = 1 - a sinθdifferentiate x with respect to θ,dx/dθ = 0 - a.d(sinθ)/dθ = - a.cosθ ------(1)y = bcos²θdifferentiate y with respect to θ,dy/dθ...
Deepak
0 0
5

Find the equations of the tangent and normal to the parabola y2 = 4ax at the point (at2, 2at).

The equation of the given parabola is y2 = 4ax. On differentiating y2 = 4ax with respect to x, we have: ∴The slope of the tangent atis Then, the equation of the tangent atis given by, y −...
Neerajchalam
0 0
5

Now ask question in any of the 1000+ Categories, and get Answers from Tutors and Trainers on UrbanPro.com

Ask a Question

Recommended Articles

Radhe Shyam is a highly skilled accounts and finance trainer with 8 years of experience in teaching. Accounting is challenging for many students and that’s where Radhe Shyam’s expertise comes into play. He helps his students not only in understanding the subject but also advises them on how to overcome the fear of accounts...

Read full article >

Raghunandan is a passionate teacher with a decade of teaching experience. Being a skilled trainer with extensive knowledge, he provides high-quality BTech, Class 10 and Class 12 tuition classes. His methods of teaching with real-time examples makes difficult topics simple to understand. He explains every concept in-detail...

Read full article >

Swati is a renowned Hindi tutor with 7 years of experience in teaching. She conducts classes for various students ranging from class 6- class 12 and also BA students. Having pursued her education at Madras University where she did her Masters in Hindi, Swati knows her way around students. She believes that each student...

Read full article >

Mohammad Wazid is a certified professional tutor for class 11 students. He has 6 years of teaching experience which he couples with an energetic attitude and a vision of making any subject easy for the students. Over the years he has developed skills with a capability of understanding the requirements of the students. This...

Read full article >

Looking for Class 12 Tuition ?

Learn from the Best Tutors on UrbanPro

Are you a Tutor or Training Institute?

Join UrbanPro Today to find students near you
X

Looking for Class 12 Tuition Classes?

The best tutors for Class 12 Tuition Classes are on UrbanPro

  • Select the best Tutor
  • Book & Attend a Free Demo
  • Pay and start Learning

Take Class 12 Tuition with the Best Tutors

The best Tutors for Class 12 Tuition Classes are on UrbanPro

This website uses cookies

We use cookies to improve user experience. Choose what cookies you allow us to use. You can read more about our Cookie Policy in our Privacy Policy

Accept All
Decline All

UrbanPro.com is India's largest network of most trusted tutors and institutes. Over 55 lakh students rely on UrbanPro.com, to fulfill their learning requirements across 1,000+ categories. Using UrbanPro.com, parents, and students can compare multiple Tutors and Institutes and choose the one that best suits their requirements. More than 7.5 lakh verified Tutors and Institutes are helping millions of students every day and growing their tutoring business on UrbanPro.com. Whether you are looking for a tutor to learn mathematics, a German language trainer to brush up your German language skills or an institute to upgrade your IT skills, we have got the best selection of Tutors and Training Institutes for you. Read more